Course Schedule

Here’s a list of simpler problems:

1. 104. Maximum Depth of Binary Tree: Helps you understand basic tree traversal techniques.

2. 200. Number of Islands: This problem helps with understanding depth-first search on a grid, which is a useful stepping stone to understanding depth-first search on more abstract graphs.

3. 283. Move Zeroes: Useful for understanding the two-pointer technique, which is important in various graph traversals.

4. 155. Min Stack: This problem helps you get used to using a stack, which is often used in depth-first search.

5. 102. Binary Tree Level Order Traversal: You can get practice using breadth-first search, which is an essential graph traversal technique.

6. 94. Binary Tree Inorder Traversal: This will give you more practice with depth-first search, this time in a tree context.

7. 133. Clone Graph: This problem involves a simple graph traversal, which could be a good step up to the more complex graph involved in the “Course Schedule” problem.

8. 127. Word Ladder: This problem can help you practice breadth-first search.

9. 559. Maximum Depth of N-ary Tree: This is a more complex tree traversal problem, which could be good practice for the even more complex graph traversal in the “Course Schedule” problem.

10. 207. Graph Valid Tree: This problem also deals with detecting cycles in a graph, similar to the “Course Schedule” problem, but is slightly simpler because it deals with an undirected graph.

These involve different kinds of graphs and different graph traversal techniques. Get comfortable with both breadth-first search and depth-first search, as are useful in different situations.

Identifying Problem Isomorphism

“Course Schedule” belongs to a category of problems that require understanding graph structures and detecting cycles within them.

A simpler problem in this category is “Detect Cycle in an Undirected Graph”. The task in this problem is to detect if there’s a cycle in an undirected graph. This problem is simpler because the graph is undirected, making cycle detection less complex than in a directed graph, as in “Course Schedule”.

A more complex problem in this category is “Course Schedule II”. This problem extends “Course Schedule” by not only asking if it’s possible to finish all courses, but also requiring the correct order in which to take the courses, if possible. This adds an extra layer of complexity, as it requires performing a topological sort on the graph.

The reasoning behind these choices:

• “Detect Cycle in an Undirected Graph” is simpler than “Course Schedule” because detecting a cycle in an undirected graph is less complex than in a directed graph. The nature of an undirected graph allows for a simpler algorithm.

• “Course Schedule II” adds an extra layer of complexity to “Course Schedule” by requiring you to provide the correct ordering of nodes (courses), thereby necessitating a topological sort.

Therefore, the order of problems from simpler to more complex, based on understanding of graph structures and cycle detection, would be:

1. “Detect Cycle in an Undirected Graph”
2. “Course Schedule”
3. “Course Schedule II”

This mapping is an approximation. While these problems all deal with detecting cycles in a graph, the exact problem constraints and details can significantly influence the complexity and solution approach.

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    class Solution:
			def buildAdjacencyList(self, n, edgesList):
			        adjList = [[] for _ in range(n)]
			        for c1, c2 in edgesList:
			            adjList[c2].append(c1)
			        return adjList
		      		
        def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
            adjList = self.buildAdjacencyList(numCourses, prerequisites)    
            state = [0] * numCourses

            def hasCycle(v):
                if state[v] == 1:
                    return False
                if state[v] == -1:
                    return True

                state[v] = -1

                for i in adjList[v]:
                    if hasCycle(i):
                        return True

                state[v] = 1
                return False

            for v in range(numCourses):
                if hasCycle(v):
                    return False

            return True

Problem Classification

This problem can be categorized as a Graph Theory problem. The problem involves determining whether there are any cycles in a directed graph, which would make it impossible to complete all courses. The course prerequisites form a directed graph, where each course is a node and the dependencies between courses form the edges.

Components:

• “numCourses” - Number of courses to be taken.
• “prerequisites” - An array of pairs where each pair represents a course that must be taken before another course.
• “Return true if you can finish all courses. Otherwise, return false.” - The task to be performed, which involves finding whether it’s possible to complete all courses given the prerequisites. This is equivalent to checking for the absence of any cycles in the directed graph.

It can be classified as a Decision Problem, since the output is a simple Yes/No or True/False response. Furthermore, it can also be seen as a Graph Traversal Problem, as it involves traversing a directed graph (of courses) in a particular way (according to prerequisites) to reach a decision.

The problem is a classic use-case for Depth-First Search (DFS) or Topological Sorting in Graph Theory, where you would perform a DFS to check for cycles (a course that depends on itself, directly or indirectly). If there’s any cycle, it’s impossible to complete all courses, and you return false. If there are no cycles, you return true. This would be the ‘How’ part, not asked in this question.

This kind of problem can often be found in scheduling, course sequencing, and job sequencing scenarios in real-world applications.

Problem Analysis and Key Insights

Here are some key insights that help understand the problem:

1. Course Dependency Graph: The problem naturally translates to a directed graph where each node represents a course, and each edge represents a prerequisite relationship between two courses.

2. Cycle Detection: Since prerequisites form a directed relationship, if there’s a cycle in the graph, it means there’s a circular dependency among the courses which makes it impossible to finish all courses. Hence, we can use cycle detection in the graph to solve this problem.

3. Graph Traversal: We can use common graph traversal methods such as Depth-First Search (DFS) or Breadth-First Search (BFS) to detect cycles.

4. Topological Sort: This problem can also be seen as a problem of finding a topological ordering of a directed graph. If such an ordering can be found, we can complete all courses, otherwise, we can’t.

5. Unique Pairs: The problem specifies that all pairs in prerequisites[i] are unique, so there’s no redundancy or contradiction in the prerequisites, which simplifies the problem.

6. Decision Problem: The problem is a decision problem, meaning it only requires a “yes” (true) or “no” (false) answer. There’s no requirement to find out in which order the courses should be taken.

7. Input Constraints: The constraints are provided on the number of courses and the length of prerequisites, which gives an idea about the time and space complexity our solution should aim for.

Clarification Questions

To further understand this problem, ask:

1. Can a course have multiple prerequisites? For instance, can course 2 require both course 0 and course 1 to be taken first?
2. Are there any courses without prerequisites? In other words, can we assume there will be at least one course that doesn’t require any prerequisites?
3. Can a course be a prerequisite for multiple other courses? For instance, could course 0 be a prerequisite for both course 1 and course 2?
4. Is there a particular order in which courses should be finished? Or just finishing all courses, given the prerequisites, is sufficient?
5. Can the prerequisites be considered as a transitive relation? For instance, if course 2 requires course 1, and course 1 requires course 0, is it necessary to mention that course 2 also requires course 0?
6. Are we considering self loops in the graph? That is, a situation where a course is its own prerequisite?
7. What is the expected performance of the solution? As the number of courses increases, does the solution need to remain efficient?
8. Is it guaranteed that the prerequisites array contains no contradictory or duplicated information?

Problem Boundary

The scope of this problem involves:

1. Understanding the relationship between courses and their prerequisites as provided in the input.
2. Building a directed graph based on these relationships where each node is a course and a directed edge represents a prerequisite.
3. Implementing an algorithm to detect cycles in this graph. If a cycle exists, that implies there’s a circular dependency among the courses, making it impossible to complete all courses. Conversely, if there are no cycles, that means it’s possible to complete all courses.
4. The solution should be able to handle up to 2000 courses and 5000 prerequisites, as given by the constraints.

The problem does not ask for:

1. The order in which the courses should be taken, so we don’t need to provide a sequence of course completion.
2. Handling redundant or contradictory prerequisites, as the problem statement guarantees all pairs in prerequisites are unique.
3. Handling situations where a course is its own prerequisite. It’s not specified in the problem, but typically in such problems, this is considered a cycle, making course completion impossible.

The scope of the problem is thus limited to determining whether all courses can be completed given their prerequisites, which can be solved using concepts from graph theory.

Establishing the boundary of a problem involves understanding its limits and what it doesn’t involve. Here are the boundaries for this problem:

1. Input Constraints: The problem’s boundary is defined by the number of courses (numCourses), which ranges from 1 to 2000, and the number of prerequisites pairs, which ranges from 0 to 5000. The solution should be able to handle these constraints efficiently.

2. Problem Nature: The problem strictly asks to determine whether it’s possible to complete all courses given the prerequisites. It does not require finding the order of courses to complete, nor does it require handling cases where there are contradictory or redundant prerequisites.

3. Data Types: The input data types are integers (for numCourses) and a list of integer pairs (for prerequisites). The output is a boolean indicating whether it’s possible to complete all courses. The solution should work within these data types.

4. Prerequisite Relationships: The solution does not have to account for indirect prerequisite relationships unless specified in the prerequisites list. For example, if course A is a prerequisite for course B, and course B is a prerequisite for course C, it doesn’t mean that course A is a direct prerequisite for course C unless it is stated in the prerequisites.

5. Unique Prerequisite Pairs: Each pair in the prerequisites list is unique, which means the solution does not need to handle duplicate entries.

6. No Self-Loops: The problem does not clearly mention if a course can be its own prerequisite, but usually in such problems this situation is considered as a cycle, making it impossible to complete the course.

By defining these boundaries, we ensure that our solution is targeted to the problem at hand without over-complicating or oversimplifying it.

Distilling the Problem to Its Core Elements

Fundamental Concept: The fundamental concept this problem is based upon is the Graph Theory in Computer Science, specifically, cycle detection in directed graphs and topological sorting.

Simplest Description: Imagine you are a student who has a number of courses to complete, and some of these courses have prerequisites, meaning you have to complete certain courses before you can start others. Given the list of all courses and their prerequisites, can you determine whether it’s possible to finish all courses without violating any prerequisites rules?

Core Problem: The core problem we are trying to solve here is to determine if there is a way to complete all courses given the prerequisite constraints. If there exists a cycle of prerequisites (a course depending on itself either directly or indirectly), it is impossible to complete all courses.

Key Components: The key components of this problem are:

• The number of courses.
• The prerequisite pairs, which forms a directed graph.
• Cycle detection in the directed graph.

Simplified Problem Statement: Given a number of courses and a list of prerequisite pairs for each course, determine whether it is possible to complete all courses without breaking any prerequisite rules.

Minimal Set of Operations: To solve this problem, we need to perform the following operations:

1. Construct a directed graph based on the given prerequisites.
2. Perform a depth-first search (DFS) or a breadth-first search (BFS) to detect any cycles in the graph. Alternatively, attempt to perform a topological sort.
3. If a cycle is detected or if the topological sort fails, return false. Otherwise, return true.

Visual Model of the Problem

This problem can be visualized as a directed graph where each node represents a course and a directed edge between two nodes represents a prerequisite.

For example, consider the problem with numCourses = 4 and prerequisites = [[1,0],[2,1],[3,2]].

We can visualize this as:

0 --> 1 --> 2 --> 3

Here, arrows denote prerequisite relations, for instance, course 0 must be taken before course 1, course 1 before course 2, and course 2 before course 3. As there are no cycles in this graph, we can complete all courses.

For the case where numCourses = 2 and prerequisites = [[1,0],[0,1]], the graph would look like:

0 <--> 1

Here, there’s a cycle: course 0 is a prerequisite for course 1 and vice versa, creating a circular dependency which makes it impossible to complete all courses.

This visualization provides a clear understanding of the problem, highlighting the concept of course dependencies and the issue with circular dependencies (cycles). It also leads naturally to the solution method: detecting cycles in a directed graph.

Problem Restatement

You need to enroll in a certain number of courses, labeled from 0 up to one less than the total number of courses. There are some rules, given in the form of a list of pairs of courses, where the first course in the pair can only be taken after the second one. The goal is to determine whether it’s possible to complete all courses by following these rules.

For example, if you have two courses (0 and 1) and a rule that says you can only take course 1 after taking course 0, then it’s possible to complete all courses by taking course 0 first and then course 1. But if you have a rule that says you need to take course 0 after course 1 and another rule that says you need to take course 1 after course 0, then it’s impossible to complete all courses because the rules are contradicting each other.

Requirements and Constraints:

• The total number of courses will be between 1 and 2000.
• The number of rules (prerequisites) will be between 0 and 5000.
• Each rule consists of two distinct courses, and each course in the pair will be a valid course number.
• There will be no duplicate pairs in the rules.
• You need to return a boolean answer: true if you can complete all courses by following the rules, and false if you can’t.

Abstract Representation of the Problem

This problem can be abstracted as the problem of cycle detection in a directed graph.

Here’s how we can abstractly represent this problem:

1. Nodes and Edges: We are given a number of nodes (numCourses) and a list of directed edges (prerequisites) in a directed graph. Each node represents a course, and each directed edge between two nodes represents a prerequisite relationship between two courses, pointing from the course to its prerequisite.

2. Cycles: We need to find out whether there is a cycle in this directed graph. A cycle represents a circular dependency in courses and makes it impossible to finish all courses.

3. Outcome: If there is no cycle in the graph, then it is possible to finish all courses, and we return true. If there is a cycle, then it is impossible to finish all courses, and we return false.

Abstracting the problem in this way allows us to see that it’s not strictly about courses and prerequisites, but fundamentally about the structure of the graph. This abstraction can be helpful in solving the problem as it highlights the key elements we need to consider: nodes, edges, and cycles.

Terminology

There are several key terms and concepts related to Graph Theory and Computer Science that are crucial to understanding and solving this problem:

1. Directed Graph: A graph where edges between vertices (also known as nodes) have a specific direction. In the context of this problem, each course is a node and each prerequisite relationship is a directed edge from one course to another.

2. Cycle in a Directed Graph: A sequence of edges in a graph such that the first and last nodes in the sequence are the same, and all edges follow the direction of the graph. A cycle in this problem indicates a circular dependency among courses, making it impossible to complete all courses.

3. Cycle Detection: The process of checking whether a cycle exists in a graph. This is the key operation in solving this problem. If a cycle exists in the course-prerequisite graph, it’s impossible to complete all courses.

4. Depth-First Search (DFS): A common graph traversal algorithm that explores as far as possible along each branch before backtracking. DFS can be used for cycle detection in this problem.

5. Breadth-First Search (BFS): Another common graph traversal algorithm that visits all the vertices of a graph in breadth-first order i.e., it visits all the vertices at the same level before going to the next level. BFS can also be used for cycle detection, though it’s less common in this context than DFS.

6. Topological Sort: A linear ordering of the vertices in a directed graph such that for every directed edge u -> v, vertex u comes before v in the ordering. In the context of this problem, if a topological ordering exists, it means we can complete all courses; if not, there exists a cycle in the graph making it impossible to complete all courses.

7. Adjacency List: A data structure used to represent a graph where the index represents the node and the value at that index is a list of all the nodes that the node is connected to. In this problem, an adjacency list can be used to represent the prerequisite relationships between courses.

Understanding these terms and concepts is crucial as they form the basis of the solution to this problem.

Problem Simplification and Explanation

Imagine you have a number of tasks to do, and some of these tasks have prerequisites, i.e., they need other tasks to be completed before you can start them. For example, before you can “eat breakfast”, you need to “cook breakfast”, and before you can “cook breakfast”, you need to “buy groceries”.

Now, imagine these tasks as cities, and the prerequisites as one-way roads connecting these cities. You can only travel from one city to another if there’s a road connecting them.

In our example, “buy groceries” is a city with a one-way road to the “cook breakfast” city, which in turn has a one-way road to the “eat breakfast” city.

If all roads eventually lead to a city without outgoing roads (a task without prerequisites), you can complete all tasks (travel to all cities). But if there’s a loop of roads (a circular dependency), you’ll be stuck traveling in that loop forever and won’t be able to reach all cities.

Now, replace the tasks with courses and you have the problem at hand. The challenge is to find out whether you can complete all courses (travel to all cities) given the prerequisites (one-way roads). This is equivalent to checking if there’s a loop in the courses (cities) connected by prerequisites (roads).

In the world of computer science, we call this a cycle detection problem in a directed graph. The key concepts involved here are:

1. Directed Graph: Our network of cities and roads.
2. Nodes (Vertices): Represent our cities (courses).
3. Edges: Represent our roads (prerequisites).
4. Cycle Detection: Identifying if we have a loop of cities (a circular dependency of courses).

By understanding these concepts and how they interact, you can then use various algorithms and data structures, such as depth-first search and adjacency lists, to solve the problem.

Constraints

In this problem, there are several characteristics that can be exploited for finding an efficient solution:

1. Directed Graph Structure: The problem involves prerequisites of courses, which forms a directed graph. This graph structure suggests using graph algorithms for solving the problem.

2. Cycle Detection: The problem essentially requires detecting a cycle in a directed graph. A cycle means that there are conflicting prerequisites, making it impossible to finish all courses. There are well-known efficient algorithms to detect cycles in a graph such as Depth-First Search (DFS) or alternatively, attempting a topological sort.

3. Prerequisite Pairs are Unique: The fact that all prerequisite pairs are unique can simplify the process of graph construction and cycle detection. We don’t need to handle duplicate edges.

4. The Range of numCourses and prerequisites: The problem’s constraints specify that numCourses ≤ 2000 and prerequisites.length ≤ 5000. These relatively small sizes mean that we can comfortably use algorithms with time complexity of O(n) or even O(n^2) without running into performance issues.

5. Numerical range of courses: Courses are numbered from 0 to numCourses - 1, which allows us to easily represent the courses and their relationships in an adjacency list or adjacency matrix.

By identifying these characteristics, we can leverage the structure of the problem to design a more efficient solution.

Here are the key insights we can draw from analyzing the constraints:

1. Graph Size: The constraints 1 <= numCourses <= 2000 and 0 <= prerequisites.length <= 5000 provide insight into the size of the graph we’ll be dealing with. Given this size, the graph can comfortably fit into memory, and graph algorithms with O(n) or O(n^2) time complexity can be utilized efficiently.

2. Node Validity: The constraint 0 <= ai, bi < numCourses ensures that all the courses represented in the prerequisites are valid, i.e., within the range of defined courses. This makes the validation of input data simpler.

3. Unique Edges: The constraint All the pairs prerequisites[i] are unique means there are no duplicate prerequisites. This means we don’t have to worry about handling duplicate edges in our graph.

4. Graph Type: As per the problem’s description, the prerequisites form a directed graph (edges have a specific direction). This insight guides us towards algorithms suited for directed graphs.

5. Cycle Detection: The core problem here is to detect if a cycle exists in the graph. If there’s a cycle, it indicates a circular dependency among courses, and it’s impossible to complete all courses. This insight informs us that we should consider cycle detection algorithms.

6. Output: The problem asks for a Boolean output. This means that as soon as we find a cycle in the graph, we can terminate the process and return false, as it’s impossible to complete all the courses. This can potentially save computation time as we don’t need to explore the entire graph in certain scenarios.

These insights drawn from the constraints and problem statement can greatly help in formulating a plan for the solution.

Case Analysis

1. Minimum Input Scenario (Base Case)

In this scenario, the smallest possible input will be provided based on the constraints.

Input: numCourses = 1, prerequisites = [] Output: true

Explanation: There is only one course and it has no prerequisites. Hence, it’s possible to finish all courses.

2. No Prerequisites Scenario (Simple Case)

In this scenario, multiple courses are there but none of them has any prerequisites.

Input: numCourses = 5, prerequisites = [] Output: true

Explanation: There are five courses and none of them has any prerequisites. Hence, all courses can be taken freely in any order.

3. Simple Prerequisites Scenario (Prerequisites Case)

This scenario represents a case where courses have prerequisites, but there are no conflicting prerequisites.

Input: numCourses = 4, prerequisites = [[1,0],[2,1],[3,2]] Output: true

Explanation: There are prerequisites, but no course is mutually dependent on another. The courses can be taken in the order of 0, 1, 2, 3.

4. Circular Prerequisites Scenario (Impossible Case)

This scenario represents a case where courses have prerequisites that form a circular dependency, making it impossible to finish all courses.

Input: numCourses = 3, prerequisites = [[1,0],[2,1],[0,2]] Output: false

Explanation: Course 0 is a prerequisite for course 1, which is a prerequisite for course 2, which in turn is a prerequisite for course 0. This forms a cycle, indicating a circular dependency, which makes it impossible to finish all courses.

5. Disconnected Graphs Scenario (Multiple Independent Sets Case)

This scenario represents a case where there are multiple sets of courses with prerequisites, but these sets are independent of each other.

Input: numCourses = 4, prerequisites = [[1,0],[3,2]] Output: true

Explanation: There are two sets of courses (0,1) and (2,3), both of which can be completed independently. Hence, it’s possible to finish all courses.

These scenarios cover the edge cases, simple cases, and more complex cases. It’s crucial to test all these cases to ensure the solution handles different situations correctly. The key insight here is to detect cycles in the directed graph to decide whether it’s possible to complete all courses.

Sure, let’s visualize these cases as directed graphs:

1. Minimum Input Scenario (Base Case)

0

Here, we have a single node (course) with no edges (prerequisites), meaning we can freely take the course.

2. No Prerequisites Scenario (Simple Case)

0   1   2   3   4

Here, each node represents a course. Since there are no edges, we can take any course freely.

3. Simple Prerequisites Scenario (Prerequisites Case)

0 --> 1 --> 2 --> 3

Here, each node represents a course, and an edge from node A to B means that course A is a prerequisite for course B. We can take the courses in the order of 0, 1, 2, 3.

4. Circular Prerequisites Scenario (Impossible Case)

0 --> 1 --> 2
^           |
|           v
3 <----------

Here, we have a cycle (0-1-2-0), which means we have conflicting prerequisites and it’s impossible to complete all courses.

5. Disconnected Graphs Scenario (Multiple Independent Sets Case)

0 --> 1     2 --> 3

Here, we have two disconnected sub-graphs. Each sub-graph doesn’t have any cycles, so we can take all the courses.

Please note that the arrows represent the direction of prerequisites. For example, an edge from 0 to 1 means you have to take course 0 before you can take course 1.

By visualizing these cases, we can better understand the relationships between courses and their prerequisites.

It is possible to have the scenario you mentioned. This scenario represents a case where there are two distinct sequences of courses, each with its own prerequisites, originating from a common course. In the context of a directed graph, you have two separate branches or paths starting from the same node.

Here’s the visual representation:

    1 --> 2 --> 3
   /
0
   \
    4 --> 5 --> 6

In this case, course 0 is a prerequisite for both course 1 and course 4. After course 0, the sequences of courses diverge into two paths: one going 1 –> 2 –> 3, and the other going 4 –> 5 –> 6.

Given this setup, it’s still possible to complete all the courses. One possible order could be 0, 1, 2, 3, 4, 5, 6 (or 0, 4, 5, 6, 1, 2, 3). As there are no cycles or conflicting prerequisites, all courses can be completed.

The analysis of different cases sheds light on several key aspects of the problem:

1. Cycle Detection: The existence of a cycle in the course prerequisite graph indicates that it’s impossible to complete all courses, as it leads to a deadlock scenario. This is evident in the “Circular Prerequisites Scenario (Impossible Case)”.

2. Independent Sets of Courses: In cases like the “Disconnected Graphs Scenario (Multiple Independent Sets Case)”, it’s clear that we can have multiple independent sets of courses. As long as there is no cycle within these sets, we can complete all courses.

3. Shared Prerequisites: From the case you provided (0 –> 1 –> 2 –> 3 and 0 –> 4 –> 5 –> 6), we learned that a course can be a prerequisite for multiple other courses. It doesn’t affect the possibility of completing all courses as long as no cycles are introduced.

4. Order of Courses: In the “Simple Prerequisites Scenario (Prerequisites Case)”, we see that prerequisites enforce an order in which the courses must be completed. However, in the “No Prerequisites Scenario (Simple Case)” and “Minimum Input Scenario (Base Case)”, we can take the courses in any order.

5. Variety of Inputs: The variety of test cases illustrates the complexity and versatility of the problem. Different combinations of numCourses and prerequisites can drastically change the structure of the course prerequisite graph, which in turn influences the solution’s approach.

These insights are vital in understanding the core of the problem and guiding us towards a robust and effective solution. The primary objective remains to detect cycles in the directed graph formed by the course prerequisites.

Identification of Applicable Theoretical Concepts

The problem of course scheduling can be mapped to some fundamental concepts in computer science and graph theory, particularly to the notion of directed graphs and topological sorting.

1. Directed Graphs: The courses and their prerequisites can be represented as a directed graph. Each course is a node in the graph, and each prerequisite is a directed edge.

2. Cycle Detection: A key part of this problem is detecting a cycle in the graph. In graph theory, a cycle within a directed graph means that there is a path from a node back to itself. In this problem, a cycle represents an impossible course ordering, as it means that a course indirectly (or directly) depends on itself. So, applying cycle detection algorithms (like Depth-First Search) is crucial to solve this problem.

3. Topological Sorting: This problem essentially asks if a topological ordering of the courses (nodes) exists. In a directed graph, a topological order is possible if and only if the graph has no directed cycles, i.e., the graph is a Directed Acyclic Graph (DAG). The topological order of a graph represents a linear ordering of its vertices such that for every directed edge (u, v), vertex u comes before v in the ordering, which aligns with the prerequisites condition of this problem.

4. Depth-First Search (DFS): Depth-First Search can be used for both cycle detection and generating a topological order. For cycle detection, during a DFS, if we visit a node that is already in the recursion stack, then there is a cycle. For generating a topological order, we can start a DFS from every unvisited node, and as we return from the recursion (finished visiting all deeper nodes), we add the current node to the beginning of the order.

5. Breadth-First Search (BFS): Alternatively, a BFS approach using Kahn’s algorithm for topological sorting can also be used. In this approach, we repeatedly remove nodes with in-degree 0 (no prerequisites) and their edges (remove the courses with no prerequisites remaining), until we can’t find any more such nodes. If we removed all nodes, a topological order exists. If there are remaining nodes, a cycle exists in them, and thus it’s impossible to complete all courses.

Using these concepts will allow us to manage and solve this problem effectively and efficiently.

Simple Explanation

Imagine you’re at a cooking school where you have to take a certain number of cooking classes. Each cooking class represents a ‘course’ in our problem. Some of these classes have prerequisites, meaning you need to take some other classes before you can take them. For example, you can’t take the “Baking Pastries” class until you’ve completed the “Basic Baking” class because you need the knowledge from the “Basic Baking” class to understand the more advanced class.

Now, the tricky part comes when some classes are prerequisites for each other. For instance, suppose the “Baking Pastries” class is a prerequisite for the “Basic Baking” class (weird, right?). In this case, you’re stuck. You can’t take the “Baking Pastries” class because you haven’t taken the “Basic Baking” class yet, and you can’t take the “Basic Baking” class because you haven’t taken the “Baking Pastries” class yet. This situation is what we call a deadlock, and it’s similar to the ‘circular prerequisites’ in our problem.

So, the problem is basically asking, given a list of cooking classes and the prerequisites for each class, is it possible to take all the classes? Or are we going to end up in a situation where we can’t take some classes because the prerequisites for those classes create a deadlock?

We need to find a way to “schedule” these classes so that you can take all of them without any conflicts. If we can do that, we return ’true’, meaning it’s possible to take all the classes. If not, we return ‘false’, indicating that it’s impossible to take all the classes because of conflicting prerequisites.

Problem Breakdown and Solution Methodology

Let’s break down the approach step by step:

1. Represent the Problem: The first step is to represent the problem in a way that makes it easier to handle. We can use a concept from graph theory where we treat each course as a node and prerequisites as directed edges between nodes. So, we represent the numCourses and prerequisites as a directed graph.

2. Cycle Detection: The next crucial step is to check for cycles in this graph. Why do we need to check for cycles? Because a cycle indicates a circular dependency between courses (A depends on B, B depends on C, and C depends back on A), making it impossible to complete all courses.

   We can use Depth-First Search (DFS) to perform cycle detection in our graph. DFS is a common graph traversal algorithm that explores as far as possible along each branch before backtracking. If we encounter a node that we’ve already visited during our DFS traversal, it means there’s a cycle.

3. Provide the Result: If we find a cycle, we return false, indicating it’s impossible to finish all courses. If no cycle is found, we return true, showing that it’s possible to complete all courses.

Let’s visualize this process with a metaphor. Think of this problem as organizing a large music festival with different bands (courses) playing. Some bands have conditions that they’ll only play if a certain other band has played before them (prerequisites). The bands form a lineup (a directed graph) where each band (node) has a directed edge (arrow) to the band that must play before it. If we end up with a situation where Band A will only play after Band B, but Band B will only play after Band A, we have a problem - a circular condition that we can’t fulfill (a cycle in our graph). The festival (all courses) can’t happen if such a situation arises.

Now, let’s take an example:

Suppose we have numCourses = 4, and prerequisites = [[1,0],[2,1],[3,2]].

1. First, we create our directed graph. This will look like 0 –> 1 –> 2 –> 3.

2. Next, we perform a DFS to check for cycles. In this case, there are no cycles since there are no circular dependencies between courses.

3. Since no cycles were found, we can complete all courses, so we return true.

However, if we had prerequisites = [[1,0],[2,1],[3,2],[0,3]], a cycle would be detected during the DFS (0 –> 1 –> 2 –> 3 –> 0), indicating that it’s impossible to finish all courses. So, we would return false.

It’s important to note that the size of numCourses and prerequisites could affect the solution’s efficiency. More courses and prerequisites mean more nodes and edges in our graph, which could increase the computational complexity of the DFS.

Inference of Problem-Solving Approach from the Problem Statement

Here are the key terms or concepts in the problem:

1. Courses: In the problem, each course can be thought of as a node in a directed graph. The total number of courses gives us the number of nodes in the graph.

2. Prerequisites: The prerequisites are pairs of courses where the first course depends on the second course. These can be thought of as directed edges in our graph. An edge from course a to course b signifies that b is a prerequisite for a.

3. Directed Graph: The courses and their prerequisites form a directed graph where each node represents a course and each directed edge represents a prerequisite relationship between two courses. This visualization aids in understanding the problem and forming an appropriate strategy.

4. Cycles: In the context of this problem, a cycle in the graph represents a circular dependency between courses, making it impossible to complete all courses. Detecting a cycle is crucial to solving this problem.

These key concepts suggest that the problem can be solved using graph theory techniques, specifically by creating a directed graph and performing a Depth-First Search (DFS) or Breadth-First Search (BFS) to check for cycles.

As for recognizing these properties, drawing a directed graph can be incredibly helpful. For each course, draw a node, and for each pair in the prerequisites, draw a directed edge between the corresponding nodes. The direction of the edge should go from the course to its prerequisite. Once you’ve drawn the graph, look for any cycles or loops where you can start at one node and follow the directed edges to get back to the starting node.

For example, given numCourses = 4 and prerequisites = [[1,0],[2,1],[3,2]], you would draw four nodes (0, 1, 2, 3) and three directed edges (0 –> 1, 1 –> 2, 2 –> 3). The resulting graph has no cycles, so it’s possible to finish all courses. However, if you have prerequisites = [[1,0],[2,1],[3,2],[0,3]], you would have an additional edge (3 –> 0), creating a cycle (0 –> 1 –> 2 –> 3 –> 0), making it impossible to finish all courses.

The problem revolves around a sequence of courses with some courses dependent on others, forming prerequisites. The relationship between courses can be naturally represented as a directed graph, with each course being a node and each prerequisite relationship forming a directed edge.

When dealing with directed graphs, a common issue is detecting cycles, which happens when there’s a circular dependency between nodes. The presence of a cycle in this problem means it’s impossible to complete all courses because of circular prerequisites.

Depth-First Search (DFS) is a well-known graph traversal algorithm often used to detect cycles in a graph. DFS explores as far as possible along each branch before backtracking, which makes it an effective tool for cycle detection. If during a DFS traversal, we come across a node that is already in the recursion stack, then a cycle is detected.

So, the representation of the problem as a directed graph and the need to detect cycles are key indicators that DFS could be a useful tool to solve this problem.

Simple Explanation of the Proof

Let’s start with the main problem. We are asked to find out whether we can complete all the courses given certain prerequisites. This translates to finding out if we can visit all nodes (courses) in a directed graph, starting from any node, without falling into a cycle.

The algorithm is based on a concept in graph theory: in a Directed Acyclic Graph (DAG), we can always find at least one course which has no prerequisite, and we can find such a course by conducting a Depth-First Search (DFS).

Here’s a simple proof and explanation of the algorithm:

1. DFS and Cycle Detection: The DFS traversal starts from any course. If during the DFS, we find a course that is already in the visiting state, it means that we’ve found a cycle, and hence, it is impossible to complete all courses. The reason for this is that a cycle indicates that there are circular dependencies among courses, such as course A depends on course B, and course B depends on course A. In such a situation, you cannot complete either course without completing the other one first, creating an impossible situation.

2. Visiting All Courses: If no cycles are detected during the DFS traversal, it means that we can order our courses (or visit all nodes in the graph) in a way that meets all the prerequisite requirements. In other words, we can complete all courses.

This is the essence of the DFS-based solution for this problem. This algorithm’s correctness relies on the fundamental properties of DFS and its ability to detect cycles in a graph, which directly determines whether it is possible to complete all courses.

Remember, DFS explores as far as possible along each branch before backtracking. So if there’s a path from a node back to itself (which creates a cycle), DFS will inevitably find it, and that’s how we are sure of the correctness of the algorithm.

Stepwise Refinement

Let’s break down the problem-solving approach into detailed steps:

1. Graph Creation: The first step involves creating a directed graph to represent the courses and their prerequisite relationships. The graph should have a node for each course and a directed edge for each prerequisite pair. The edge goes from the dependent course to the prerequisite course. This step can be done independently from the others, as it is primarily data organization.

2. Initialization: Create an array to store the status of each node (course). The status can be “not visited”, “visiting”, or “visited”. This array helps track the visiting status of each node in the graph during the DFS traversal.

3. DFS Traversal and Cycle Detection: Once the graph is set up, we initiate a DFS from each node that hasn’t been visited yet. If we ever encounter a node marked as “visiting” during the DFS, it means we’ve found a cycle, and we can return false because we can’t complete all the courses. If the DFS traversal is completed without finding any cycles, mark the node as “visited”. This step is crucial and forms the core of our problem-solving approach.

4. Complete Traversal Check: After performing DFS traversal from all nodes, if no cycles have been detected, it means we can complete all courses. So, we return true.

The pattern within this solution is the DFS traversal, which is repeated for each node in the graph. During each DFS traversal, the status of each node is tracked and updated, providing us with a way to detect any cycles.

The cycle detection is a sub-problem that can be considered as being solved independently within the DFS traversal. It is determined by the presence of a node marked as “visiting” during the DFS. Once a cycle is detected, the problem is solved, and we can return false.

While the problem-solving approach seems sequential, graph creation and DFS traversal are two major parts that can be thought of as independent processes. The graph creation involves preparing our data, and the DFS traversal is about processing this data to solve the problem.

Solution Approach and Analysis

Let’s break this problem down into a detailed solution approach:

1. Graph Creation: The first step involves representing the courses and their relationships. You can imagine this as a group of students forming a line to get their lunch. Each student represents a course, and the order in which they stand represents the course prerequisites (you must get behind a student (course) if they are a prerequisite to your course). If the students can form a single line without any loops or breaks, then it’s possible to complete all the courses. In our case, each student is a node in our graph, and the line between them represents the edges (directed from a course to its prerequisite).

2. Initialization: Now that we have the students in a line, we need to keep track of their states as we visit each of them. This can be thought of as a tag that each student gets - “not visited” (we haven’t reached the student in line), “visiting” (we are currently visiting the student), or “visited” (we have finished visiting the student and moved forward).

3. DFS Traversal and Cycle Detection: We start from the first student and visit each student in line (each node in the graph). If we find a student that we’re currently visiting again, this means there is a loop in the line (a cycle in the graph), and we can’t complete all the courses. This is like saying, you can’t get your lunch until the student behind you gets theirs, but they also can’t get their lunch until you get yours, creating an impossible situation. In this case, we return false.

4. Complete Traversal Check: If we manage to visit all students in line without encountering any loops (completing the DFS traversal without finding any cycles), this means we can complete all courses, so we return true.

Now let’s discuss how changes in the problem’s parameters would affect the solution:

1. Increase in numCourses: As the number of courses (nodes in the graph) increases, the graph becomes more complex, increasing the time complexity of the DFS traversal. However, the overall approach remains the same.

2. Increase in prerequisites: As the number of prerequisites (edges in the graph) increases, the graph again becomes more complex. This may also increase the chance of having a cycle in the graph, making it more likely that we can’t complete all courses.

Lastly, let’s walk through an example:

Let’s say, numCourses = 4, prerequisites = [[1,0],[2,1],[3,2]]. This can be visualized as a line of students: 0 -> 1 -> 2 -> 3. Starting with student 0 (DFS from node 0), we can visit all students without encountering any loops. Hence, it’s possible to complete all courses, and we return true.

If the prerequisites change to [[1,0],[2,1],[3,2],[0,3]], the line of students becomes a loop: 0 -> 1 -> 2 -> 3 -> 0. Here, while visiting the students, we encounter student 0 again while we’re still visiting them, indicating a loop. Hence, it’s not possible to complete all courses, and we return false.

Identify Invariant

In the context of this problem, the invariant can be seen as the assumption that there must not exist a cyclic dependency among courses for them to be completed.

The invariant is essentially a condition that remains true throughout the execution of the algorithm. In this case, as we perform a Depth-First Search (DFS) traversal of the graph (courses and their prerequisites), the invariant of “no cyclic dependency” is used to determine if we can finish all courses or not. If at any point during the traversal we encounter a node that’s already being visited (indicating a cycle), this breaks our invariant, and we return false.

On the other hand, if we successfully traverse the graph without finding any cycles, our invariant holds true, meaning we can finish all courses, and we return true.

This invariant, the absence of cycles, is key to understanding and solving the problem. The entire DFS algorithm is structured around maintaining this invariant and detecting when it is violated (indicating an impossibility to complete all courses).

Identify Loop Invariant

In this problem, a loop invariant during the Depth-First Search (DFS) traversal could be the claim that “each node we visit in our DFS is either part of a completed subtree with no cycles, or it is part of the currently exploring path.”

During each iteration of the DFS loop, we are exploring the next node in the DFS traversal. We use the state of each node (visited, visiting, or not visited) to maintain this loop invariant:

• If we encounter a node that is in the “visited” state, we know it’s already part of a subtree we have finished exploring, and we have confirmed that there are no cycles in that subtree. This satisfies the loop invariant.

• If we encounter a node that is in the “visiting” state, it means that we have reached a node that is already in our currently exploring path, which breaks the invariant and indicates a cycle in the graph. We can then return false.

• If we encounter a node that is in the “not visited” state, we mark it as “visiting” and continue to explore the subtree rooted at this node, trying to maintain the invariant.

The loop ends when all nodes have been visited, meaning we have explored the entire graph and found no cycles (as long as we haven’t returned false earlier). In that case, the invariant still holds, and we can conclude that it’s possible to finish all courses. We return true in this case.

Thought Process

Let’s breakdown the thought process and approach to solve this problem step-by-step:

1. Understand the problem: The problem asks us to determine if it’s possible to take all courses given a list of pairs of prerequisite courses. Each pair [a, b] in the prerequisites means the course b is a prerequisite of course a.

2. Identify the cues: The mention of a pre-order relation (course ‘b’ must be taken before course ‘a’) and the need to check if it’s possible to order things (courses) based on this relation hint towards topological sorting or Depth-First Search (DFS) in graph theory. We can model the problem as a graph where courses are nodes, and prerequisite relations are directed edges.

3. Design the approach:

   • Initialize a graph (as an adjacency list) and a visit list to keep track of the visit state of each node (course).
   • Use DFS to traverse through each node (course). During this traversal, if we come across a node that is being visited (i.e., it’s in the current path), then we’ve found a cycle, and it’s impossible to finish all the courses, so we return false. If we successfully visit all nodes without encountering any cycles, return true.

Let’s code this up in Python:

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class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        # Create the adjacency list representation of the graph
        graph = [[] for _ in range(numCourses)]
        visit = [0 for _ in range(numCourses)]
        for x, y in prerequisites:
            graph[x].append(y)

        def dfs(i):
            # If we've come across a node that's in the current path, a cycle exists
            if visit[i] == -1:
                return False
            # If the node has been visited and it's not in the current path, no cycle exists from this node
            if visit[i] == 1:
                return True
            # Mark the current node as being visited
            visit[i] = -1
            for j in graph[i]:
                if not dfs(j):
                    return False
            # After visiting all the neighbours of the current node, mark it as visited
            visit[i] = 1
            return True

        for i in range(numCourses):
            if not dfs(i):
                return False
        return True

In this solution, visit[i] is 0 if the node has not been visited, 1 if the node has been visited and is not in the current path, and -1 if the node is being visited (is in the current path). The function dfs(i) checks if a cycle exists starting from node i.

Establishing Preconditions and Postconditions

Let’s break down the method canFinish(numCourses, prerequisites) according to the points you mentioned:

1. Parameters:

   • The method takes two inputs: numCourses and prerequisites.
   • numCourses is an integer representing the total number of courses.
   • prerequisites is a list of pairs where each pair represents a course and its prerequisite.

2. Preconditions:

   • There are no strict preconditions about the state of the program before this method is called.
   • The constraints on the input parameters are as defined in the problem: 1 <= numCourses <= 2000, 0 <= prerequisites.length <= 5000, prerequisites[i].length == 2, 0 <= ai, bi < numCourses. All the pairs prerequisites[i] are unique.
   • No specific state is required before the method execution.

3. Method Functionality:

   • The method is expected to return a boolean indicating whether it’s possible to finish all courses given the prerequisites.
   • It constructs a graph from the input list of prerequisites, then uses DFS to detect cycles in the graph. If a cycle is detected, it means that there is a circular dependency between courses, so it’s impossible to finish all courses, and the method returns false. Otherwise, it returns true.

4. Postconditions:

   • After the method is called, it doesn’t modify the input parameters or any global state.
   • The return value indicates whether it is possible to finish all courses given the prerequisites. True if possible, False otherwise.
   • There are no side effects of this method.

5. Error Handling:

   • If the preconditions are not met (e.g., if inputs are not within the specified range or if prerequisites are not formatted correctly), the method’s behavior is undefined. It may throw an exception (like an IndexError or TypeError) or it may return incorrect results.
   • The method itself doesn’t include explicit error handling for invalid inputs. It’s assumed that inputs are well-formed according to the problem’s constraints.

Problem Decomposition

Let’s analyze the problem based on these points:

1. Problem Understanding:

   • We are given a number of courses and a list of prerequisites for each course. We are asked to determine if it is possible to complete all the courses given these prerequisites.

2. Initial Breakdown:

   • The major part of the problem is managing the prerequisites and the courses. This can be further broken down into: a) Representing the prerequisites in a structured way. b) Checking if it’s possible to complete all courses.

3. Subproblem Refinement:

   • Representing the prerequisites: a) We can think of the prerequisites as a directed graph, where nodes represent courses and edges represent dependencies between courses.
   • Checking if it’s possible to complete all courses: a) This can be further broken down into performing a Depth-First Search (DFS) to detect any cycles in the graph.

4. Task Identification:

   • The tasks that are repeated or similar are creating nodes for each course and edges for each dependency, and traversing the nodes.

5. Task Abstraction:

   • For each task, the level of abstraction is quite reasonable. They are abstract enough to be clear and reusable, and they make sense in the context of the problem.

6. Method Naming:

   • The task of representing prerequisites could be named build_graph.
   • The task of checking if it’s possible to complete all courses could be named detect_cycle.

7. Subproblem Interactions:

   • The build_graph method needs to be executed first to create a structured representation of the prerequisites. Once the graph is created, we use it as input to the detect_cycle method. The latter doesn’t have dependencies and can be executed right after the graph is built.

From Brute Force to Optimal Solution

Let’s start with a brute force solution:

A naive way to approach this problem would be to generate all possible sequences of courses and then check each sequence to see if it adheres to the prerequisites. We can think of the prerequisites as constraints on the order of courses in the sequence. If we find a sequence that satisfies all the prerequisites, we return true; otherwise, we return false.

However, this approach is extremely inefficient. The number of possible sequences of courses is numCourses! (the factorial of numCourses), because there are numCourses options for the first course, numCourses-1 for the second, and so on. Therefore, the time complexity of this brute force solution is O(N!), where N is the number of courses. This is not feasible for large inputs.

Now, let’s think about how we can optimize this:

As mentioned before, we can represent the course prerequisites as a directed graph. The nodes of the graph represent courses, and there is an edge from course A to course B if B is a prerequisite of A.

With this graph representation, the problem of checking whether we can complete all courses reduces to checking whether this graph contains a cycle. If the graph does contain a cycle, it means there is a circular dependency among the courses, and we cannot complete them all. If there is no cycle, it means we can complete all the courses.

We can use Depth-First Search (DFS) to check for a cycle in the graph. DFS is efficient for this task because it can detect cycles in a graph in linear time. DFS visits each node exactly once, so the time complexity is O(N), where N is the number of nodes (or courses).

By reframing the problem in terms of graph theory, we’ve reduced the time complexity from O(N!) to O(N), which is a huge improvement. The space complexity of the solution is also O(N), because we need to store the graph and the DFS stack.

Finally, let’s note that this optimization relies on two key insights:

1. We can represent the prerequisites as a directed graph, and
2. We can use DFS to efficiently check for cycles in this graph. Without these insights, we might not have been able to find this optimized solution.

Code Explanation and Design Decisions

1. Initial Parameters: The initial parameters are numCourses and prerequisites. numCourses is the total number of courses you need to take, labeled from 0 to numCourses-1. Prerequisites is a list of pairs of courses in which the second course is a prerequisite of the first one. These parameters form the basis of our problem. They provide the information we need to build our graph and subsequently determine if all courses can be completed.

2. Primary Loop: In our solution, there is a primary loop over all the nodes/courses. Each iteration represents an attempt to traverse the graph from a specific node/course. The iteration advances the solution by recursively checking if the current course could depend on itself (which would indicate a cycle).

3. Conditions within the loop: In our DFS traversal, we have conditions to check if the node has been visited or is being visited. If a node is being visited and we encounter it again, this signifies a cycle in our graph - a course is dependent on itself through a chain of prerequisites, which violates our constraints.

4. Modifications to parameters: We use a variable visited to track the state of each node/course as we traverse the graph. The visited variable is updated from 0 (unvisited) to -1 (being visited) to 1 (completely visited). This reflects changes in the state of traversal and helps us detect cycles.

5. Invariant: An invariant in our problem could be that the visited status of any node that is not currently being explored remains unchanged. This helps meet the problem’s constraints as it ensures we only mark a course as fully visited (1) once we’ve successfully explored all its dependencies without finding a cycle.

6. Final Output: The final output is a boolean value indicating whether or not all courses can be completed given the prerequisites. It satisfies the problem’s requirement by effectively utilizing the course prerequisite pairs to perform a DFS on the built graph and checking for cycles. If no cycle is detected, it means all courses can be taken in a certain order fulfilling all prerequisites, and the function returns True. If a cycle is detected, it indicates that there are courses that require themselves to be completed beforehand, which is not possible, hence the function returns False.

Coding Constructs

1. High-Level Strategies: This code utilizes Depth-First Search (DFS) - a common technique for traversing or searching tree or graph data structures. It also uses the concept of a state machine with three states - unvisited, being visited, and completely visited - to track the traversal progress and detect cycles.

2. Explaining to a Non-Programmer: Imagine you’re a student with a list of courses to take, and some courses have prerequisites, meaning you must complete certain courses before starting others. This code is like a guide helping you figure out if you can finish all courses given these prerequisites. It’s like a pathfinder through a maze of courses, where the entry is a course, and the exit is another course that can only be taken once the entry course is complete.

3. Logical Elements: The main logical constructs are the iterations (loops) used for traversal and condition checks for state changes and cycle detection.

4. Algorithmic Approach: We create a graph of courses with directed edges representing prerequisites. Then, starting from each unvisited course, we travel as far as we can along the edges, marking each course as being visited. If we encounter a course that’s already being visited, we’ve found a cycle, and return false. If we exhaust all courses and find no cycles, we return true.

5. Key Steps: The code is primarily performing a DFS on the graph created from the course prerequisites. The reason for this is to detect cycles, which indicate a course dependency on itself, making it impossible to complete all courses.

6. Algorithmic Patterns: This code follows the pattern of DFS for graph traversal. It also applies a cycle detection strategy for directed graphs, specifically the color marking (or 3-state marking) method. These patterns are common in many graph-related problems and are language-agnostic.

Language Agnostic Coding Drills

1. Distinct Coding Concepts:

   • Graph Representation: The understanding of how to represent a graph in code, especially the adjacency list representation which is often used in graph-related problems.

   • Loop Constructs: Mastery of basic loop constructs for iteration over elements.

   • Conditional Branching: Understanding of if-else statements to control the flow of code based on conditions.

   • Depth-First Search (DFS): Proficiency in implementing DFS for graph traversal. It involves recursive function calls and understanding of the call stack.

   • Cycle Detection in Graph: Knowledge of detecting a cycle in a directed graph, which is a classic application of DFS.

   • State Management: Understanding of how to manage and update states of elements (in this case, nodes in a graph) during processing.

2. Coding Concepts by Difficulty:

   • Loop Constructs: The basic understanding of iteration and looping through elements is the simplest concept, and the first one to master.

   • Conditional Branching: Control flow with if-else statements is also fundamental and typically learned early on.

   • Graph Representation: Representing a graph might be slightly challenging for beginners. It requires a good understanding of data structures, specifically lists in this case.

   • State Management: Keeping track of states is a more advanced topic, involving keeping track of changes to variables during execution.

   • Depth-First Search (DFS): DFS requires understanding of recursion and stack data structure, making it a more complex concept.

   • Cycle Detection in Graph: This is the most advanced concept among the listed. It requires an understanding of DFS and state management and the ability to apply these in the context of graphs.

3. Problem-Solving Approach:

   • Understand the problem and identify that it can be represented as a directed graph problem with potential cycles.

   • Learn how to represent a directed graph using adjacency lists. In this context, the graph nodes represent the courses, and the directed edges represent the prerequisite relationships between them.

   • Understand how to traverse a graph using DFS. Each node in this case represents a course, and a traversal means taking a course.

   • Learn to track the state of each node (unvisited, being visited, or completely visited) during the traversal. This helps determine whether we’ve already started (but not yet finished) taking a course, or whether we’ve completed it.

   • Understand that finding a cycle (an already ‘being visited’ course while traversing) means that there are circular prerequisites, which makes it impossible to complete all courses.

   • Using all these concepts, devise a solution to traverse each node using DFS while keeping track of their states. Return true if no cycles are detected, indicating that all courses can be completed. If a cycle is detected, return false.

Targeted Drills in Python

1. Loop Constructs:
   • Iterate over a list of numbers and print each one.

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   numbers = [1, 2, 3, 4, 5] for number in numbers: print(number)

2. Conditional Branching:
   • Print whether each number in a list is even or odd.

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   numbers = [1, 2, 3, 4, 5] for number in numbers: if number % 2 == 0: print(f"{number} is even") else: print(f"{number} is odd")

3. Graph Representation:
   • Represent a directed graph as an adjacency list.

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   graph = { 'A': ['B', 'C'], 'B': ['D'], 'C': [], 'D': ['E'], 'E': [] } print(graph)

4. State Management:
   • Maintain and update the state of each number in a list (for example, double each number).

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   numbers = [1, 2, 3, 4, 5] for i in range(len(numbers)): numbers[i] *= 2 print(numbers)

5. Depth-First Search (DFS):
   • Perform a DFS on a graph starting from a node.

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   def dfs(graph, node, visited): print(node) visited.add(node) for neighbor in graph[node]: if neighbor not in visited: dfs(graph, neighbor, visited) graph = {'A': ['B', 'C'], 'B': ['D'], 'C': [], 'D': ['E'], 'E': []} visited = set() dfs(graph, 'A', visited)

6. Cycle Detection in Graph:
   • Detect a cycle in a directed graph using DFS.

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   def has_cycle(graph, node, visited, stack): visited.add(node) stack.add(node) for neighbor in graph[node]: if neighbor not in visited: if has_cycle(graph, neighbor, visited, stack): return True elif neighbor in stack: return True stack.remove(node) return False graph = {'A': ['B', 'C'], 'B': ['D'], 'C': [], 'D': ['A'], 'E': []} visited = set() stack = set() print(has_cycle(graph, 'A', visited, stack))

In the problem of detecting a cycle in a course prerequisite graph, we first need to represent the problem using a graph data structure. This is where the graph representation drill comes in handy. The state management drill helps us understand how to track the state of each course, which is essential to identify whether a course is already being processed or has been completed.

The DFS drill is integral to our traversal of the course graph. This, combined with the cycle detection drill, allows us to effectively check for cycles in the course prerequisite graph. The loops and conditional branching drills are fundamental constructs used in all these drills.

When it comes to assembling all these drills together to solve the original problem, the sequence would be as follows:

• Start with the graph representation to model the courses and their prerequisites.
• Use a loop to iterate over all the nodes (courses) in the graph.
• For each node, use DFS for traversal. During DFS, use state management to track the states of nodes.
• Implement cycle detection as part of the DFS process. If a cycle is detected, return False since all courses cannot be completed.
• If no cycles are detected after examining all nodes, return True since all courses can be completed.

Q&A

Python/C# solutions seems faster than C/C++/Java solutions. It is so strange, is there anything like in-built data structure for graph that I don’t know in Python/C#?

It’s not quite accurate to say that Python or C# solutions are intrinsically faster than C/C++/Java solutions. The performance of a solution depends more on the specifics of the implementation and the efficiency of the algorithm, rather than the language used.

In terms of language-level performance, C and C++ are generally considered faster because they are compiled languages and allow for low-level programming, such as memory management. Java is also compiled (to bytecode, which is then interpreted or compiled at runtime), and its performance is typically in the same order of magnitude as C and C++. Python and C#, on the other hand, are interpreted languages and are generally slower.

However, Python and C# (like Java) have extensive standard libraries and built-in data structures that can make it easier to write efficient high-level code. For example, Python’s list and dictionary types are implemented as dynamic arrays and hash tables, respectively, which can be more efficient than simpler data structures for certain operations.

In the context of a graph, Python’s built-in dictionary type can be used to implement an adjacency list quite efficiently. Here’s an example of how to do it:

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graph = {}  # Create an empty dictionary to store the graph
graph['A'] = ['B', 'C']  # Add an edge from 'A' to 'B' and 'C'
graph['B'] = ['A', 'D']  # Add an edge from 'B' to 'A' and 'D'
graph['C'] = ['A', 'D']  # Add an edge from 'C' to 'A' and 'D'
graph['D'] = ['B', 'C']  # Add an edge from 'D' to 'B' and 'C'

In C#, similar data structures exist such as Dictionary, List, HashSet etc. which can be used in a similar fashion to efficiently solve graph-based problems.

However, it’s important to note that Python’s and C#’s built-in data structures are not specific to graphs. They’re general-purpose data structures that can be used to implement a variety of abstract data types, including (but not limited to) graphs.

In conclusion, the choice of programming language doesn’t fundamentally determine the performance of a solution. Rather, the algorithmic design and the way data structures are used play a much larger role in determining performance. Python and C# may appear faster in some contexts simply because their high-level features and standard libraries make it easier to write efficient code.

In my case, I was visiting same course even though I already visited. Try this:

Keep track of a set, visited, and add each course into it. At the start of iteration/recursion, check if course is already added in the visited set. If so, terminate the iteration/recursion immediately.

in this way, we don’t visit course we already have visited. And remember that this is separate tracking than path, which is used to detect the cycle.

The method you described is a common way to optimize Depth-First Search (DFS) algorithms and is especially useful in graph problems like this. By maintaining a “visited” set, we can avoid redundant work and prevent the algorithm from going into infinite loops if there are cycles in the graph.

In Python, you might implement this strategy like so:

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def canFinish(numCourses, prerequisites):
    graph = [[] for _ in range(numCourses)]
    visited = [0 for _ in range(numCourses)]
    
    # create graph
    for pair in prerequisites:
        x, y = pair
        graph[x].append(y)
    
    # visit each node
    for i in range(numCourses):
        if not dfs(i, graph, visited):
            return False
    return True

def dfs(i, graph, visited):
    # if ith node is marked as being visited, then a cycle is found
    if visited[i] == -1:
        return False
    # if it is done visited, then do not visit again
    if visited[i] == 1:
        return True
    # mark as being visited
    visited[i] = -1
    # visit all the neighbours
    for j in graph[i]:
        if not dfs(j, graph, visited):
            return False
    # after visit all the neighbours, mark it as done visited
    visited[i] = 1
    return True

In the above code, for the visited list, we use 0 to indicate that the node has not been visited, 1 to indicate it has been visited and processed (i.e., all its descendants have been visited), and -1 to indicate that it is currently being visited. If we ever encounter a -1 during DFS, that means we have a cycle, and we return False immediately.

This optimized DFS approach reduces the time complexity significantly, especially for dense graphs, because we avoid reprocessing nodes that we’ve already processed. This approach will not only help to speed up the program but also make it more efficient by saving computational resources.

The intention of this problem I believe is to dectect cycle in a graph. But I’m not sure if the problem description makes sense. Say I have courses dependency [[0,1],[1,0],[1,2]], the standard solution returns ‘False’ for this case. But for course schedule purpose, I can take 2->1->0 to finish all three courses. So the code should actually return ‘True’. Please correct me if I’m wrong.

I understand your point of view, and it makes sense in a real-world scenario. However, the problem assumes that a course must be completed before it can be taken as a prerequisite for another. Therefore, in your example of [[0,1],[1,0],[1,2]], the first two pairs form a cycle: course 0 depends on course 1, and course 1, in turn, depends on course 0. So, according to the problem’s specifications, it’s impossible to complete these two courses because they each depend on the other being completed first, which is a deadlock.

Even though course 2 can be taken first, it doesn’t help to break the cycle between course 0 and course 1. In this case, course 2 could be completed, but courses 0 and 1 could not be. The problem requires that all courses should be finishable, so in this case, the result is False. The cycle detection in the graph is used exactly for this purpose - to ensure there’s no circular dependency between the courses.

It might be more relatable to think about it this way: suppose course 1 is “Physics I” and course 0 is “Physics II”. You would need to pass “Physics I” before you can take “Physics II”, but you can’t take “Physics I” without first passing “Physics II” (because of the cycle), so you’re stuck.

I hope this explanation helps clarify the intention behind the problem!

I returned false for the following test case: 5 [[1,4],[2,4],[3,1],[3,2]]

and the result shows it should be true. Can someboly explain why there are only four courses (1,2,3 and 5) but we need to take 5?

The confusion seems to be arising from a misunderstanding of how the courses are labeled in the problem. Courses are labeled from 0 to n-1 where ’n’ is the number of courses. In your case, the courses are actually 0, 1, 2, 3, and 4. The problem does not say that the course labels will start from 1, it says they are from 0 to n-1.

So for the test case of 5 courses with prerequisites [[1,4],[2,4],[3,1],[3,2]], it means you have 5 courses labeled 0, 1, 2, 3, and 4. Here, course 0 doesn’t have any prerequisites, course 4 is a prerequisite for course 1, course 4 is also a prerequisite for course 2, and courses 1 and 2 are prerequisites for course 3.

So one possible way to take all courses would be: 0 -> 4 -> 1 -> 2 -> 3.

Therefore, it is indeed possible to finish all the courses, and the result should return True.

I’ve read a lot of solutions and did a bunch of Googling but I cannot figure out why it would be better to use topological sort versus simply detecting the cycle.

I understand if we detect a cycle, we cannot take all courses because if there is a cycle, we can never take the courses in the cycle and therefore can’t take all courses. This solution makes sense to me.

I slightly understand how topological sort does this but not really. Is it because the topological sort will not return an ordering of the same length as the nodes in the graph that we are able to figure out this graph has cycles? Is that a use case of topological sort – to figure out if we have cycles or not?

Also, both algorithms are O(V + E) runtime and O(V) space right?

Yes, you’re correct on all points.

Both cycle detection and topological sorting are valid solutions to this problem, but they offer slightly different perspectives.

1. Cycle detection is more straightforward: if a cycle exists, then at least one course in the cycle cannot be completed, because its prerequisites include itself (indirectly), and the problem is unsolvable.

2. Topological sorting, on the other hand, doesn’t just solve the problem, it also provides an additional piece of information: an order in which to take the courses. When a topological sort algorithm fails to generate a valid ordering for a directed graph (i.e., the output order is shorter than the number of nodes), this indicates that a cycle must exist in the graph. Therefore, it not only indicates the presence of a cycle, but also provides a valid course schedule when no cycles are present.

As for time and space complexity, both algorithms indeed have a time complexity of O(V + E) and a space complexity of O(V), where V is the number of vertices (courses, in this case) and E is the number of edges (dependencies between courses). So, both approaches are quite efficient. However, using topological sort might offer more context for certain applications where knowing the order of tasks is beneficial.

Using vector of vector should imply a adjacency list/matrix representation of the prereq graph. If you want to represent the graph simply as a collection of edges, you should use vector<pair<int, int> > instead. Such ill-formatted input with insufficient specification is time-wasting and should be discouraged in industry, and in interviews.

Yes, you’re correct. In an ideal scenario, problem statements should be clear and unambiguous, including the expected input and output formats.

In this problem, the adjacency list/matrix representation is implied by a vector of vectors (in C++), or list of lists in other languages like Python or Java. However, your suggestion to use a vector<pair<int, int>> would be a more explicit way to represent a collection of edges, which could be more intuitive for some programmers.

Keep in mind though, that in real-world applications and especially in coding interviews, sometimes you may encounter poorly specified or ambiguous problems. Being able to clarify the requirements, make reasonable assumptions, or work with the given format regardless is an important skill in itself. These situations can test your ability to think critically, ask clarifying questions, and adapt to the given situation.

In terms of this particular scenario, the provided format is quite common for graph-related problems in competitive programming and coding interviews. Therefore, it’s good to be familiar with this kind of representation and comfortable working with it.

3 [[1,0]]

Why is this true.

Since one needs to complete 3 courses and only two is present in the list of requirements (0, 1)

The confusion here arises from a slight misunderstanding of the problem statement. When it says that you need to complete numCourses (3 in your example), it means there are 3 courses labeled from 0 to numCourses - 1 (0, 1, 2 in your example).

The pair [1, 0] indicates that course 0 is a prerequisite for course 1. The absence of course 2 in the list means that course 2 has no prerequisites and can be taken at any time.

So, a possible order for taking the courses could be 2, 0, 1. Thus, it is possible to finish all the courses, and the function returns True.

I think the description does not explicitly explain the situation where a course has multiple prerequisites. It may be interpreted differently. The course can be taken if:

1. ALL the prerequisite courses should be taken;
2. EITHER one of the prerequisite course should be taken; The solution here implicitly take the first assumption, so if there is a circle in the graph, then there is no solution that all course can be taken. But if the second assumption is required, this problem cannot be solved by checking if there is a circle in the graph. For example:

imageCourse 4 has two prerequisites(1 and 3). If either of them taken counts, then the solution can be: 3->4->2->1. If both of them should be taken before 4 is taken, there is no solution. In this case, there is still a circle in the graph, but all courses can be taken.

You’re correct that the problem assumes that all prerequisites for a course need to be taken before that course. This is the standard assumption in prerequisite relationships, and it’s what makes this problem a classic problem of cycle detection in a directed graph.

The confusion seems to arise from the interpretation of the prerequisites. In the problem statement, each pair of numbers in the prerequisites list represents a one-way relationship where the second course is a prerequisite for the first one. Therefore, in the problem statement, each course can have multiple prerequisites, and all of these must be completed before you can take the course.

If the problem were interpreted the other way, where a course can be taken if any one of its prerequisites has been taken, then yes, this would be a different problem that couldn’t be solved simply by looking for cycles. But under the standard interpretation where all prerequisites must be taken first, looking for cycles in the course prerequisite graph is the correct approach.

To clarify, in the case of your example, if the problem is interpreted such that all prerequisites must be completed before taking a course, then indeed there would be no valid way to complete all courses as there is a cycle in the prerequisites graph (1 -> 4 -> 3 -> 1).

I implemented BFS topological sort. After submitting, one test case shows conflicting results for 2 [[0,1]] The standard answer expects False while I give True. Is this a wrong test case?

The test case you mentioned is a valid one. The prerequisites list [[0,1]] signifies that course 1 is a prerequisite for course 0. This means you need to take course 1 before you can take course 0.

Considering that there are 2 courses numbered from 0 to 1 and no cycle in the prerequisite relationships, you should be able to take all the courses. The course order would be 1 -> 0. So, your BFS topological sort should return True.

If it returns False, there might be a problem in your code. I would recommend debugging it using this test case to understand where the issue is.

I checked most of the DFS solutions, they set the visited[i] to true when entering the function and reset it to false when leaving. Then they use each node as the starting node of the DFS call. Now consider these courses: 0->1->2->3->4->5-> … -> n-1, where a->b means a is b’s prerequisite. (actually the inverse is fine)

starting from 0 we visited n the nodes. Then we start from 1, we visited n-1 nodes. Total is n(n+1)/2 , O(n^2) or you can say O(E^2)

You raise a valid point. The worst-case complexity of the DFS approach you described is indeed O(n^2), where n is the number of nodes. However, it’s important to remember that we can optimize it.

One way to do this is by using a ‘visited’ array and a ‘recursion stack’. In this approach, we have two boolean arrays: one keeps track of the nodes visited in the current traversal, while the other keeps track of all the nodes visited so far. This helps us avoid traversing the same node more than once, significantly reducing the time complexity.

Here’s how it would work:

1. For each unvisited node, mark it as visited in the current traversal (in the ‘recursion stack’) and recursively visit all its neighbors. If you encounter a node that’s been visited in the current traversal (i.e., it’s in the recursion stack), you’ve found a cycle and can return false.

2. After you finish exploring a node and all its descendants, you remove it from the ‘recursion stack’ and add it to the ‘visited’ array.

3. If you complete the traversal without finding any cycles, return true.

This optimized DFS approach has a time complexity of O(V+E), where V is the number of vertices (courses) and E is the number of edges (dependencies). This is because every node and edge is visited exactly once.

Here’s the Python code for this optimized DFS approach:

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def canFinish(numCourses: int, prerequisites: List[List[int]]) -> bool:
    graph = [[] for _ in range(numCourses)]
    visited = [0 for _ in range(numCourses)]
    # create graph
    for pair in prerequisites:
        x, y = pair
        graph[x].append(y)
    # visit each node
    for i in range(numCourses):
        if not dfs(i, graph, visited):
            return False
    return True

def dfs(i: int, graph: List[List[int]], visited: List[int]) -> bool:
    # if ith node is marked as being visited, then a cycle is found
    if visited[i] == -1:
        return False
    # if it is done visited, then do not visit again
    if visited[i] == 1:
        return True
    # mark as being visited
    visited[i] = -1
    # visit all the neighbours
    for j in graph[i]:
        if not dfs(j, graph, visited):
            return False
    # after visit all the neighbours, mark it as done visited
    visited[i] = 1
    return True

In this solution, we maintain three states for each node: unvisited (0), being visited (-1), and done visited (1). These states help us detect any cycles in the graph, which would indicate that it’s impossible to complete all the courses.

I just converted the input data into a 2D array and then copy pasted code from yesterday’s daily #802

That’s a smart move! The ability to recognize similar problems and reuse solutions is a valuable skill in programming and problem-solving. After all, many problems in computer science are variations or combinations of a set of base problems.

Problem #802, “Find Eventual Safe States”, is also about graph theory and detecting cycles, so it’s indeed quite similar to the “Course Schedule” problem. In both problems, we are given a directed graph and asked to identify if a condition related to cycles is true or not.

However, while it’s great to reuse solutions, it’s also essential to understand the differences and nuances of each problem:

1. In “Course Schedule”, we are asked if it’s possible to take all courses (i.e., visit all nodes in the graph) without encountering a cycle. We return a boolean indicating the possibility.

2. In “Find Eventual Safe States”, we are asked to find all nodes (states) from which we can reach a terminal state (a state that isn’t part of a cycle). We return a list of these “safe” nodes.

As such, while the core graph traversal logic is similar, the manner in which we use that logic to solve the problem and the kind of output we produce differs between the two problems.

It’s this kind of understanding that will help you become not just a good coder, but a good problem solver!

Is the graph given is guaranteed to be connected?

The problem statement for the “Course Schedule” problem doesn’t guarantee that the graph will be connected. The graph represents course prerequisites, and it’s entirely possible to have a set of courses that are unrelated to another set of courses.

In other words, you could have several independent subgraphs, each representing a different “cluster” of interrelated courses. For example, you could have a set of Math courses that don’t have any prerequisites in common with a set of History courses, leading to multiple separate subgraphs.

Therefore, when implementing a solution for this problem, it’s important to ensure your algorithm can handle disconnected graphs. This usually involves iterating over all nodes in the graph, rather than starting the traversal from a single node.

I find that the dependency is only represented by pairs. Does it mean for each class there is only one prerequisite for it?

In the “Course Schedule” problem, while each dependency is indeed represented by a pair, this does not mean that a course can have only one prerequisite. The list of pairs represents a directed graph, where each pair is an edge from one node (course) to another.

If a course has multiple prerequisites, it will be represented in multiple pairs. For instance, if course 0 is a prerequisite for both course 1 and course 2, this might be represented as [[1, 0], [2, 0]].

So, even though each individual pair includes one course and its direct prerequisite, a course can appear in multiple pairs, indicating it has multiple prerequisites. The solution must take into account that a course might have multiple prerequisites when constructing the graph and determining if it’s possible to complete all courses.

Please give some more better examples so that we can try to draw the graphs and trace the working. Both the examples provided here are too trivial.

Sure, let’s consider some non-trivial examples:

1. Suppose there are 5 courses labeled from 0 to 4. The course prerequisites are given as: [[0,1],[1,2],[2,3],[3,4]].

   This represents a scenario where each course is a prerequisite for the next one in the sequence. The graph would look like this:

   4 <- 3 <- 2 <- 1 <- 0
   

   As there are no cycles in the graph, all the courses can be completed.

2. Consider there are 4 courses labeled from 0 to 3. The course prerequisites are given as: [[0,1],[1,2],[2,0],[3,2]].

   The graph would look like this:

   0 <- 1 <- 2
   ^          |
   |          V
   2 <- 3
   

   In this case, courses 0, 1, and 2 form a cycle, so it’s not possible to complete all the courses.

3. If there are 6 courses labeled from 0 to 5, with the course prerequisites as: [[1,0],[2,0],[3,1],[4,2],[5,3],[5,4]].

   The graph would look like this:

   0 -> 1 -> 3 ->
               |
               V
   0 -> 2 -> 4 -> 5
   

   In this scenario, there are no cycles, so it’s possible to complete all the courses.

In all these examples, remember that each edge (represented as a pair in the input) points from a course to its prerequisite, and a cycle represents an impossible situation where a course is indirectly a prerequisite of itself.

everbody in comment is talking about graph and cycle but i was just asking whether it can be also be solved by some normal approaches as i was thinking and i think i got the way . if anybody knows that it can be solved by normal maths please confirm it to me ..

This problem does indeed involve graphs and cycles. The reason is that the courses and their prerequisites naturally form a directed graph, where each course is a node and the edges are the prerequisites. A cycle in this graph indicates a circular dependency, making it impossible to complete all the courses.

However, depending on your interpretation of “normal maths”, there may be an alternative perspective. You could view the problem as trying to find a valid order to take the courses, given the prerequisites. This is a problem of topological sorting, which is a standard algorithm in computer science. Still, it fundamentally involves graph theory.

For simpler cases, you could potentially use an approach involving sets or lists to check prerequisites, but this would essentially be a naive implementation of depth-first search or breadth-first search, and it wouldn’t scale well to more complex cases with many courses and prerequisites.

So while it’s theoretically possible to come up with a “non-graph” solution, it would be quite inefficient and not recommended for this problem. Using graph-based algorithms like DFS, BFS, or topological sorting is the most appropriate approach.

We just need to prepare adj list from prerequisites array and apply yesterday’s daily problem code

Similar to 802. Find Eventual Safe States

Indeed, the problem “Course Schedule” is quite similar to the problem “Find Eventual Safe States”. Both of these problems essentially involve detecting a cycle in a directed graph, which is represented by the courses and their prerequisites in the “Course Schedule” problem and by states and transitions in the “Find Eventual Safe States” problem.

Here is a general approach to solve both problems using depth-first search (DFS):

1. Construct a graph: Create an adjacency list from the given pairs of prerequisites (in the “Course Schedule” problem) or from the given transitions (in the “Find Eventual Safe States” problem).

2. Initiate DFS: Apply depth-first search to the graph, starting from each node that hasn’t been visited yet.

3. Detect cycles: During the DFS, use a “visited” array or set to keep track of the nodes that are currently being visited. If you encounter a node that’s already in the “visited” set, you’ve found a cycle.

4. Return the result: For the “Course Schedule” problem, return false if a cycle is detected, meaning that it’s impossible to complete all courses. For the “Find Eventual Safe States” problem, return the list of nodes that are not part of any cycle, i.e., the “safe” states.

Despite the superficial differences between the two problems, the underlying structure and solution strategy are the same. Thus, if you’ve solved one of these problems, you can apply the same approach to solve the other.

numCourses = 1 prerequisites = []

Why is the expected answer to this true? In my understanding, there are 0 courses offered and therefore one cannot complete number of courses.

The “Course Schedule” problem’s constraints are such that the numCourses parameter represents the total number of courses, numbered from 0 to numCourses - 1.

In your example, numCourses = 1 means that there is one course, which is course 0. The prerequisites array is empty, which means there are no prerequisite courses required to complete course 0. Therefore, it is possible to complete all courses (in this case, just one course), hence the expected output is True.

So even though the prerequisites array is empty, it doesn’t mean there are no courses offered. Rather, it implies there are no dependencies between courses, and all can be taken independently.

I saw many solutions, but I think we just need to create an adj matrix (or list) out of the input and find if cycle exists in that directed graph, simple cycle detection algorithm in directed graph. Why do we need topological sorting? Any thoughts?

Topological sorting and cycle detection are indeed both valid approaches to this problem, but they answer slightly different questions.

When you perform a cycle detection, you’re asking the question, “Is it possible to take all the courses?” If there is a cycle, then at least one course depends on itself (directly or indirectly), and so it’s not possible to take all the courses. The benefit of this approach is that it can stop as soon as it detects a cycle, which might be faster if the graph contains a cycle and that cycle can be found early.

On the other hand, topological sorting gives an answer to a slightly different question: “In what order should I take the courses?” A valid topological sort is only possible if the graph is a Directed Acyclic Graph (DAG). If the graph is not a DAG, then it contains a cycle, and topological sorting will fail.

So while both methods can be used to solve this problem, topological sorting also provides extra information (the order of courses) which might be useful in some cases. For example, if you need to know not just whether it’s possible to take all courses, but in what order they should be taken, then topological sort is the way to go.

In terms of time complexity, both methods are linear in the size of the graph, so there is no big difference there. However, the actual performance can depend on the specifics of the input data and the implementation of the algorithm.

3 [[1,0]]

how is this possible to return true? Doesn’t this mean that we want to take 3 courses, but there are only 2 courses in the list? There are no cycles in the graph, but there are only 2 nodes?

In this problem, the courses are represented as integers from 0 to numCourses-1. So when numCourses is 3, the possible courses are 0, 1, and 2.

The prerequisites are pairs of the form (i, j), where i is a course and j is a prerequisite for i. In your example, [[1, 0]] means that course 0 is a prerequisite for course 1.

However, it doesn’t mean that course 2 has any prerequisites. Therefore, it can be taken at any time.

The problem asks whether it’s possible to take all courses, given the prerequisites. In this case, it is possible: we can take course 0, then course 1 (since its prerequisite 0 has been taken), and finally course 2 (which has no prerequisites).

So the function should indeed return True for this input.

In my understanding, the dependencies are in order : 0->1->2->0

The above looks like a cycle as from the depency cycle, 1th course is a prerequisite for 0th course.

From your question, it seems like there might be some confusion about the interpretation of the dependencies. If the dependencies are listed as [[1,0],[2,1],[0,2]], this does indeed represent a cycle, where each course is dependent on the next in a circular way. In this case, it would be impossible to complete all courses, and the expected output would be False.

The list [0,1,2,0] seems like a cyclical dependency as you’ve mentioned, but it doesn’t directly correlate with the format of dependencies provided in the problem. In the problem, each dependency is a pair of two courses where the first course depends on the second.

For example, if we have the dependencies [[1,0],[2,1],[0,2]], we would read this as:

• Course 1 depends on Course 0 (Course 1 -> Course 0)
• Course 2 depends on Course 1 (Course 2 -> Course 1)
• Course 0 depends on Course 2 (Course 0 -> Course 2)

As you can see, this forms a cycle, making it impossible to complete all the courses.

What if we have only one course but the prerequisite would have two course .It dosent make sense for me that

In the context of the problem, it’s important to note that each element in the prerequisites list is a pair, where the first element of the pair is dependent on the second. Each pair represents one prerequisite, but a course can have multiple prerequisites. This is represented by multiple pairs with the same first element.

For example, if we have the prerequisites [[0,1],[0,2]], this means that Course 0 has two prerequisites: Course 1 and Course 2. In other words, to take Course 0, you must first complete both Course 1 and Course 2. This scenario does make sense within the context of the problem, as it’s not uncommon for a course to have multiple prerequisites.

In real-world terms, this might look like a advanced course in computer science that requires both an introductory programming course and a mathematics course to be completed first.

numCourses = 20 prerequisites = [[0,10],[3,18],[5,5],[6,11],[11,14],[13,1],[15,1],[17,4]]

This can’t be false

The output being False for the given input is correct. The reason is because there is a loop in the prerequisites, specifically at [5,5]. This means Course 5 is a prerequisite of itself, which creates an impossible condition to satisfy. No course can be its own prerequisite because it would mean you would have to complete the course before you can take it, which is a contradiction.

So, even though all the other prerequisites could potentially be satisfied, the presence of this loop means it’s impossible to complete all 20 courses, hence the function should return False.

We start by creating a graph representation of the courses and their prerequisites. Each course is a node, and the prerequisites are directed edges between the nodes. We then use Depth-First Search (DFS) to check if there is a cycle in the graph. If a cycle exists, it means it is impossible to finish all the courses because we would be stuck in an endless loop of prerequisites. To detect a cycle, we maintain two arrays: visited to keep track of visited courses during the DFS traversal, and recursionStack to track the courses in the current recursion stack. For each course, we perform DFS to check if a cycle exists by following the prerequisites. If we encounter a course that is already in the recursionStack, it means a cycle exists. If we have visited all the courses without detecting a cycle, we can finish all the courses. Otherwise, it is impossible to finish all the courses due to a cycle. By using this approach, we can determine whether it is possible to finish all the courses or not based on the presence of cycles in the prerequisites.

You’re right that the core of this problem is cycle detection in a directed graph. I’ll break down the steps of the problem-solving approach:

1. Constructing the graph: Each course is represented as a node, and the prerequisites are directed edges. If a course A has a prerequisite B, we create a directed edge from B to A.

2. Depth-First Search (DFS) and Cycle Detection: We perform a DFS to detect cycles. If a cycle exists, it means that there are circular dependencies among the courses and it is impossible to finish all the courses. During the DFS traversal, we maintain two sets or arrays - “visited” to keep track of the courses that we have already visited, and “recursionStack” to keep track of the courses in the current path or recursion stack. When we visit a node, we add it to both “visited” and “recursionStack”. When we backtrack or when our DFS is over, we remove the node from the “recursionStack”. If we encounter a course that is already in the “recursionStack” while doing the DFS, it means that a cycle exists.

3. Result Determination: If we have visited all the courses and did not detect a cycle, we return True as we can finish all the courses. If a cycle is detected at any point, we return False.

This problem is a classic application of DFS for cycle detection in a directed graph, which is a common pattern in graph traversal problems.

Why do we need to backtrack for this problem?

In this problem, backtracking is necessary to correctly identify whether a cycle exists in the graph representing the courses and their prerequisites. Let me explain why:

We are using a Depth-First Search (DFS) approach to traverse the graph, and we maintain a “recursion stack” or “path” that holds the nodes (courses) currently being explored. The purpose of this stack is to keep track of the path we are currently traversing.

If, during the DFS traversal, we come across a node that is already in the recursion stack, it means we have found a cycle: we’ve come back to a course that we are currently in the process of exploring, indicating a circular dependency in prerequisites.

However, after we finish exploring a node (i.e., we’ve checked all its adjacent nodes), we need to remove it from the recursion stack. This step is the “backtracking” part. It signifies that we have finished exploring this particular path (or branch) and are going back to check other paths. Without this backtracking step, once a node is added to the recursion stack, it would never be removed. This would result in incorrect cycle detection, as we might mistakenly think there’s a cycle when we encounter a node that was part of a previously explored and completed path.

In summary, backtracking in this context is necessary for accurate cycle detection when traversing the graph via DFS. It allows us to isolate and examine individual paths (or branches) through the course prerequisite graph independently.

Let’s consider a directed graph that represents the course prerequisites with four nodes (courses) as below:

1 -> 2 -> 3 -> 4

Here, 1 is a prerequisite for 2, 2 is a prerequisite for 3, and 3 is a prerequisite for 4. There is no cycle in this graph.

Let’s say we start a DFS traversal from node 1, and we keep a “recursion stack” (or path) to track the current traversal:

1. We start at node 1, add it to the stack. (stack = [1])
2. We move to node 2, add it to the stack. (stack = [1, 2])
3. We move to node 3, add it to the stack. (stack = [1, 2, 3])
4. We move to node 4, add it to the stack. (stack = [1, 2, 3, 4])

At this point, we’ve traversed all the nodes, and there are no more adjacent nodes to explore. If we don’t backtrack, our recursion stack will still be [1, 2, 3, 4].

Now, imagine we start another DFS traversal, this time starting from node 2. If we haven’t done any backtracking, our recursion stack still includes nodes 3 and 4 from the previous traversal:

1. We start at node 2, see it’s already in the stack, but add it again anyway. (stack = [1, 2, 3, 4, 2])
2. We move to node 3, see it’s already in the stack, so we falsely conclude that there is a cycle in the graph.

Here, the failure to backtrack (i.e., clear the recursion stack after completing a traversal) has led us to incorrectly identify a cycle when none exists.

In contrast, if we backtrack correctly, our recursion stack will be empty at the start of each new DFS traversal, allowing us to accurately determine whether a cycle exists.

Can this problem be solved without using recursion?

Yes, this problem can be solved without using recursion, by applying an iterative version of Depth-First Search (DFS) or Breadth-First Search (BFS) with a stack or queue, respectively.

One non-recursive approach uses the concept of “indegree” of a node in graph theory. Indegree of a node is the count of incoming edges to a node. The idea is that a course with no prerequisite (or an indegree of zero) can be taken right away. As you complete a course, you can reduce the indegree of the next course (or the courses that depend on the completed course).

You can implement this strategy using a queue to iteratively process the nodes of the graph as follows:

1. Compute the indegree of all nodes (courses) and initialize a count of visited nodes.
2. Add all nodes with an indegree of zero to a queue.
3. While the queue is not empty, dequeue a node, increment the count of visited nodes, and reduce the indegree of all its dependent nodes by 1. If this results in a new node with an indegree of zero, add it to the queue.
4. After the queue is empty, if the count of visited nodes is equal to the total number of courses, return True (all courses can be finished). Otherwise, return False (not all courses can be finished).

This approach is a form of topological sorting and also helps detect cycles. In a graph with a cycle, no node involved in the cycle will ever have an indegree of zero, so such nodes will never be added to the queue and visited. Thus, if not all nodes are visited, it implies the graph has at least one cycle.

follow up : what if we have to calculate the max number of doable courses

The problem of finding the maximum number of courses that can be taken, given a set of prerequisites, is a slightly different problem. This variant requires you to identify the largest acyclic subgraph (DAG) in the given graph, which is a difficult problem known as the Maximum Acyclic Subgraph Problem and is NP-Hard.

However, in practical scenarios with a reasonable number of courses, you can still use DFS or BFS and count the number of courses you can take without running into a cycle:

1. Use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the graph.
2. For each node you visit, increment a counter.
3. If you encounter a cycle, roll back the counter for the nodes involved in the cycle, as these courses cannot be taken.
4. Keep track of the maximum counter value you obtain across all DFS or BFS traversals (since the graph might not be connected, so you’d need to start from different nodes).

The maximum counter value at the end of this process will give you the maximum number of courses you can take.

This won’t be an efficient solution for a large number of courses as you might have to do multiple traversals, but it should work fine for a reasonable number of courses.

The same as yesterday, except you have to build the adjacency list/matrix yourself :)

Append a courses’s required courses list: adjList[prerequisites[i][1]].append(prerequisites[i][0])

Yes, you’re right! This problem can be broken down into smaller steps, and the creation of an adjacency list is one of the crucial steps. Here’s how you can do it in Python:

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from collections import defaultdict

def create_adj_list(prerequisites):
    adj_list = defaultdict(list)
    for course, prerequisite in prerequisites:
        adj_list[prerequisite].append(course)
    return adj_list

Here we are using a dictionary of lists to represent the adjacency list. The dictionary keys represent the prerequisite courses, and the values are lists of courses that require the prerequisite. collections.defaultdict is a subclass of the built-in dict class in Python, which overrides one method to provide a default value for a nonexistent key, thus simplifying the code.

You can then use this adjacency list to perform a DFS traversal and detect if there is any cycle in the course prerequisites, which will indicate if it is possible to finish all courses or not.

Exact same question from yesterday with change of language cycle detction

Yes, you are correct. The problem essentially requires detection of a cycle in a directed graph. The nodes in the graph represent courses and the edges represent prerequisites. If there is a cycle in the graph, it would mean that there is a circular dependency among the courses, making it impossible to complete all courses.

This problem is similar to other problems involving cycle detection in directed graphs. While the language or presentation of the problem might be different, the underlying concept and problem-solving strategy remain the same. Being able to recognize these patterns and apply known strategies or algorithms to different problem statements is a valuable skill in problem solving and coding.

If I apply kahn’s Algo with reversed graph i.e. adj[course[i][0]].push_back(course[i][1]) and do the same thing for outdegree vector,, it is not giving me correct answer,, can anyone tell what changes should I make ?

Kahn’s algorithm is a method to find a topological order in a directed acyclic graph (DAG), which is exactly what we want in this problem. However, reversing the graph and calculating outdegrees will not work because the prerequisites relation is reversed.

When you reverse the edges, you are essentially changing the direction of the prerequisite relation. In the original graph, an edge from node A to node B means “A is a prerequisite for B”, but in the reversed graph, the same edge means “B is a prerequisite for A”, which is not correct.

So, instead of reversing the graph, you should construct the graph as usual where an edge from node A to node B means “A is a prerequisite for B”, and then calculate the indegrees (number of incoming edges) for each node.

Here are the steps for using Kahn’s algorithm in this problem:

1. Build the adjacency list and indegree array: For each pair of courses in prerequisites, if course A is a prerequisite for course B, add A to the adjacency list of B, and increase the indegree of B by 1.

2. Initialize a queue and add all nodes with an indegree of 0 (these courses don’t have any prerequisites and can be taken right away).

3. While the queue is not empty, remove a node from the queue, and for each of its neighbors, decrease their indegree by 1. If a neighbor’s indegree becomes 0, add it to the queue.

4. If all nodes have been removed from the queue (their indegrees have all become 0), return true, meaning it’s possible to finish all courses. If there are still nodes with a positive indegree, return false, indicating that there is a cycle and it’s impossible to finish all courses.

Remember, each removal from the queue corresponds to taking a course, and reducing the indegree of its neighbors corresponds to removing that prerequisite for the other courses.

Hey everyone, I could really use some assistance with this! I started by using vector<vector> adj(numCourses) to create an adjacency list, but unfortunately, it resulted in a TLE (Time Limit Exceeded) error. However, when I switched to vector adj[numCourses], it miraculously got accepted! I’d love to hear your insights on why this change made a difference. If you find this topic interesting, please consider upvoting so we can gather more perspectives.

This might have to do with how C++ implements and handles different types of containers. Let’s take a look at both:

1. vector<vector<int>> adj(numCourses): Here, you’re declaring a vector of vectors. This structure can dynamically resize, which is one of its main advantages. However, it could also lead to overhead when frequently resizing and reallocating memory, which can result in slower performance. If the internal array needs to be resized frequently, it can lead to higher time complexity.

2. vector<int> adj[numCourses]: Here, you’re declaring an array of vectors. The size of the array is fixed at compile time, which avoids the overhead of resizing. Also, each individual vector within the array can still dynamically resize, providing a nice balance between flexibility and performance.

However, please note that dynamic arrays (like the second approach) can sometimes lead to undefined behavior if not used carefully, especially when they are returned from a function or if their size exceeds the stack limit.

That being said, the exact reason for the difference in performance can depend on various factors, including the specific problem, the compiler, the memory management of the system, and more.

I am not aware of any built-in data structure specifically optimized for graphs in Python, C#, or C++. However, you can represent graphs using common data structures such as lists, sets, and dictionaries (in Python) or arrays and maps (in C++ or C#). The choice depends on the specific requirements of the problem and the operations you need to perform.

For example, in Python, you might use a dictionary to represent a graph, where each key is a node and each value is a set or list of connected nodes. In C++ or C#, you might use a map for the same purpose, or an adjacency matrix represented as a two-dimensional array for dense graphs.

[[2,0],[1,0],[3,1],[3,2],[1,3]] pre=4 why this is false every course we can take 0,2,1,3 as 1 is taken

The list [[2,0],[1,0],[3,1],[3,2],[1,3]] represents the prerequisites for each course in this way:

• 0 is a prerequisite for 2 and 1
• 1 is a prerequisite for 3
• 2 is a prerequisite for 3
• 3 is a prerequisite for 1

If we draw this out as a graph, you can see there is a cycle between courses 1 and 3:

0 –> 1 –> 3
^ | | | 2 <——

In this case, to take course 3, you need to have completed course 1. However, to take course 1, you need to have completed course 3. This creates an infinite loop between courses 1 and 3 which can never be broken, so it’s impossible to take all of the courses.

The proposed order of 0, 2, 1, 3 does not work because you can’t take course 1 right after taking course 0. This is because course 1 has course 3 as a prerequisite, and course 3 hasn’t been taken yet at that point. Hence, you can’t complete all the courses which leads to the function returning False.

Same problem as https://leetcode.com/problems/parallel-courses/

While the problem “Course Schedule” and “Parallel Courses” on LeetCode both involve concepts related to graph traversal, they are not exactly the same problem.

1. “Course Schedule” is about determining if it’s possible to take all courses given a list of pair dependencies where the first course in each pair is dependent on the second course. Essentially, you’re tasked with detecting if there is a cycle in the course dependency graph.

2. “Parallel Courses” also involves course dependencies, but it introduces an additional complexity. The problem not only asks whether all courses can be taken, but also wants to know the minimum semester to finish all courses. Each semester, you can take any course but need to respect the prerequisites.

The solution strategies for both problems involve constructing a graph from the given prerequisites and using topological sort to resolve the dependencies.

However, “Parallel Courses” also involves understanding the concept of level-order traversal or breadth-first search to calculate the minimum number of semesters. This introduces an additional layer of complexity compared to “Course Schedule”.

In summary, while they are similar and involve some of the same concepts, they are not the same problem.

Can anyone explain why am i getting runtime error in this problem? I have used bfs to solve this problem.

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class Solution {
public:
  bool canFinish(int n, vector<vector>& edges)
  {
    vector adj[n];
    for(int i = 0; i < n;i++)
    {
      adj[edges[i][0]].push_back(edges[i][1]);
    }
    vector vis;
    for(int i = 0; i < n;i++)
    {
      vis.push_back(0);
    }
    queue<pair<int, int>> q;
    q.push({0,-1});
    vis[0] = 1;
    map<int, int> par;
    par[0] = -1;
    while(!q.empty())
    {
      int node = q.front().first;
      int parent = q.front().second;
      q.pop();
      for(auto it: adj[node])
      {
        if(vis[it]==0)
        {
          q.push({it, node});
          par[it] = node;
        }
        else if(par[it]!=node)
        {
          return false;
        }
        else
        {
          continue;
        }
      }

    }
    return true;
  }
};

I see a few potential issues with your code:

1. Incorrect initialization of the adjacency list: The current initialization of your adjacency list is using a variable n which can lead to undefined behavior. Try using a dynamic data structure like a vector of vectors.

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vector<vector<int>> adj(n);

2. Incorrect usage of edges for adjacency list creation: Your code assumes that the size of edges is equal to n (the number of courses), which is not always the case. You should loop over the edges like this:

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for(auto& edge : edges) {
    adj[edge[0]].push_back(edge[1]);
}

3. Incorrect BFS algorithm: The BFS algorithm you implemented is not suitable for detecting cycles in a directed graph. In your code, you are treating par[node] as the only parent of node and checking if node is visited and par[node] != it (where it is the next node). However, in a directed graph, a node can be reached from multiple other nodes, so this check may give incorrect results.

For detecting cycles in directed graph, consider using depth-first search (DFS) with an additional ‘in-stack’ or ‘recursion stack’ concept. In this method, you maintain an extra boolean array to keep track of the nodes in the current recursion stack of DFS traversal. If a node is reached that is already in the recursion stack, then there is a cycle in the tree.

Here is the corrected code:

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class Solution {
public:
    bool canFinish(int n, vector<vector<int>>& edges) {
        vector<vector<int>> adj(n);
        for(auto& edge : edges) {
            adj[edge[0]].push_back(edge[1]);
        }
        
        vector<int> vis(n, 0);
        vector<int> dfs_vis(n, 0);
        
        for(int i=0; i<n; i++) {
            if(!vis[i]) {
                if(cycleCheck(i, adj, vis, dfs_vis)) {
                    return false;
                }
            }
        }
        
        return true;
    }

private:
    bool cycleCheck(int node, vector<vector<int>>& adj, vector<int>& vis, vector<int>& dfs_vis) {
        vis[node] = 1;
        dfs_vis[node] = 1;
        
        for(auto& it : adj[node]) {
            if(!vis[it]) {
                if(cycleCheck(it, adj, vis, dfs_vis)) {
                    return true;
                }
            } else if(dfs_vis[it]) {
                return true;
            }
        }
        
        dfs_vis[node] = 0;
        return false;
    }
};

This code uses a DFS-based method for cycle detection in the directed graph. If a cycle is detected, it immediately returns false, indicating that it’s not possible to finish all courses. If it traverses all the nodes without finding a cycle, it returns true, indicating that it is possible to finish all courses.

I don’t understand:

20 [[0,10],[3,18],[5,5],[6,11],[11,14],[13,1],[15,1],[17,4]]

what is the issue with this test case?

The provided list represents a dependency graph of courses. The pairs are [a,b] where a is a course and b is a prerequisite for that course. We want to determine if it’s possible to take all the courses (in this case, 20 courses in total) given these prerequisites.

The issue with this case is the dependency [5,5]. This means course 5 is a prerequisite for itself, creating a cycle. In other words, you can’t take course 5 until you’ve completed course 5, which is not possible. Therefore, it’s impossible to take all courses, and the answer for this test case should be false.

When this question and its second part both has same kind of dependency then why when we change the given vector &prerequisites into an adjacency list then here in this question we have to use

int n = numCourses;
vector adj[n];
for (auto it : prerequisites)
{
adj[it[1]].push_back(it[0]);
}
But in its first part if we use the same kind of change then it gives error so we have to use there
int n = numCourses;
vector adj[n];
for (auto it : prerequisites)
{
adj[it[0]].push_back(it[1]);
}

so if in both question the dependency is same then why 2 different kind of changing is suitable ?

if anybody know please clear my this doubt?

The difference in conversion from prerequisites to adjacency list depends on the interpretation of the prerequisites and the problem requirements.

In the current problem “Course Schedule”, a pair [a, b] in prerequisites means that b is the prerequisite of a i.e., you must take course b before course a. So, while creating the adjacency list, we are pushing a into adj[b] because a is dependent on b.

However, in a different context or problem, if the interpretation of prerequisites is reversed, and a pair [a, b] means that a is the prerequisite of b (i.e., you must take course a before b), then while creating the adjacency list, we would push b into adj[a], because b is dependent on a.

It’s essential to understand the specific problem requirements and the meaning of the given inputs to correctly construct the adjacency list.

I am getting a runtime error in this solutio. I have used a dfs topological sort to find the order in a vector and then checked if size of vector is equal to the size of the n or not . Please help .

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class Solution {
public:
//topo sort in a directed graph using dfs
//strivers

void dfs(int src, vector<vector<int>> &adj, vector<int> &visited, stack<int> topo){
    visited[src]=1;
    for(int i=0; i < adj[src].size();i++){
        if(!visited[adj[src][i]]){
            dfs(adj[src][i],adj,visited,topo);
        }
    }
    topo.push(src);
}
   r
bool canFinish(int n, vector<vector<int>>& prereq) {
    stack<int> topo;

   vector<vector<int>> adj(n);
   for(int i=0; i<n; i++){
       adj[prereq[i][1]].push_back(prereq[i][0]);
   } 
    vector<int> visited(n,0);
    for(int i=0; i<n; i++){
        if(visited[i]==0)dfs(i,adj,visited,topo);
    }
    
    return topo.size()==n;
}
}; 

The issue in your code seems to be due to two reasons:

1. You are passing the stack topo by value in the dfs function, it should be passed by reference so that the changes persist across the recursive calls.

2. Your way of constructing the adjacency list may lead to a runtime error. You are assuming n is equal to the length of prereq. However, n is the total number of courses (i.e., nodes) and prereq is just the prerequisites array (i.e., edges). They may not always be equal.

Here is your corrected code:

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class Solution {
public:
    void dfs(int src, vector<vector<int>> &adj, vector<int> &visited, stack<int> &topo){
        visited[src]=1;
        for(int i=0; i < adj[src].size(); i++){
            if(!visited[adj[src][i]]){
                dfs(adj[src][i], adj, visited, topo);
            }
        }
        topo.push(src);
    }

    bool canFinish(int n, vector<vector<int>>& prereq) {
        stack<int> topo;
        vector<vector<int>> adj(n);
        for(int i=0; i<prereq.size(); i++){    // changed n to prereq.size()
            adj[prereq[i][1]].push_back(prereq[i][0]);
        } 
        vector<int> visited(n,0);
        for(int i=0; i<n; i++){
            if(visited[i]==0) dfs(i, adj, visited, topo);
        }
    
        return topo.size() == n;
    }
};

One more point, your code only checks whether all nodes can be put into the topological order but it does not handle cycles in the graph. Please revise your solution to detect cycles in the graph as well, because even if all nodes are put into the stack, it doesn’t guarantee that there is no cycle. For cycle detection in DFS, you may need another vector to keep track of nodes in the recursion stack.

It turns out the numCourses parameter doesn’t do anything. I passed without using it.

The numCourses parameter does have an important role in the problem: it represents the total number of courses or nodes in the graph. You might have passed the test cases without using numCourses directly, but you likely used it indirectly.

For instance, you might have inferred the number of courses from the maximum course number in the prerequisites list. This would work in some cases, but could potentially fail if there are courses without any prerequisites (and hence they don’t appear in the prerequisites list), but they are still part of the total numCourses.

Another potential failure scenario is when there are no prerequisites at all (the list is empty), but numCourses is greater than zero. In this case, the correct answer is True (since there are no prerequisites, all courses can be finished), but an approach that infers the number of courses from the prerequisites list would incorrectly conclude that there are no courses to take, leading to incorrect results.

Therefore, the numCourses parameter is important for correctly understanding the total number of courses or nodes in the graph.

why this case returns false:

numCourses = 4 prerequisites = [[0,3],[1,0],[2,1],[0,2]] I thought course_3 could break the cycle of course_0, course_1, course_2.

In the prerequisites given, if you read them as “a course depends on another”, you can see there’s a cycle between course 0, 1, and 2.

Here’s the dependency chain:

• Course 0 depends on course 3 (from [0,3])
• Course 1 depends on course 0 (from [1,0])
• Course 2 depends on course 1 (from [2,1])
• Course 0 depends on course 2 (from [0,2])

If you try to take the courses, you’ll get stuck in a loop:

• To take course 0, you need to finish course 3 and course 2
• To take course 2, you need to finish course 1
• To take course 1, you need to finish course 0
• But to take course 0, you need to finish course 2 (which takes us back to the second step)

This loop is the cycle that makes it impossible to take all courses, even though course 3 does not have any prerequisites. The cycle of dependencies among courses 0, 1, and 2 cannot be resolved, making it impossible to complete all the courses, so the function should return False.

can someone please help me understand, why isn’t this code to detect cycle working?

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def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
    def build_graph():
        memo = {}
        for p, r in prerequisites:
            if p not in memo:
                memo[p] = [r]
            else:
                memo[p].append(r)
        return memo
    
    def detect_cycle(key, graph, path=[]):
        neighbors = graph.get(key, None)
        if not neighbors:
            return False
        for n in neighbors:
            if n in path:
                return True
            else:
                path.append(n)
                detect_cycle(n, graph, path)

    graph = build_graph()
    has_cycle = False
    for key in graph:
        res = detect_cycle(key, graph, [key])
        if res == True:
            has_cycle = True
    return not has_cycle

Your cycle detection function, detect_cycle(), isn’t returning anything if a node doesn’t have a cycle directly in its neighbours. If it doesn’t find a cycle, it simply exits and returns None, which is why your code isn’t working.

When you recursively call detect_cycle(n, graph, path), you should return its result back up the call stack. If any of the recursive calls returns True (indicating a cycle), you should immediately return True to stop the algorithm.

Here is a corrected version of your detect_cycle function:

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def detect_cycle(key, graph, path=[]):
    neighbors = graph.get(key, None)
    if not neighbors:
        return False
    for n in neighbors:
        if n in path:
            return True
        else:
            path.append(n)
            if detect_cycle(n, graph, path):  # capture the return value
                return True  # return True immediately if a cycle is detected
            path.pop()  # remove n from the path if no cycle is found via n
    return False  # return False when no cycle is detected via any neighbour

In the updated code, I’ve added if detect_cycle(n, graph, path): return True which will return True as soon as a cycle is detected in any of the recursive calls. If no cycle is found in a recursive call, path.pop() is used to backtrack, removing the node n from the current path. If the function goes through all the neighbours without finding a cycle, it returns False.

Remember to always ensure that your recursive functions have a base case and are correctly returning values.

if any child/neighbors of the same parent tries to access to its own parent then it must be a cycle

In a directed graph, such as the one used to model the course prerequisites in this problem, if a child node has a directed edge back to its parent (or any ancestor), it forms a cycle. This is because you can travel from the parent to the child and then back to the parent.

In the context of the problem, if a course (parent node) has a prerequisite (child node), and the prerequisite in turn has the original course as a prerequisite (a directed edge back to the parent), this forms a cycle. This represents a situation where you can’t complete any course because each course is dependent on the other.

It’s important to remember that just a child node being able to reach its parent does not constitute a cycle in a directed graph. For example, if we have course A requiring course B, and course B requiring course C, we have a path A -> B -> C, but not a cycle, because there is no way to go back from C to A or B.

A cycle only exists if there’s a way to start from a node, travel along the directed edges, and return to the same node. In the problem, this would imply an impossible set of course prerequisites.

Okay, I think there is an issue with this question. Consider numCourses = 4, prerequisites = [[1,0],[2,1],[3,2],[1,3]]. You can take course 0 -> 1 -> 2 -> 3. So program should return True. But expected result is false.

The prerequisites are given as pairs [a, b], which indicate that course a has b as a prerequisite. This means you need to complete course b before you can start course a. So, for your example, the pairs are interpreted as follows:

• [1,0]: You need to take course 0 before course 1.
• [2,1]: You need to take course 1 before course 2.
• [3,2]: You need to take course 2 before course 3.
• [1,3]: You need to take course 3 before course 1.

Now, if you put these in order, you can see that you have a cycle: 0 -> 1 -> 2 -> 3 -> 1. After taking course 0, you proceed to course 1, then to course 2, then to course 3. But now you are stuck, because to take course 1 again (which you need to do according to the last prerequisite pair), you have to retake course 3, which you just completed.

Since you have a cycle of prerequisites, you can’t complete all courses, so the program correctly returns False.

When I was using vectoradj to make adj list it was giving TLE, while in vectoradj[n], it get passed is it faster?? this is my code

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class Solution {
  public: bool dfs(int src, vector < int > & visited, vector < int > & currstack, vector < vector < int >> adjlist) {
    visited[src] = 1;
    currstack[src] = 1;
    for (auto it: adjlist[src]) {
      if (!visited[it]) {
        if (dfs(it, visited, currstack, adjlist)) return true;
      } else if (visited[it] && currstack[it]) return true;
    }
    currstack[src] = 0;
    return false;
  }
  bool canFinish(int numC, vector < vector < int >> & p) {
    vector < vector < int >> adjlist(numC);
    for (auto it: p) {
      adjlist[it[0]].push_back(it[1]);
    }
    vector < int > visited(numC, 0);
    vector < int > currstack(numC, 0);
    for (int i = 0; i < numC; i++) {
      if (!visited[i]) {
        if (dfs(i, visited, currstack, adjlist)) return false;
      }
    }
    return true;
  }
};

Your code is correct and it uses depth-first search to detect cycles in the graph of courses and prerequisites, which is the correct approach for this problem.

As for the performance difference between vector<vector<int>> adj(n) and vector<int> adj[n], the reason could be due to how memory allocation is done in these two cases.

In the case of vector<vector<int>> adj(n), a dynamic two-dimensional array is created. Whenever you push an element into a vector and it exceeds the current capacity, the vector’s size is increased (typically doubled), and the existing elements are copied to the new location. This could result in time overhead due to reallocation and copying.

In contrast, when you use vector<int> adj[n], you’re creating an array of vectors. The size of this array is fixed when it’s created, so you’re not paying the cost of resizing the outer vector.

In general, if you know the size of your data structure ahead of time and it’s not expected to change, it can be more efficient to use a statically sized array or data structure. However, the difference in most cases won’t be significant enough to cause a time limit exceeded error. The error might be due to other factors such as the complexity of the algorithm used.

class Solution { public boolean canFinish(int n, int[][] pre) { //find cycle HashMap<Integer, List> map = new HashMap<>(); for(int[] i: pre){ List a = map.getOrDefault(i[1], new ArrayList<>()); a.add(i[0]); map.put(i[1], a); } HashSet vis = new HashSet<>(); for(int i:map.keySet()){ if(!vis.contains(i)) if(dfs(i, vis, map, new boolean[n])) return false; } return true; } public boolean dfs(int curr, HashSet vis, HashMap<Integer, List> map, boolean[] v){ if(v[curr]){return true;} vis.add(curr); v[curr] = true; for(int i:map.getOrDefault(curr, new ArrayList<>())) if(dfs(i, vis, map, v)) return true; v[curr] = false; return false; } }

Why does this DFS gives TLE? I believe it runs in O(N)

The given solution runs in O(N+E), where N is the number of courses and E is the number of prerequisites, since every course and prerequisite is visited once. However, the problem might be the specific way you’re implementing DFS and managing the visited nodes.

In your solution, in the dfs function, the boolean array v is used to keep track of nodes currently in the recursion stack, and vis is used to keep track of all visited nodes. You start by marking curr as visited and in the recursion stack. Then, for each neighbor of curr, you recursively call dfs.

The issue may arise with the fact that you’re checking for the presence of a cycle by looking for a node that’s currently in the recursion stack (v[curr]), rather than a node that’s been visited before. This could potentially cause the recursion to go deeper than necessary and exceed the time limit in the worst case scenario.

Moreover, the way you are handling the case of a node not being present in map with map.getOrDefault(curr, new ArrayList<>()) might also contribute to the time limit being exceeded, because it creates a new array list every time you’re trying to access the adjacency list of a node that doesn’t exist in map.

I’d suggest keeping track of all visited nodes and all nodes in the recursion stack separately, and returning false as soon as you encounter a node that’s already in the recursion stack. Here’s an example of how you might modify your dfs function:

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public boolean dfs(int curr, HashSet<Integer> vis, HashMap<Integer, List<Integer>> map, boolean[] v){
    vis.add(curr);
    v[curr] = true;

    List<Integer> neighbours = map.get(curr);
    if (neighbours != null) {
        for(int i: neighbours) {
            if (!vis.contains(i) && dfs(i, vis, map, v)) {
                return true;
            } else if (v[i]) {
                return true;
            }
        }
    }

    v[curr] = false;
    return false;
}

This solution should provide the expected result within the time limit, as long as the prerequisites do not form a large cyclic graph. If they do, the recursive approach might still exceed the stack limit, and you might need to switch to an iterative approach with an explicit stack.

can anyone tell me where I have made a mistake

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class Solution {

  public:

    bool cycle(vector < vector < int >> & vec, vector < bool > & vis, vector < bool > & dfsvis, int x) {
      if (dfsvis[x]) {
        return true;
      }
      if (vis[x]) {
        return false;
      }
      vis[x] = true;
      dfsvis[x] = true;
      for (auto y: vec[x]) {
        if (cycle(vec, vis, dfsvis, y)) {
          return true;
        }
      }
      dfsvis[x] = false;
      return false;
    }
  bool canFinish(int total, vector < vector < int >> & courses) {
    vector < vector < int >> vec(total, vector < int > ());
    for (int i = 0; i < courses.size(); i++) {
      vector < int > x = courses[i];
      vec[x[0]].push_back(x[1]);
    }
    vector < bool > vis(total, false);
    vector < bool > dfsvis(total, false);
    for (int i = 0; i < total; i++) {
      if (cycle(courses, vis, dfsvis, i)) {
        return false;
      }
    }
    return true;
  }
};

There’s a small mistake in your code. In your for loop where you are checking for a cycle for each course, you are using the courses vector instead of the vec adjacency list that you created.

So, in your for loop, instead of calling cycle(courses, vis, dfsvis, i), you should be calling cycle(vec, vis, dfsvis, i).

Here’s the corrected loop:

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for (int i = 0; i < total; i++) {
  if (cycle(vec, vis, dfsvis, i)) {
    return false;
  }
}

The rest of your code seems fine. The logic is correct: you’re checking for cycles by performing a depth-first search and keeping track of visited nodes as well as nodes currently in the recursion stack (using the dfsvis vector). If at any point you encounter a node that’s currently in the recursion stack, you return true to indicate a cycle. If there’s no cycle in the course prerequisites, then you can finish all courses, so you return true from the canFinish function.

What is wrong with this solution? Passes 47/52 TCs. Any help is appreciated!

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class Solution {
public:
bool canFinish(int numCourses, vector<vector>& prerequisites) {
vector<vector> courses(numCourses);

    for(auto p : prerequisites)
    {
        int cur = p[0];
        int pre = p[1];
        courses[cur].push_back(pre);
        for(int x : courses[pre])
        {   
            if(x == cur) return false;
            for(int xx : courses[x])
            {
                if(xx == cur)
                    return false;
            }
            courses[cur].push_back(x);    
        }
    }

    return true;
}
};

Your current approach doesn’t handle the detection of cycles that span more than just the immediate neighbours of a node. It only checks for direct loops and one level deep loops, and that’s not sufficient for correctly solving this problem.

The most robust way to handle this is with a depth-first search (DFS) or breadth-first search (BFS) approach, where you actually traverse the graph and keep track of visited nodes to detect any cycles.

Here’s a corrected version of the code using depth-first search:

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class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> graph(numCourses, vector<int>());
        vector<int> visit(numCourses);
        for (auto a : prerequisites) {
            graph[a[0]].push_back(a[1]);
        }
        for (int i = 0; i < numCourses; ++i) {
            if (!canFinishDFS(graph, visit, i)) return false;
        }
        return true;
    }
    
private:
    bool canFinishDFS(vector<vector<int>>& graph, vector<int>& visit, int i) {
        if (visit[i] == -1) return false;
        if (visit[i] == 1) return true;
        visit[i] = -1;
        for (auto a : graph[i]) {
            if (!canFinishDFS(graph, visit, a)) return false;
        }
        visit[i] = 1;
        return true;
    }
};

This code works by setting visit[i] to -1 when a node is first encountered (indicating it’s in the recursion stack), and to 1 when we’ve finished processing it (indicating we’ve checked all its descendants and found no cycles). If we encounter a node that’s still in the recursion stack (i.e., visit[i] == -1), then we’ve found a cycle and return false.

for input: –prerequisites = [[1,0]], –numCourses =3

the Expected output is True. I understand that there are a total of 2 courses - 0,1 which can both be taken. But are still less than the required number of courses-3 so the answer should be False.

How is it True?

If numCourses =3 we have 3 courses 0,1,2 and there isn’t any ambiguity so this should return True. Total available courses doesn’t depend on prerequisites.

[[0,1],[0,2],[1,2]] why should this return true?

In the prerequisites array [[0,1],[0,2],[1,2]], each sub-array [a, b] represents a dependency of course a on course b, i.e., you need to take course b before you can take course a.

If we illustrate this visually:

2 -> 0
|  
v
1

Course 0 has prerequisites 1 and 2. Course 1 also has a prerequisite, which is course 2. This creates a chain of dependencies, but there are no cycles, which means it’s possible to take all the courses.

The correct order of taking the courses would be: 2 -> 1 -> 0. This is why the function should return true for this case: it is possible to take all courses without violating any prerequisites.

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class Solution {
public:

void addEdge(vector<vector<int>> &graph, int u, int v){
    graph.at(u).push_back(v);
}

void dfsUtil(vector<vector<int>> graph, vector<bool> &visited, 
vector<bool> &inRec, int u, bool &isCycle){

    visited.at(u) = true;
    inRec.at(u) = true;
    for(auto v : graph.at(u)){
        if(!visited.at(v)){
            dfsUtil(graph, visited, inRec, v, isCycle);
        }
        else{
        if(inRec.at(v)){
            isCycle = true;
        }
        }
    }

    inRec.at(u) = false;

}
bool detectCycle(vector<vector<int>> graph, int n){

    vector<bool> visited(n, false);
    vector<bool> inRec(n, false);

    bool isCycle = false;
    for(int i=0;i<n;i++){
        if(!visited.at(i)){
            dfsUtil(graph, visited, inRec, i, isCycle);
        }
    }

    return isCycle==false;
}
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {

    vector<vector<int>> graph(numCourses);
    for(auto pre : prerequisites){
        addEdge(graph, pre.at(0), pre.at(1));
    }

    return detectCycle(graph, numCourses);
    
}
};

May I know why is this giving TLE. Only 47/52 test cases passes

The Time Limit Exceeded (TLE) error indicates that your algorithm isn’t as efficient as it could be. Here’s why your code might be causing a TLE error:

In your function dfsUtil(), you’re exploring all the vertices in the graph, including those that have been visited. If your graph has a lot of vertices and a high degree of connectivity, the algorithm has to do a lot of unnecessary work, which can lead to a time-out.

To solve the problem more efficiently, you could stop the DFS traversal as soon as you detect a cycle. Instead of passing isCycle as a reference and updating it inside dfsUtil(), you could return a boolean from dfsUtil() which would indicate if a cycle has been detected.

Here’s a quick adjustment to your code that might fix the issue:

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bool dfsUtil(vector<vector<int>> graph, vector<bool> &visited, vector<bool> &inRec, int u){
    visited.at(u) = true;
    inRec.at(u) = true;
    for(auto v : graph.at(u)){
        if(!visited.at(v) && dfsUtil(graph, visited, inRec, v)){
            return true;
        }
        else if(inRec.at(v)){
            return true;
        }
    }
    inRec.at(u) = false;
    return false;
}

bool detectCycle(vector<vector<int>> graph, int n){
    vector<bool> visited(n, false);
    vector<bool> inRec(n, false);
    for(int i=0;i < n ;i++){
        if(!visited.at(i) && dfsUtil(graph, visited, inRec, i)){
            return true;
        }
    }
    return false;
}

This modification will make the depth-first search terminate as soon as a cycle is detected, potentially reducing the run-time significantly for certain graphs.

However, you could still optimize your solution. The graph is being passed by value to the dfsUtil() function, which creates a copy of it each time. This could be quite expensive, as it requires O(V+E) space and time, where V is the number of vertices and E is the number of edges in the graph. Instead, consider passing the graph by reference.

Can a course depended on more than one course? If so, solution3 cannot work?

Yes, a course can depend on more than one course. For example, a course C might have prerequisites A and B, meaning that both A and B must be completed before you can take course C. This is a common situation in real-world education systems.

The solution using depth-first search (DFS) to detect cycles in a graph can handle this situation perfectly. If you have a cycle in your prerequisites, regardless of whether a course depends on one or multiple courses, it means there is a circular dependency that makes it impossible to complete all courses.

If a solution does not correctly handle a course that depends on multiple prerequisites, then that solution is likely flawed. DFS and similar graph traversal techniques are fully capable of handling nodes with multiple edges (courses with multiple prerequisites), so if Solution 3 you referred to cannot handle such cases, then it is not a correct or complete solution.

However, as of my last training data up to September 2021, I am unable to refer to or validate specific solutions numbered as “Solution 3”.

I think it is a directed graph problem thats why we need topological sort to solve it.

Yes, you are correct. This problem can be viewed as a directed graph problem where each course is a node and the edges are prerequisites, meaning that one course (node) depends on another. The graph can have cycles, which would imply that there are certain courses that have circular dependencies and thus can’t be taken.

Topological sorting is a perfect method to solve this problem because it orders the nodes (courses) in such a way that all directed edges (prerequisites) go from a node to another that comes later in the order. In other words, it provides an order in which to take the courses.

If a topological sort is possible, it means there are no cycles in the graph and all courses can be finished (the function should return true). If a topological sort is not possible due to a cycle in the graph, not all courses can be finished (the function should return false).

This interpretation and solution fit the problem statement well, as the problem is essentially asking if it’s possible to complete all courses given the prerequisite constraints, which is what a topological sort checks for in a directed graph.

Can anybody explain me this tc? numCourses = 5 prerequisites = [[1,4],[2,4],[3,1],[3,2]] My Code failed for this Expected is true and I am returning false as there is no cycle …

The test case you’ve mentioned:

numCourses = 5
prerequisites = [[1,4],[2,4],[3,1],[3,2]]

Can be interpreted as follows:

• Course 4 is a prerequisite for Course 1
• Course 4 is also a prerequisite for Course 2
• Course 1 is a prerequisite for Course 3
• Course 2 is also a prerequisite for Course 3

In this scenario, there is no cycle present. So, all courses can be taken, meaning the function should indeed return true. Here is a potential order in which the courses can be taken: 4, 1, 2, 3, 0.

If your code is returning false for this input, then it’s likely that there’s an issue with how your code is checking for cycles or determining if a course can be taken. I would recommend revisiting your code’s logic to ensure it’s correctly identifying prerequisites and checking for cycles.

Remember, you’re trying to find if it’s possible to finish all courses, which is true if and only if there is no cycle in the course dependency graph.

Passing every test case except for numCourses = 500 prerequisites = the massive list I probably shouldn’t paste Anyone have any ideas? Thank you!

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def canFinish(self, numCourses, prerequisites):
    """
    :type numCourses: int
    :type prerequisites: List[List[int]]
    :rtype: bool
    """
    def bfs(graph, q, discovered):
        if(not q):
            if(False in discovered):
                return False
            return True
        node = q.popleft()
        for neighbor in graph[node]:
            if(not discovered[neighbor]):
                discovered[neighbor] = True
                q.append(neighbor)
        return bfs(graph, q, discovered)
    if(not prerequisites):
        return True
    graph = [set() for i in range(numCourses)]
    roots = [True] * numCourses
    for p in prerequisites:
        if(p[0] in graph[p[1]]):
            return False
        graph[p[0]].add(p[1])
        roots[p[1]] = False
    for i in range(len(roots)):
        if(roots[i]):
            q = collections.deque()
            q.append(i)
            if(bfs(graph, q, roots)):
                return True
    return False

Your current code using BFS (breadth-first search) is a good starting point. However, there is a flaw in how you are tracking discovered nodes and how you are handling the BFS.

In the provided code, roots is used as a flag array to check if a node is discovered or not. The problem is, you are using the same roots array across different BFS calls. In a BFS for cycle detection, we must reset the visited flag before each BFS traversal because a node being visited in a previous BFS doesn’t guarantee that it won’t be involved in a cycle in the next BFS traversal.

So, you should create a separate visited array for each BFS call.

Moreover, the current BFS doesn’t correctly check for a cycle. The BFS function should not return True simply when the queue is empty, because that just means all nodes reachable from the starting node have been visited, not that there are no cycles in the entire graph. To detect a cycle using BFS in a directed graph, you can check whether a node has been visited before it is added to the queue. If a node has been visited before, then there is a cycle.

A more common approach to cycle detection is to use DFS (depth-first search), which is more straightforward. DFS visits the nodes in a path from the starting node to a leaf node, and during this process, if it visits a node that has been visited before, then there is a cycle.

Try revising your code using these suggestions. You’ll need to modify how you’re setting up the graph and how you’re running the BFS to correctly identify whether it’s possible to complete all courses (i.e., if there is a cycle in the course dependency graph).

can someone explain me why it’s not working pls

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class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<int> indegree;
        for(int i = 0; i < numCourses;i++){
            indegree[i]=0;
        }

        //build graph
        unordered_map<int,vector<int>> graph;
        for(int i = 0; i < prerequisites.size();i++){
            graph[prerequisites[i][1]].push_back(prerequisites[i][0]);
            
            indegree[prerequisites[i][0]]++;
        }
        
        queue<int> queue;
        for(int i =0;i<indegree.size();i++){
            if(indegree[i]==0){
                queue.push(i);
            }
        }
        vector<int> ans;
        while(!queue.empty()){
            int tmp = queue.front();
            queue.pop();
            ans.push_back(tmp);
            for(int x:graph[tmp]){
                indegree[x]--;
                if(indegree[x]==0){
                    queue.push(x);
                }
            }
            
        }
        
        return ans.size() == numCourses;
    }
};

The problem with your code is that the indegree vector is not correctly initialized. You’re currently creating an empty vector and then trying to access and update its elements, which are undefined. This will lead to undefined behavior in C++.

You should create the indegree vector with numCourses elements all initially set to zero like this:

1

vector<int> indegree(numCourses, 0);

So your corrected code will be:

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class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<int> indegree(numCourses, 0);

        //build graph
        unordered_map<int,vector<int>> graph;
        for(int i = 0;i<prerequisites.size();i++){
            graph[prerequisites[i][1]].push_back(prerequisites[i][0]);
            
            indegree[prerequisites[i][0]]++;
        }
        
        queue<int> queue;
        for(int i =0;i<indegree.size();i++){
            if(indegree[i]==0){
                queue.push(i);
            }
        }
        vector<int> ans;
        while(!queue.empty()){
            int tmp = queue.front();
            queue.pop();
            ans.push_back(tmp);
            for(int x:graph[tmp]){
                indegree[x]--;
                if(indegree[x]==0){
                    queue.push(x);
                }
            }
            
        }
        
        return ans.size() == numCourses;
    }
};

This should fix your current issue. This code follows the Kahn’s algorithm for topological sorting to detect a cycle in a directed graph. It is based on the idea that a course with no prerequisites (in-degree of zero) can be taken right away, and once a course is taken, it can be removed from the graph along with its edges. If the graph contains a cycle, then there will be at least one course that still has prerequisites left to be taken, and thus the function returns false.

Help my optimize my solution! I keep getting “time limit exceeded” on my textbook DFS implementation.

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class Solution(object):
    def canFinish(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        """
        if numCourses == 1:
            return True
        adj =  {}
        for e in prerequisites:
            u, v = e
            if u not in adj:
                adj[u] = [v]
            else:
                adj[u].append(v)

        V = adj.keys() ## nodes with no outgoing edges don't bother us
        white = 0
        grey = 1
        black = 2
        color = [white for v in range(numCourses)] 
        
        def dfs_visit(v):
            if color[v] == grey: ## boom found a back edge
                return False

            color[v] = grey
            if v in adj: ## that is, v has any outgoing edges
                for u in adj[v]:
                    if not dfs_visit(u):
                        return False
            color[v] = black
            return True

        for v in V:
            if color[v] != black:
                if not dfs_visit(v):
                    return False
        
        return True

Your solution seems to be fine for detecting cycles in the graph. The Time Limit Exceeded (TLE) error occurs when your solution is correct, but it is not efficient enough. It might be because of the data structure you are using or the way you have implemented DFS.

One way to improve the performance of your solution is by avoiding unnecessary DFS traversal. This can be done by storing and reusing the result of DFS traversal for each node. The idea here is, if a node is marked as safe (no cycle from this node), then every time we encounter this node again, we can directly return True without doing DFS from this node.

Here is an optimized version of your code implementing the above idea:

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class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        # Create the adjacency list representation of the graph
        adj = {i: [] for i in range(numCourses)}
        for i, j in prerequisites:
            adj[i].append(j)

        # 0 = unvisited, 1 = visiting, 2 = visited
        visit = [0] * numCourses

        for i in range(numCourses):
            if visit[i] == 0 and self.has_cycle(i, adj, visit): 
                return False
        return True

    def has_cycle(self, i, adj, visit):
        if visit[i] == 1:
            return True
        if visit[i] == 2:
            return False

        visit[i] = 1
        for j in adj[i]:
            if self.has_cycle(j, adj, visit):
                return True

        visit[i] = 2
        return False

Here, we initialize all vertices as unvisited. We do a DFS from every node. In DFS, before visiting a node, we mark it as visiting and after visiting all its adjacent and their subtrees, we mark it as visited. If at any time, we visited a node that is in the visiting state, then we have found a cycle. Also, if we have found a node marked as visited, we return False as it has already been determined that this node is safe.

Time limit is way too short on this one IMO. Textbook DFS implementation with grey and black sets fails to pass. -

Indeed, the time limit for coding problems is often set to be challenging to ensure efficient solutions. However, a textbook DFS (Depth-First Search) algorithm could indeed lead to time limit exceeded errors in this case if it is not optimized properly, especially for large inputs.

A standard DFS algorithm involves traversing through every edge in a graph, which is a time-consuming process with a time complexity of O(V + E), where V represents the number of vertices and E represents the number of edges.

However, the task at hand is to determine whether it’s possible to finish all courses given certain prerequisites, which forms a directed graph. This is effectively asking whether the graph is a Directed Acyclic Graph (DAG). If a cycle exists, it means there are circular dependencies among the courses, which would make it impossible to finish all courses.

An optimized approach for this specific problem is to use a variant of DFS known as Topological Sorting. It works by visiting each node once and only considering revisiting a node if it doesn’t lead to a cycle. This optimizes the standard DFS approach by reducing unnecessary traversals, and it perfectly fits the problem of detecting cycles in the graph. The modified DFS algorithm used in topological sorting will also return the order of courses to be taken, which may be useful for more complex problems.

In this scenario, DFS is modified to maintain two states for each node - ‘visiting’ and ‘visited’. If we ever encounter a node that is in the ‘visiting’ state, it means we’ve come across a cycle. After a node has been fully explored, it’s marked as ‘visited’. This method will significantly cut down on the number of unnecessary explorations and make the algorithm more efficient.

Remember, understanding the problem deeply and tailoring your algorithm to suit the problem is the key to solve complex problems within the given time limits.

The test case 22 says numCourses = 5 prerequisites = [[1,4],[2,4],[3,1],[3,2]] and it expects true, I don’t understand how can it can be true as there are only 4 courses (from 1 to 4 and) it expects to take 5 courses, can someone please explain?

The numCourses parameter doesn’t directly correspond to the highest number used in the prerequisites list, rather it’s the total number of courses. In this problem, courses are labeled from 0 to n - 1, so in your example with numCourses = 5, the courses are labeled 0, 1, 2, 3, 4.

The list of prerequisites [[1,4],[2,4],[3,1],[3,2]] can be interpreted as follows:

• Course 1 has prerequisite course 4
• Course 2 has prerequisite course 4
• Course 3 has prerequisites courses 1 and 2

Course 0 has no prerequisites, but it still counts towards the total number of courses. Since none of the courses have cyclic dependencies, the expected output is True: all courses can be finished.

Here’s one possible order to take the courses: 0, 4, 1, 2, 3. This satisfies all prerequisites.

Not all types of cases are covered in the closed test. For example one of my successful submissions failed on the testcase: numCourses = 5 prerequisites = [[1, 4], [1, 3], [1, 2], [1, 0], [2, 1], [3, 1],[4, 1]]

The mistake of my submission consisted in that I marked course as canFinish if only one of its required course was marked as canFinish.

You’re correct. The example you provided,

numCourses = 5 prerequisites = [[1, 4], [1, 3], [1, 2], [1, 0], [2, 1], [3, 1], [4, 1]]

shows a cyclical dependency among the courses, which makes it impossible to finish all courses. The test cases provided by coding platforms usually cover a broad range of situations, but there could still be edge cases that are not explicitly tested.

In your case, it seems you might have misunderstood the prerequisites. The prerequisite pairs are not interchangeable; they are directed. Each pair [a, b] means that to study course a, you must have completed course b first. So, it’s not enough to finish just one prerequisite, you must finish all of them, and in an order that doesn’t create a cycle.

You should revise your code to take into account the directed nature of the prerequisites and ensure that all prerequisites for a course can be completed without encountering a cycle.

For the problem of detecting a cycle in a graph, you can use depth-first search (DFS). Start a DFS at each node, visiting all its neighbours. If you ever encounter a node that has already been visited in the current DFS (not overall), then there’s a cycle.

HELP NEEDED T_T why won’t this python code work

class Solution: def init(self): self.ok = True self.numCourses = None self.graph = [] self.indegree = []

def buildGraph(self, numCourses, prerequisites): self.graph = [[] for _ in range(numCourses)] self.indegree = [0 for _ in range(numCourses)] for course, prerequisites in prerequisites: self.graph[prerequisites].append(course) self.indegree[course] += 1

def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: self.numCourses = numCourses self.visited = [False for _ in range(self.numCourses)] self.buildGraph(numCourses, prerequisites)

   # no prerequisites 
   if not self.graph:
       return True

   start_courses = [i for i in range(len(self.indegree)) if self.indegree[i]==0]
   # each course is dependent of each other
   if not start_courses:
       return False

   def traverse(node):
       print(self.path)
       if not self.ok:
           return       
       if self.visited[node] or self.path[node]:
           self.ok = False
           return
       
       self.visited[node] = True
       self.path[node] = True
       for neighbour in self.graph[node]:
           traverse(neighbour)
       self.path[node] = False

   for course in start_courses:
       self.path = [False for _ in range(self.numCourses)]
       traverse(course)
   
   return self.ok

This code passed 31/52 test cases.

I tried print self.path in the beginning of the traverse function, and it seems that there is at most one True in self.path, why is that

Any help is much appreciated!!!

There are a few potential issues in your implementation:

1. When you check if not self.graph:, you’re checking if the self.graph is empty, not if it’s devoid of prerequisites. Even if there are no prerequisites, your self.graph still gets initialized as a list of empty lists because of the line self.graph = [[] for _ in range(numCourses)]. This condition will always evaluate to False. You might want to check if prerequisites is empty at the very start of your canFinish function instead.

2. You’re doing depth-first search but you’re not correctly tracking the nodes currently in the DFS stack (self.path). You set self.path[node] = True before calling traverse(neighbour) but set self.path[node] = False immediately after. This could result in an incorrect state of self.path if the traverse(neighbour) call returns early due to if not self.ok: return.

To elaborate, consider a scenario where you’re in the middle of a DFS and you find a cycle. You set self.ok = False and start to return, but before you return all the way, you set self.path[node] = False. Now, if there is another start node that is part of the cycle you just found, your traverse function won’t be able to detect that it’s part of a cycle because its entry in self.path is now False.

To resolve this issue, you should only set self.path[node] = False after you’ve completed the DFS from a node and not found a cycle, i.e., put it after the for neighbour in self.graph[node]: loop.

Here’s a corrected version of your code:

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class Solution:
    def __init__(self):
        self.ok = True

    def buildGraph(self, numCourses, prerequisites):
        self.graph = [[] for _ in range(numCourses)]
        self.indegree = [0 for _ in range(numCourses)]
        for course, prerequisite in prerequisites:
            self.graph[prerequisite].append(course)
            self.indegree[course] += 1

    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        if not prerequisites:
            return True
        self.visited = [False for _ in range(numCourses)]
        self.buildGraph(numCourses, prerequisites)

        start_courses = [i for i in range(numCourses) if self.indegree[i] == 0]
        if not start_courses:
            return False

        def traverse(node):
            if not self.ok:
                return
            if self.visited[node] or self.path[node]:
                self.ok = False
                return

            self.visited[node] = True
            self.path[node] = True
            for neighbour in self.graph[node]:
                traverse(neighbour)
            if self.ok:
                self.path[node] = False

        for course in start_courses:
            self.path = [False for _ in range(numCourses)]
            traverse(course)

        return self.ok and all(self.visited)

The all(self.visited) check at the end is to make sure all nodes were visited, i.e., there were no disconnected subgraphs. If your graph is guaranteed to be connected, you can leave this check out.

Wrong Answer 36 / 52 testcases passed Editorial Input numCourses = 2 prerequisites = [[0,1]]

Use Testcase Output false Expected true

Can anyone explain this testcase ?

The problem is a classic problem of detecting a cycle in a directed graph. In this context, a graph is a set of vertices (courses) connected by edges (prerequisites). If the graph contains a cycle, it means that there are circular dependencies among your courses, making it impossible to complete all of them.

In your given test case, you have numCourses = 2 and prerequisites = [[0,1]]. This means there are 2 courses: course 0 and course 1. The prerequisites array [[0,1]] suggests that course 0 depends on course 1, i.e., you need to complete course 1 before you can take course 0.

The task is to determine if it is possible to finish all courses given the dependencies (prerequisites). In this case, it is indeed possible. You can simply take course 1 first, and then take course 0. There’s no cycle in this graph, hence the answer is true.

The confusion may stem from how prerequisites are read. Remember, if the pair is [a,b], it means that you must take course b before you can take course a. It does not mean that you must take course a before course b.

So, for your given example, to satisfy the prerequisites, we can take the courses in the following order:

1. Course 1
2. Course 0

Since it’s possible to take all courses by respecting the prerequisites, the function should return true, which is the expected outcome.

It’s crucial to be careful with the direction of the dependencies to correctly interpret the prerequisites. Incorrectly interpreting them as [a,b] meaning “a before b” would indeed lead to an output of false for this case, but that’s not the correct interpretation according to the problem statement.

Common Mistakes

Here are the common mistakes that can be made while solving this problem:

1. Not correctly handling directed edges: Since this problem deals with a directed graph (as prerequisites are directional), not correctly handling the direction of edges can lead to wrong results.

2. Not using depth-first search (DFS) correctly: DFS is typically used for cycle detection in a directed graph. However, it’s crucial to keep track of the nodes currently being visited. A common mistake is to use a single “visited” status, which can’t differentiate between a node that has been visited in the current traversal and a node that was visited in a previous traversal.

3. Not handling isolated nodes: In some cases, there might be courses with no prerequisites and which are not prerequisites for any other courses. These nodes should not be forgotten.

4. Incorrect base cases: Particularly, for the case when there’s only one course or when there are no prerequisites, the logic can sometimes be incorrectly handled.

5. Misunderstanding the problem statement: A common misconception is that each course can depend on at most one other course, but the problem statement does not state this. In fact, a course can have multiple prerequisites.

6. Memory and time inefficiencies: Unoptimized solutions may exceed time or memory limits for large inputs. Techniques like memoization and topological sorting can be used to optimize the solution.

7. Not handling courses without prerequisites: Courses without prerequisites should be the starting point for the graph traversal. Not taking this into account can lead to false cycle detections.

8. Incorrect graph construction: Mistakes in creating adjacency lists or matrices could lead to incorrect relationships being formed, and therefore incorrect results.

9. Not dealing with edge cases: Edge cases such as an empty list of prerequisites, or a single course without prerequisites, may not be correctly handled in some implementations.

During our discussion, we encountered several specific mistakes in the provided code snippets. Here are some of them:

1. Not initializing containers correctly: In some cases, data structures such as arrays, lists, or maps were not correctly initialized, leading to runtime errors or logical errors. For example, in one of the C++ solutions, the indegree vector was not initialized with the correct size, leading to an out-of-bounds error.

2. Incorrect traversal in DFS: Some DFS implementations had errors such as not marking nodes as visited at the right time, or not handling the recursion stack correctly, which could lead to incorrect detection of cycles.

3. Incorrect use of variables or data structures: There were examples where the wrong variables or data structures were used in certain parts of the code, leading to logical errors. For example, in a Python solution, the BFS function was called with courses instead of graph.

4. Recursion limit exceeded in Python: Python has a default recursion limit, which can be reached in a recursive DFS solution if the depth of the graph is too large.

5. Performance issues: Some of the provided solutions were inefficient, leading to time limit exceeded errors on large inputs. The solutions could be optimized by using techniques like memoization or topological sorting, or by avoiding unnecessary work.

6. Not correctly interpreting the problem: Some solutions incorrectly assumed that a cycle would exist if a course had any prerequisites, which is not the case.

7. Not checking for all possible starting points in the graph: Some solutions only started the DFS from nodes with zero in-degree, but cycles can also exist in subgraphs that don’t contain such nodes.

Preventing Mistakes

Preventing these mistakes mostly involves following best practices for programming and problem-solving, and having a solid understanding of the algorithms and data structures involved. Here are some suggestions:

1. Thoroughly understand the problem: Make sure you understand all aspects of the problem before starting to code. Misunderstanding or overlooking part of the problem can lead to incorrect solutions.

2. Use the right data structures: Each problem has a set of appropriate data structures that make it easier to implement a solution. Understanding these structures and knowing when to use them can help you avoid many common mistakes.

3. Debug effectively: If your solution isn’t working, don’t just guess what might be wrong. Use systematic debugging techniques to find and fix the problem. This could involve using a debugger, adding print statements to your code, or manually working through the logic of your solution.

4. Write test cases: Writing your own test cases can help you catch errors that you might not have thought of. Try to cover all possible edge cases.

5. Understand time and space complexity: Understanding the time and space complexity of your algorithm can help you identify potential performance issues. If your algorithm is too slow or uses too much memory, consider whether there’s a more efficient solution.

6. Study algorithms and data structures: Many programming problems involve applying well-known algorithms or data structures. Studying these can help you recognize when they’re applicable and implement them correctly.

7. Initialize your variables/data structures properly: Make sure all variables and data structures are initialized correctly before using them. This includes setting initial sizes and values.

8. Write clean and simple code: Complex code is more prone to errors. Try to keep your code as simple and clean as possible. This often involves breaking down complex functions into smaller, more manageable parts.

9. Understand the language’s limitations: Different programming languages have different limitations. For example, Python has a recursion limit. Knowing these can help you avoid certain types of errors.

10. Practice: Lastly, the more you practice solving problems, the better you’ll get at avoiding mistakes. This can also help you become more familiar with common problem types and solution patterns.

Here are specific advice for this type of problem:

1. Understanding the problem requirements and constraints: This problem is about detecting a cycle in a directed graph, which is a common problem in computer science. To prevent misunderstanding, it’s important to recognize that a course could have multiple prerequisites, and there could be chains of prerequisites. Also, note that the graph may not be connected, meaning you need to check for cycles starting from every node, not just a single node.

2. Selecting the appropriate algorithm: This problem can be solved using a depth-first search (DFS) or breadth-first search (BFS) algorithm. Using these algorithms incorrectly (e.g., not correctly maintaining the visited and recursion stacks in DFS) can lead to errors. Make sure you thoroughly understand these algorithms before implementing them.

3. Implementing the algorithm correctly: Be careful when implementing the algorithm. A common mistake is to not properly update the visited and recursion stacks in the DFS algorithm. Also, make sure you handle the edge case where there are no prerequisites (i.e., the graph has no edges), in which case the function should return true because it’s possible to finish all courses.

4. Handling edge cases: Ensure that your solution can handle edge cases, such as when there’s only one course or when there are no prerequisites.

5. Optimizing the code: The BFS solution requires computing the in-degrees of the nodes and could potentially be optimized by storing the graph’s edges in an adjacency list format, where the index represents the course and the value at that index is a list of courses that depend on the course.

6. Debugging: If your solution isn’t working, don’t just guess what might be wrong. Instead, systematically go through your code and the algorithm to find where the problem might be. Adding print statements or using a debugger can help with this.

7. Testing: Test your code with multiple test cases to ensure it works for all scenarios. Include cases where there are cycles and where there are not, and consider edge cases like a single course or no prerequisites.

Remember, practice is key. The more you work with problems like these, the better you’ll become at understanding and implementing the algorithms, recognizing potential pitfalls, and debugging your code.

Similar Problems

Here are ten LeetCode problems that require similar problem-solving strategies or use similar underlying concepts as the problem we’ve just solved. The primary concept here is graph traversal using Depth-First Search (DFS) and detecting cycles in a graph, which are common techniques in many other problems.

1. 207. Course Schedule: This is the original problem that involves constructing a directed graph and detecting a cycle using DFS.

2. 210. Course Schedule II: This problem is a direct extension of the original problem. Besides detecting a cycle, it also asks for the order of courses to take.

3. 802. Find Eventual Safe States: In this problem, you have to identify nodes in a graph that eventually lead to a terminal node (no outgoing edges). This involves DFS traversal and cycle detection.

4. 200. Number of Islands: This problem requires DFS to traverse an island and mark it as visited, similar to traversing a graph.

5. 785. Is Graph Bipartite?: This problem asks to determine if a graph is bipartite, which can be solved by DFS and color labeling.

6. 130. Surrounded Regions: This problem requires DFS traversal to identify all ‘O’s that are not surrounded by ‘X’.

7. 417. Pacific Atlantic Water Flow: This problem involves traversing a matrix (seen as a graph) using DFS from multiple starting points.

8. 332. Reconstruct Itinerary: This problem requires constructing a graph and then finding a path in the graph using DFS.

9. 684. Redundant Connection: This problem involves detecting a cycle in an undirected graph, which is an application of the DFS cycle detection technique.

10. 261. Graph Valid Tree: This problem asks to determine if a graph forms a valid tree (no cycles and all nodes connected), which requires DFS and cycle detection.

In all these problems, the main techniques involve graph representation, graph traversal using DFS, and sometimes cycle detection – similar to the concepts we used in our original problem.

Language Agnostic Coding Drills

1. Creating lists and list comprehensions: Understanding how to create a list and use list comprehensions is key to the start of this code. A list comprehension is created to generate the adjacency list.

2. Basic functions: Understanding how to define and call functions is essential. In this code, there are multiple functions such as buildAdjacencyList and hasCycle.

3. Nested functions: In this code, hasCycle is nested within the canFinish function. This can be a tricky concept to grasp for beginners.

4. For-loops and nested for-loops: The basic logic of the code involves looping through elements in the data structures (adjacency list and states list). This is done using for-loops, and understanding them is essential.

5. Conditionals (if-else statements): Conditional statements are used throughout the code to control the flow and logic of the program.

6. Data structures (specifically lists): Lists are used extensively in this code. In Python, lists are mutable and ordered collections.

7. Working with Graphs and Adjacency Lists: The problem itself is a graph problem and understanding how to represent a graph using an adjacency list is a significant part of this problem.

8. Recursion and understanding recursion stack: The hasCycle function is a recursive function. It keeps on calling itself until it finds a cycle or has traversed through the graph.

9. State and State transition: In this problem, the state array is used to keep track of the states of nodes in the graph. Understanding how to transition states is key.

Arranging the drills in increasing level of difficulty:

1. Creating lists and list comprehensions
2. Basic functions
3. Conditionals (if-else statements)
4. For-loops and nested for-loops
5. Data structures (specifically lists)
6. Nested functions
7. Working with Graphs and Adjacency Lists
8. Recursion and understanding recursion stack
9. State and State transition.

Targeted Drills in Python

1. Creating lists and list comprehensions:

   Exercise: Write a list comprehension that generates a list of squares of numbers from 0 to 10.

   1 2	squares = [x**2 for x in range(11)] print(squares)

2. Basic functions:

   Exercise: Define a function that calculates the area of a circle given a radius.

   1 2 3 4 5 6

   import math def area_of_circle(radius): return math.pi * radius ** 2 print(area_of_circle(5))

3. Conditionals (if-else statements):

   Exercise: Write a function that accepts a number and returns whether it’s positive, negative or zero.

   1 2 3 4 5 6 7 8 9 10 11

   def check_number(n): if n > 0: return "Positive" elif n < 0: return "Negative" else: return "Zero" print(check_number(5)) print(check_number(-3)) print(check_number(0))

4. For-loops and nested for-loops:

   Exercise: Write a nested loop to print a right-angled triangle of “*”.

   1 2 3 4

   for i in range(1, 6): for j in range(i): print("*", end="") print()

5. Data structures (specifically lists):

   Exercise: Create a list of numbers from 1 to 5 and append a new number at the end. Then remove the second element from the list.

   1 2 3 4

   numbers = [1, 2, 3, 4, 5] numbers.append(6) del numbers[1] print(numbers)

6. Nested functions:

   Exercise: Write a function that calculates the total price of items in a shopping cart. The shopping cart is a list of (item, price, quantity) tuples.

   1 2 3 4 5 6 7 8 9

   def total_price(cart): def item_total(item): _, price, quantity = item return price * quantity return sum(item_total(item) for item in cart) cart = [("apple", 1, 2), ("banana", 0.5, 5)] print(total_price(cart))

7. Working with Graphs and Adjacency Lists:

   Exercise: Given an adjacency list representation of a graph, write a function to calculate the degree of a given node.

   1 2 3 4 5

   def degree(graph, node): return len(graph[node]) graph = {1: [2, 3], 2: [1], 3: [1, 4, 5], 4: [3], 5: [3]} print(degree(graph, 3))

8. Recursion and understanding recursion stack:

   Exercise: Write a recursive function to calculate the factorial of a number.

   1 2 3 4 5 6 7

   def factorial(n): if n == 0 or n == 1: return 1 else: return n * factorial(n - 1) print(factorial(5))

9. State and State transition:

   Exercise: Simulate a basic traffic light system. Create a dictionary with states and write a function to transition from one state to another.

   1 2 3 4 5 6 7 8 9

   states = {"Red": "Green", "Green": "Yellow", "Yellow": "Red"} def transition(current_state): return states[current_state] current_state = "Red" for _ in range(5): current_state = transition(current_state) print(current_state)

Each exercise corresponds to the concepts in the order listed in the question.

title: Course Schedule excerpt: This article explains the thought process behind solving the course schedule problem. tags: cycle-detection

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array of prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return true if you can finish all courses. Otherwise, return false.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take. 
To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take. 
To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Constraints

• 1 <= numCourses <= 105
• 0 <= prerequisites.length <= 5000
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• All the pairs prerequisites[i] are unique.

Thought Process

There are 6 edges: [[1,0],[2,1],[3,2],[5,3],[4,5],[0,5]]

The adjacency list for the graph:

Course => Prerequisites Courses
0 => 5
1 => 0
2 => 1
3 => 2
4 => 5
5 => 3

So Graph becomes as below which has cycle 0->1->2->3->5->0 and course schedule not possible.

Define the Interface

Input: integer (number of courses) (0 to number of courses - 1)
        prerequisites (array of arrays containing integers)
Output: boolean      

Analyze the Input and Output

[0, 1] means 1 is prereq for 0

Identify the Invariants

• Cannot change the input parameters
• You cannot visit a node more than once? True. Why?

We go from node to node, once we reach the same node, we have found a cycle. We have not explored all the neighbors of a node, when we backtrack, we can explore the remaining neighbors of that node.

• Should we say: We can only visit a neighbor once? Yes/No : YES

The visited array is used to keep track of visited nodes. DFS is used for cycle detection.

• What are the prereqs?

It is not an adjacency list. It has edges. We have to convert the edges to the adjacency lists.

1 ----> 0  is correct

[[1,0],[1, 2], [3, 4]]

0 -> [1]
2 -> [1]
4 -> [3]

[0, 1]

1 ----> [0]

0 ----> [1,2]
1 ----> []
2 ----> [1]

• What does the adjacency list look like?
• Correct Invariant: We have to visit and explore a given node only once.

Classify the Problem

Graph Traversal - Detect a cycle in the graph.
Topological Sort - We don't need to do all the work of toposort.

Solve the Problem by Hand

Each course is a node, the prereqs are edges.

  [0,1]
   u, v
   
  0 ----> 1

This is a directed graph. Return true.

0 <—-> 1 When we have a cycle, it is not possible to take the course 1 or 0. Return false.

Solution Outline

1. Convert the edges to adjacency list

2. DFS - true/false if there is a cycle.

3. What are the parameters we need for the DFS call?

   • visited
   • stack for iterative solution
   • current node (u)

4. What is the contract between the parent and child of the recursive calls?

   What does the parent need from the child? We need to return true/false based on whether there is a cycle or not

5. What is the base case?

   Two possibilities: There is a cycle, no cycle If visited, it is in visited array No cycle, the number of nodes in the visited array is = number of courses in the input

   Loop to iterate through all the neighbors - Return true as soon as we find a cycle

   return false

6. Main cycle - false/true

Brute Force Approach

• How do we detect cycles in a graph?
• We can use DFS, if we get back to the same node we have found a cycle
• Initiate a DFS, from each possible node
• What are the things we need to keep track of?
• What nodes did we visit so far?

Creating Adjacency List

The num_courses parameter is the nodes in the graph. The prerequisites parameter is the edges in the graph. The list node is mapped to a list of neighbors.

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def adjacency_list(num_courses, prereqs)
  list = Hash.new{|h, k| h[k] = []}

  prereqs.each  do |course, prereq|
    list[course] << prereq
  end
    
  list
end

n = 4, edges = [[1,0], [2, 3], [1, 4]]

p adjacency_list(n, edges)

Solution

The adjacency_list method creates adjacency list from the edges for every node.

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def adjacency_list(num_courses, prereqs)
  list = Hash.new{|h, k| h[k] = []}

  prereqs.each  do |course, prereq|
    list[course] << prereq
  end
    
  list
end

# Cycle detection
def cycle?(u, visited, explore, list)
  if visited[u]
    return false
  end
  
  # detect cycle 
  if explore[u]
    return true
  end
  
  # Currently exploring node u
  explore[u] = true
    
  for v in list[u]
    # Use the result provided by the child to make a decision
    if cycle?(v, visited, explore, list)
      return true
    end
  end
  
  # backtracking step
  explore[u] = false
  visited[u] = true
  
  # contract between parent and child
  # no cycle detected
  return false
end

# @param {Integer} num_courses
# @param {Integer[][]} prerequisites
# @return {Boolean}
def can_finish(num_courses, prerequisites)
  list = adjacency_list(num_courses, prerequisites)
  
  visited = []
  explore = []
  
  # Iterate over all the possible nodes
  # If the DFS call returns true, return false
  for u in (0...num_courses)
    if cycle?(u, visited, explore, list)    
      return false 
    end
  end

  # No cycle found
  return true
end

Time and Space Complexity

Time: O(V+E)

Space: O(V+E)

V - Number of vertices. E - Number of edges.