Jump Game

  1. We will have at least one element. Return true

  2. We start at the first index.

    Can we reach the last index? We don’t care if we took one jump or several jumps.

  3. Maximum: It is not an optimization problem.

  4. From the first element, we have 1, 2, … value of the element I will have search the entire search space (feasible and infeasible solutions). Only one feasible solution is required to return true.

  5. Brute Force

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

    possible to reach array: stores whether we can reach a position or not. the value represents the maximum number of jumps we can have

    for (over elements) for (number of jumps current) if (condition) – needed for any element other than at index 0 flip the boolean array 1 or 2 (starting from index 0)

    If the last element is true, then return true, otherwise false.

    T: O(N^2) S: O(N)

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

    Many paths, you are taking all of them.

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

    If we are at index 0, we will try:

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

  6. What are the choices that we have at each index? I can either make 1 jump

  7. Imagine a loop that starts from 0, the loop goes over the number of steps we could take.

Is reachable array from the start, initialize all of them to false. From start we can reach the start

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

0 1 2. 3. 4 [f, f, t, f, t]

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

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

Time: O( ) Space: O( )

Key Takeaways

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# @param {Integer[]} nums
# @return {Boolean}
def can_jump(nums)
  max_jump = 0
  
    for i in (0..nums.size-1)
       return false if max_jump < i

        max_jump = [max_jump, i + nums[i]].max
    end
    return true
end

This problem can be tackled by iterating through the array and keeping track of the farthest index that can be reached at each step. Here’s how you can understand and solve this problem:

  1. Start at the First Index: You are initially positioned at the first index of the array.

  2. Maximum Jump Length: Each element in the array represents the maximum jump length from that position.

  3. Iterate Through the Array: As you iterate, you calculate the farthest index that can be reached from the current position.

  4. Check for Unreachable Positions: If at any point, the current index is greater than the farthest reachable index, it means there is a position that cannot be reached, and the function returns false.

  5. Reach the Last Index: If you can iterate through the entire array without finding an unreachable position, the function returns true.

Here’s a Python code to accomplish this:

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class Solution:
    def canJump(self, nums: List[int]) -> bool:
        max_reach = 0  # Keeps track of the farthest reachable index
        for i in range(len(nums)):
            if i > max_reach:  # If the current index is greater than the maximum reachable index
                return False
            max_reach = max(max_reach, i + nums[i])  # Update the maximum reachable index
        return True  # If you have iterated through the entire array, return true

The code iterates through the array once, so the time complexity is (O(N)), where (N) is the length of the array. The space complexity is (O(1)), as we only use a constant amount of extra space.

This solution provides a clear and concise answer to whether you can reach the last index of the given array based on the jump lengths provided in the array.

Identifying Problem Isomorphism

“Jump Game” has a similar structure to “Can Place Flowers”.

In “Jump Game”, the challenge is to determine if it’s possible to reach the last index of an array where each element represents your maximum jump length at that position. The “Can Place Flowers” problem also presents a similar array-based challenge where you need to determine if a certain number of flowers can be placed in a flowerbed represented as an array without violating the no-adjacent-flowers rule.

Both problems involve traversing an array and making decisions based on the current and sometimes future elements. They both require a form of ‘greedy’ strategy to solve, where you make the locally optimal choice at each step with the hope that these local solutions lead to a global solution.

It’s an approximate mapping based on the structure of the problems. Of the two problems, “Can Place Flowers” is simpler as it requires less complex decision-making at each step. “Jump Game” is more complex due to the need to calculate jumps and track the furthest reachable index.

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class Solution:
    def canJump(self, nums: List[int]) -> bool:
        # Take curr variable to keep the current maximum jump...
        curr = nums[0]
        # Traverse all the elements through loop...
        for i in range(1,len(nums)):
            # If the current index 'i' is less than current maximum jump 'curr'...
            # It means there is no way to jump to current index...
            # so we should return false...
            if curr == 0:
                return False
            curr -= 1
            # Update the current maximum jump...
            curr = max(curr, nums[i])       # It’s possible to reach the end of the array...
        return True

Problem Classification

The problem falls under the domain Arrays and Dynamic Programming or Greedy Algorithms.

What

  1. Input: An integer array nums, where each element represents the maximum jump length at that position.
  2. Initial Position: You start at the first index of the array.
  3. Objective: Determine if you can reach the last index of the array.
  4. Output: Return true if you can reach the last index, and false otherwise.
  5. Constraints:
    • Array length: ( 1 \leq \text{nums.length} \leq 10^4 )
    • Element values: ( 0 \leq \text{nums}[i] \leq 10^5 )

The problem can be classified as:

  1. Decision Problem: You’re required to make a decision - can you reach the last index or not.
  2. Optimization Aspect: While not explicit, solving the problem efficiently will likely involve optimizing the path taken.
  3. Traversal Problem: You have to traverse the array based on the jump values, making it a kind of traversal problem.
  4. State-based Problem: The ability to reach the last index can be based on the state at each index, like maximum reach from that point, making it a candidate for Dynamic Programming or Greedy Algorithms.

It’s an algorithmic problem requiring an efficient traversal of the array, with a decision to be made at the end.

Clarification Questions

  1. Starting Point: Is the initial position always at the first index of the array?

  2. Negative Numbers: Can the array contain negative numbers, or is it limited to the constraints mentioned?

  3. Array Length: What is the minimum and maximum length of the input array?

  4. End Behavior: Do we have to land exactly on the last index, or is it sufficient to jump past it?

  5. Duplicate Numbers: Can the array contain duplicate numbers?

  6. Empty Array: Is an empty array a valid input?

  7. Multiple Solutions: Is it possible to have multiple solutions, and are we interested in finding just one or all possible paths?

  8. Zero Jump: What happens if the jump length at a particular position is zero?

  9. Output Format: Is a Boolean output sufficient, or is there a need for additional information, like the path taken?

  10. Time Complexity: Are there any time complexity constraints we should be aware of?

  11. Space Complexity: Are there any space complexity constraints?

  12. Real-world Context: Is the problem a simplified version of a more complex real-world problem that has additional constraints?

Asking these questions helps in understanding edge cases, constraints, and the exact requirements of the problem.

Problem Analysis and Key Insights

  1. Greedy Approach Feasibility: Since you’re always starting from the first index and each element gives you the maximum jump length, a greedy approach could work well. You can keep track of the furthest you can reach at each step.

  2. Dynamic Programming Alternative: Another insight is that Dynamic Programming (DP) could be used, especially if variations of this problem require more complex considerations. However, DP might be an overkill for this specific problem statement as it stands.

  3. Boolean Output: The problem doesn’t ask for the path to reach the last index, just a true or false. This simplifies the output and allows you to focus solely on the feasibility of reaching the last index.

  4. Initial and Terminal States: You start at index 0 and aim to reach the last index. This means you can ignore any jumps that won’t either improve your current maximum reach or get you to the goal.

  5. Element as State Information: Each element serves as state information that helps you decide the next move. This makes it important to only consider the most beneficial next steps.

  6. Zero Jump Block: An element with a value of zero is crucial. It can potentially block your path unless you can jump over it from a previous index.

  7. Constraints Indicate Efficiency Needs: The constraints on array length ((10^4)) and element values ((10^5)) indicate that the algorithm needs to be fairly efficient. A simple brute-force solution is likely not sufficient.

  8. Single Objective: The single objective of reaching the last index makes the problem straightforward and helps focus the algorithmic approach.

By analyzing these key aspects, we can more effectively strategize our approach to solving the problem.

Problem Boundary

The scope of this problem is well-defined and confined to the following:

  1. Algorithmic Solution: The primary focus is on developing an algorithm to determine whether one can reach the last index of a given integer array based on jump lengths.

  2. Input Constraints: The input is an integer array with specific constraints on its length and the values it can hold. This limits the problem to certain types of data.

  3. Output Requirement: The output is binary—either true or false, indicating whether reaching the last index is possible or not.

  4. Efficiency: Given the constraints, the algorithm needs to be efficient. However, there’s no explicit requirement on time or space complexity, suggesting a focus on correctness over optimization.

  5. No Path Requirement: The problem doesn’t ask for the sequence of jumps to reach the end, only whether it’s possible or not. This narrows the scope considerably.

  6. Initial Condition: The starting point is always the first index of the array, and the objective is to reach the last index.

  7. Single Scenario: The problem doesn’t account for variations like starting from a different index, having constraints on the number of jumps, or other real-world complexities.

  8. Element-wise Decisions: The problem is built on making decisions at each array element based on its value, which serves as a state.

  9. No External Factors: There are no external variables or conditions affecting the problem. It is self-contained.

In summary, the scope is narrow, focusing on algorithmic problem-solving within well-defined constraints and a singular objective.

To establish the boundary of this problem, we need to consider the following aspects:

  1. Input Constraints:

    • The array’s length is between (1) and (10^4).
    • The elements are between (0) and (10^5).

  2. Initial Conditions:

    • You always start at the first index of the array.

  3. Output:

    • The output is strictly boolean: either true or false.

  4. Algorithmic Focus:

    • The objective is to find a correct and efficient algorithm to solve the problem.

  5. State Information:

    • The maximum jump length at each position is the key state information used to make decisions.

  6. Decision-making:

    • At each index, you decide how far to jump based on the maximum jump length at that index and the maximum reachable index so far.

  7. End Condition:

    • The algorithm terminates when you either reach the last index or determine that it’s impossible to do so.

  8. Exclusions:

    • There is no need to find the actual path.
    • No additional scenarios like multiple starting points or multiple end goals are considered.
    • No other data types or data structures apart from integer arrays are in scope.

  9. Complexity Constraints:

    • While not explicitly stated, the size constraints on the array and elements suggest that the algorithm should be fairly efficient.

By understanding these aspects, you define the boundary of the problem, separating it from variations or extensions that could complicate or broaden the scope.

Distilling the Problem to Its Core Elements

Fundamental Concept

The fundamental concept this problem is based on is “reachability” in a constrained space. You start at one end of an array, and each element specifies how far you can jump from that point. The question is whether these constraints allow you to reach the other end.

Simplest Description

Imagine you are hopping on a series of stones across a river. Each stone has a number written on it that tells you the maximum number of stones you can skip to make your next jump. You start on the first stone. Can you make it to the last one?

Core Problem

The core problem is determining whether you can jump from the start to the end of an array based on the maximum jump length at each position.

Simplified Problem Statement

Can you get from the start to the end of an array if each element tells you the furthest you can jump from there?

Key Components

  1. Starting Point: First element in the array.
  2. Jump Lengths: The numbers in the array.
  3. Current Position: Where you are at any point in time.
  4. Maximum Reachable Index: The furthest index you can reach based on your current position and jump lengths.
  5. Goal: The last index of the array.

Minimal Set of Operations

  1. Initialize: Start at the first index and initialize a variable to keep track of the maximum index you can reach (maxReach).
  2. Iterate: Loop through the array.
    • Update maxReach: At each index, update maxReach if the current index plus the jump length at that index is greater than maxReach.
    • Check Goal: If maxReach is greater than or equal to the last index, return true.
    • Dead End: If at any point maxReach is equal to your current index and you haven’t reached the end, return false.
  3. Return: Once the loop ends, return the result.

By focusing on these aspects, you can succinctly and effectively tackle the problem.

Visual Model of the Problem

Visualizing this problem can be particularly helpful for understanding its dynamics. Here are a few ways to do that:

Number Line

  1. Step 1: Draw a number line representing the array indices, from 0 to ( n-1 ) where ( n ) is the length of the array.
  2. Step 2: Place the numbers from the array above their respective indices on the number line.
  3. Step 3: Start at index 0. Draw an arrow indicating the maximum distance you can jump from that index.
  4. Step 4: Move to the next index that falls within the arrow’s range. Repeat Steps 3 and 4.
  5. Step 5: Observe if you can reach or surpass the last index.

Graphs or State Diagram

  1. Nodes: Each array index becomes a node.
  2. Edges: Draw directed edges from one node to another based on the maximum jump length from that node. You may end up with multiple edges from one node.
  3. Start and End: Mark the first and last nodes.
  4. Path: Try to find a path from the starting node to the ending node.

Grid Visualization

  1. Columns: Represent each array index as a column.
  2. Cell Values: Write the maximum jump length in the cells.
  3. Arrows: Draw arrows from one cell to the reachable cells based on the jump length.
  4. Traversal: Visually traverse from the first column to the last, following the arrows.

Pseudocode Annotations

  1. Annotate Code: As you iterate through the array in your pseudocode, make notes or draw arrows on your number line or grid visualization to indicate where you can jump to from each index.
  2. Max Reach: Use a different color or symbol to indicate the furthest you can reach at any point.

By visualizing the problem in these ways, you make the abstract more concrete, thereby gaining insights into possible solution strategies.

Problem Restatement

You have an array of integers where each number specifies the maximum distance you can jump forward from that position. Starting at the first element of the array, you need to determine if it’s possible to reach the last element by jumping according to these distances.

Requirements:

  1. Start Point: You always begin at the first element of the array.
  2. Jump Lengths: The integers in the array dictate how far you can jump from each position.
  3. End Point: The goal is to determine if you can reach the last element of the array.
  4. Output: Your answer should be a simple ‘yes’ or ’no’, represented as true or false.

Constraints:

  1. The array length will be between (1) and (10^4).
  2. The integers in the array will range from (0) to (10^5).

By paraphrasing the problem like this, we focus on its essential elements, making it easier to then dive into solving it.

Abstract Representation of the Problem

Let’s abstract the problem to focus on its core structure and key elements.

Abstract Representation

  1. Set of States (S): Let ( S ) represent a set of states, where each state ( s_i ) corresponds to an index ( i ) in the array. ( S = { s_0, s_1, …, s_{n-1} } ).

  2. Transition Function (T): Define a transition function ( T: S \rightarrow 2^S ) that takes a state ( s_i ) and returns a set of states reachable from ( s_i ). The set depends on the value ( v ) at ( s_i ) in the array and is ( { s_{i+1}, s_{i+2}, …, s_{i+v} } ).

  3. Initial State (I): The initial state is always ( s_0 ), corresponding to the first index of the array.

  4. Goal State (G): The goal state is ( s_{n-1} ), corresponding to the last index of the array.

  5. Feasibility Function (F): The task is to find a function ( F: S \rightarrow { \text{true}, \text{false} } ) that returns true if there exists a sequence of transitions ( T ) leading from ( I ) to ( G ), and false otherwise.

Constraints:

  1. ( |S| ) is bounded by ( 1 ) and ( 10^4 ).
  2. The value ( v ) at any state ( s_i ) is bounded by ( 0 ) and ( 10^5 ).

Problem Statement:

Determine ( F(I) ), i.e., can you reach ( G ) starting from ( I ) given the transition function ( T ) and the constraints.

This abstract representation removes specific details and focuses on the structural aspects, making it easier to reason about possible approaches to solve the problem.

Terminology

  1. Array: A data structure that stores elements in a contiguous block of memory. In this problem, the array contains integers that specify maximum jump lengths at each index.

  2. Index: A position in the array, starting from 0 and going up to ( n-1 ), where ( n ) is the length of the array. The index plays a crucial role as both the state in our abstract representation and the position from which you make jumps.

  3. Jump Length: The integer value at a given index in the array. It dictates how far you can jump forward from that index.

  4. Initial State: The state from where the process begins, corresponding to the first index of the array.

  5. Goal State: The state you are trying to reach, corresponding to the last index of the array.

  6. Reachability: The property of being able to get from one state to another within given constraints. The main task is to determine the reachability of the goal state from the initial state.

  7. Transition Function: A function that takes the current state and returns the possible next states based on the jump lengths. In this problem, it tells you where you can go from a given index.

  8. Boolean Output: A type of output that is either true or false. In this problem, it represents whether you can reach the goal state or not.

  9. Traversal: The act of moving through states, or indices in the array, according to the transition function. Traversal methods are key to solving this problem.

  10. Maximum Reachable Index: The furthest index you can reach at any point, updated based on the current index and its jump length. This concept is central to optimizing the solution.

Each of these terms plays a specific role in either defining the problem or structuring its solution. Understanding them is key to grasping both the problem and potential ways to solve it.

Problem Simplification and Explanation

You have a list of numbers. Each number tells you the maximum steps you can move forward from that position. You start at the first number. Your goal is to figure out if you can reach the last number by following these step rules.

Key Concepts and Interactions

  1. Starting Point: You always start at the first number.
  2. Step Rules: Each number tells you how far you can move from that spot.
  3. Current Position: The number you’re at right now.
  4. Farthest You Can Go: The maximum distance you can reach based on the numbers you’ve seen so far.
  5. End Goal: The last number in the list.

You start at the first number and look at its value to see how far you can go. As you move, you keep track of the farthest point you can reach. If this point is ever at or beyond the last number, you win. If you get stuck at a point where you can’t move forward, you lose.

Metaphor

Imagine you’re playing a game of hopscotch where the squares aren’t equally distant. Each square has a number written on it, and that number tells you the maximum number of squares you can hop forward. You win if you can land on or hop over the last square. If you land on a square that says “0” and it’s not the last one, you’re stuck, and you lose.

By understanding these simplified terms and the hopscotch metaphor, you can get a grasp of the problem’s core elements and how they interact.

Constraints

Here are some characteristics that can be leveraged for an efficient solution:

  1. Array Length Constraint: The array length is limited to (10^4). An algorithm with time complexity (O(n)) or (O(n \log n)) should be efficient enough.

  2. Value Range: Each element in the array is between 0 and (10^5). No negative numbers mean we don’t have to consider backward movement, simplifying the problem.

  3. Start Always at Index 0: Since you always start at the first index, there’s no need to consider other starting points. This reduces the problem space.

  4. Goal is Last Index: You only need to reach the last index, so once you’ve determined that it’s reachable, you can terminate the algorithm early. This allows for possible optimizations.

  5. Boolean Output: The output is simply true or false. This means that you don’t have to keep track of the path you took to reach the end, only whether or not it’s possible to do so.

  6. Maximum Reachable Index: As you traverse the array, you can keep track of the furthest index you can reach from the current index. This allows you to skip checking some indices, thus potentially speeding up the algorithm.

  7. Zero Values: A zero value is only problematic if it’s the maximum reachable index. If you encounter a zero but your maximum reachable index is greater, you can safely continue.

  8. Sequential Scan: Given that you’re starting at the first index and aiming to reach the last one, you can solve this problem with a single pass through the array, updating the maximum reachable index as you go.

  9. Early Termination: If at any point the maximum reachable index is greater than or equal to the last index, you can terminate the algorithm early, as you’ve confirmed reachability.

By recognizing and exploiting these specific characteristics, you can design an efficient algorithm to solve the problem.

From analyzing the constraints, we gather the following key insights:

  1. Linear Time Complexity: The array length constraint of (10^4) suggests that a linear time solution ((O(n))) would be efficient enough for this problem.

  2. Unidirectional Movement: All elements are non-negative, eliminating the need to consider backward jumps. This simplifies the problem space.

  3. Limited Problem Space: You always start at index 0 and aim to reach the last index, which constrains the problem to specific start and end points.

  4. Simple Output: The boolean output (true or false) informs us that the problem is decision-based. We don’t need to find a path, just confirm its existence or absence.

  5. Potential for Early Termination: Since the goal is to reach the last index, as soon as we determine that this is possible, the algorithm can terminate, providing an opportunity for optimization.

  6. Zero-Value Handling: A zero value isn’t necessarily a blocker unless it becomes the maximum reachable index, which simplifies the traversal logic.

  7. Sequential Traversal: The nature of the problem allows for a single pass through the array, implying that we can solve it without having to revisit elements.

Understanding these aspects of the constraints guides us toward focusing on efficient, single-pass algorithms and also highlights opportunities for optimization.

Case Analysis

Absolutely, let’s cover a range of test cases including edge and boundary conditions.

Edge Cases

  1. Minimum Length Array

    • Input: [1]
    • Output: true
    • Analysis: The array has only one element, so you are already at the last index. This tests the lower limit of array size.

  2. All Zeros Except First Element

    • Input: [1, 0, 0, 0]
    • Output: false
    • Analysis: The first element allows a jump of length 1, but subsequent zeros make it impossible to proceed further. This tests how zeros can block progress.

  3. Single Non-Zero Element

    • Input: [2]
    • Output: true
    • Analysis: Similar to the first case but with a different value. You’re already at the last index.

Typical Cases

  1. Ascending Order

    • Input: [1, 2, 3, 4, 5]
    • Output: true
    • Analysis: Each jump can easily take you to the next index, and you’ll ultimately reach the end.

  2. Descending Order

    • Input: [5, 4, 3, 2, 1]
    • Output: true
    • Analysis: The first element is enough to take you straight to the end.

  3. Zeros in Between

    • Input: [2, 0, 0, 3, 0]
    • Output: true
    • Analysis: The first 2 is enough to skip over the zeros, and the 3 takes you to the end.

  4. Zero at the End

    • Input: [1, 2, 3, 0]
    • Output: true
    • Analysis: A zero at the end is irrelevant as long as you can reach or cross it.

  5. Blocked by Zero

    • Input: [1, 1, 0, 3]
    • Output: false
    • Analysis: You’re stuck at the third index with a zero and can’t move further.

Boundary Cases

  1. Maximum Length, All Ones

    • Input: [1, 1, 1, ..., 1] (10^4 elements)
    • Output: true
    • Analysis: This tests the upper limit of the array size constraint. Even though it takes (10^4) operations, it’s still efficient.

  2. Maximum Value Element

    • Input: [10^5, 0, 0, ..., 0]
    • Output: true
    • Analysis: The first element is large enough to take you straight to the end. This tests the upper limit of individual elements.

By analyzing these cases, we understand better the pitfalls of zeros, the irrelevance of the last element’s value, and the implications of the constraints on array size and element value. These insights will guide us in creating a robust solution.

Visualizing these cases can help to better understand the nature of the problem. Here’s a simple way to visualize using text characters:

Notation

  • S: Start point
  • E: End point
  • →: Forward jump direction
  • x: Blocker (zero that you can’t bypass)

Edge Cases

  1. Minimum Length Array

    S-E

  2. All Zeros Except First Element

    S-→x→x→E

  3. Single Non-Zero Element

    S-E

Typical Cases

  1. Ascending Order

    S-→-→-→-→E

  2. Descending Order

    S------------→E

  3. Zeros in Between

    S-→→-x-x-→→→E

  4. Zero at the End

    S-→-→-→-E

  5. Blocked by Zero

    S-→-→-x→E

Boundary Cases

  1. Maximum Length, All Ones

    S-→-→-→...→E

  2. Maximum Value Element

    S--------------------------------------------------→E

In each of these visualizations, you can trace the “path” from S to E. If you can make it to E, the output is true; otherwise, it’s false.

This visualization helps to quickly grasp what each test case represents and what the jump paths look like. It also makes it easier to identify blockers and points where the algorithm can terminate early.

Analyzing the different cases yields several key insights:

  1. Minimum Array Length: For arrays with a single element, you are already at the end, making the answer always true.

  2. Effect of Zeros: Zeros act as potential blockers. They are problematic only when they become the furthest reachable index.

  3. Irrelevance of Last Element: The value of the last element doesn’t matter as you just need to reach or cross it, not jump from it.

  4. Early Termination: As soon as you determine that the last index is reachable, you can terminate the algorithm, making it more efficient.

  5. Single Pass Sufficiency: You can determine reachability with a single pass through the array by updating the maximum reachable index.

  6. Maximum Length and Values: The problem scales well within the constraints of (10^4) elements and element values up to (10^5), allowing for linear time complexity solutions.

  7. Decision-based Output: The problem requires only a true or false answer, simplifying what you need to keep track of.

These insights are critical for designing an efficient and robust algorithm. They help you understand what to focus on and what conditions to check for as you traverse the array.

Identification of Applicable Theoretical Concepts

There are a a few mathematical and algorithmic concepts:

  1. Greedy Algorithm: This problem can be effectively solved by always making the locally optimal choice, i.e., updating the furthest reachable index as you traverse the array.

  2. Dynamic Programming: Though not as efficient for this particular problem, it can be approached using dynamic programming to check if each index is reachable from the starting index.

  3. Boolean Logic: The problem is a decision problem requiring a true or false output, simplifying what we need to keep track of during algorithm execution.

  4. Early Termination: Due to the decision-based nature, the algorithm can be designed to terminate as soon as a conclusion is reached.

  5. Element-wise Maximization: You can maintain a running maximum of reachable indices, which is a basic operation in optimization theory.

  6. Graph Theory: Conceptually, this can be modeled as a directed graph where each index is a node and each jump a directed edge. This view might be useful for understanding the problem, although it doesn’t necessarily lead to a more efficient solution in this case.

  7. Complexity Analysis: Given the problem constraints, the focus will be on linear time complexity ((O(n))) algorithms, making it computationally efficient.

These concepts can guide the development of a robust and efficient algorithm to solve the problem. Greedy algorithms stand out as the most appropriate for providing an efficient and simple solution.

Simple Explanation

Let’s say you’re playing a video game. In this game, you start at the beginning of a road and your goal is to get to the end of it. Along the road, there are various spots where you can jump. Each spot has a number on it, and that number tells you how far you can jump from that spot. The catch is, you can jump fewer steps if you want, but not more.

Now, the question is: can you make it to the end of the road by jumping from spot to spot, following the numbers?

For example, if the road has spots with numbers [2, 3, 1, 1, 4], you can jump 2 steps from the first spot, then 3 steps from the second spot to reach the end.

But if the road has spots with numbers [3, 2, 1, 0, 4], you’ll get stuck at the spot with the number 0 because you can’t move any further.

So, can you figure out if you can reach the end of the road by just looking at the numbers? That’s the problem we’re trying to solve.

Problem Breakdown and Solution Methodology

Certainly. Solving this problem is like making a series of jumps across stepping stones to cross a river. Each stone has a number written on it, which tells you the maximum number of stones you can skip to make your next jump.

Steps to Solve:

  1. Start at the Beginning: Imagine standing on the first stone, or the first element of the array.

  2. Check Reachability: Keep track of the furthest stone you can reach from your current position. Initially, this is the same as the first stone’s number.

  3. Iterate Through Stones: Walk through each stone one by one.

    a. Update Farthest Reach: At each stone, update the farthest you can reach if that stone allows you to go further than your current maximum.

    b. Check for Blockers: If you encounter a stone that you can’t pass, then you know you can’t make it to the end.

    c. Early Termination: If your farthest reach includes the last stone, you can end the game—you’ll make it across!

How Parameters Affect the Solution

  • Array Length: Longer arrays may require more time to traverse, but they don’t fundamentally change the approach.

  • Element Values: High numbers in the array could allow you to skip many elements, making the process faster. But zeros can be blockers unless bypassed.

  • Initial Element: If the first element is zero and the array length is more than 1, you’re stuck right away—the answer is false.

Example Case: nums = [2, 3, 1, 1, 4]

  1. Start at index 0: Farthest reach is 2 (from the first element).

  2. Move to index 1: Farthest reach is now 4 (2nd element value 3 + index 1).

  3. Move to index 2: Farthest reach remains 4 (3rd element value 1 + index 2 is not greater than 4).

  4. Move to index 3: Farthest reach remains 4 (4th element value 1 + index 3 is not greater than 4).

  5. Termination: Before we reach index 4, we see that our farthest reach is 4, which is the last index. So we terminate early.

In this case, you can reach the end of the array, so the answer is true.

By following this approach, you’re making the most of each stone’s “jump power” to ensure you can make it to the end of the river, or the array in our problem.

Inference of Problem-Solving Approach from the Problem Statement

Here are the key terms or concepts in the problem and how they inform the approach:

  1. Integer Array (nums): The main data structure. It dictates a linear, one-pass traversal strategy for efficiency.

  2. Initial Position: You start at the array’s first index. This sets the boundary for the starting point.

  3. Maximum Jump Length: Each array element represents this. It guides the greedy approach, as you’ll always try to extend your reach as far as possible at each step.

  4. Last Index: The final destination. It informs the termination condition for the algorithm.

  5. Return True/False: The problem asks for a boolean answer, simplifying what you need to keep track of.

  6. Constraints: They define the input space and allow us to make assumptions on the computational complexity of the solution.

Each of these terms points towards a simple, linear time complexity solution:

  • Integer Array and Constraints suggest that a one-pass, O(n) algorithm is possible and sufficient.

  • Maximum Jump Length pushes us towards a greedy algorithm. At each index, you aim to extend your ‘reach’ as far as possible.

  • Initial Position and Last Index define the boundaries of our problem and guide when the algorithm starts and stops.

  • Return True/False tells us we don’t need to keep track of the path, just whether reaching the end is possible.

By understanding these key terms and concepts, we’re guided towards a greedy, one-pass algorithm as the most straightforward and efficient approach to solving this problem.

Visualizing this problem can be very helpful for understanding its properties. Here are ways you can use tables or diagrams:

1. Index Table

Create a table with two columns: one for the index and another for the corresponding array element. A third column can represent the “Farthest Reach” at each step.

IndexElementFarthest Reach
022
134
214
314
44

This table can help you keep track of your position and your maximum reach as you walk through the array.

2. Reachability Graph

Imagine a horizontal line where each point on it represents an index in the array. Mark each point with the maximum distance you can reach from it. Use arrows to represent possible jumps.

0 ----> 2
1 ----------> 4
2 ----> 3
3 ----> 4
4

The lengths of the arrows depict how far you can go from each index. You can visually trace a path to see if reaching the last index is possible.

3. Visual Steps

On a piece of paper or a board, draw circles for each index and put the number (element value) inside it. Then draw directed lines (arrows) connecting each circle to the furthest circle it can reach.

0(2) ----> 1(3) -----------> 4(4)
       \----> 2(1)
             \----> 3(1)

Walk through these directed lines to simulate the jumps, updating the “Farthest Reach” at each step.

By drawing these tables or diagrams, you can physically see the properties of “Farthest Reach,” “Maximum Jump Length,” and termination conditions, making it easier to understand the problem and find a solution.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

High-Level Steps

  1. Initialize variables for the current position and the farthest reachable index.
  2. Iterate through the array, updating the farthest reachable index.
  3. Check termination conditions to return true or false.

Refined Steps

  1. Initialize current_position = 0 and farthest_reachable = nums[0].
  2. Loop through the array from index 1 to n-1:
    • If current_position > farthest_reachable, return false.
    • Update farthest_reachable = max(farthest_reachable, current_position + nums[current_position]).
    • Move current_position one step forward.
  3. After the loop, if farthest_reachable >= n-1, return true. Otherwise, return false.

Granular, Actionable Steps

  1. Declare and initialize two integer variables, current_position and farthest_reachable, both set to nums[0].
  2. Use a for loop to iterate through the array starting from index 1:
    • Inside the loop, first check if current_position > farthest_reachable. If true, break the loop and proceed to step 3.
    • Calculate farthest_reachable = max(farthest_reachable, current_position + nums[current_position]).
    • Increment current_position by 1.
  3. After the loop, check if farthest_reachable >= last index. If true, return true. Otherwise, return false.

Independent Parts

The problem is inherently sequential because each step depends on the previous state of farthest_reachable. So, it’s difficult to solve parts of it independently.

Repeatable Patterns

The core pattern is the update of the farthest_reachable variable at each step. This is a greedy approach, as you take the best possible step at each stage without worrying about future steps.

The for loop encapsulates this pattern, iterating through the array and updating farthest_reachable based on the current array value. This pattern repeats until you reach the end of the array or find that you can’t proceed further.

Solution Approach and Analysis

Approach to Solving the Problem

Conceptual Overview

Think of the array as a series of stepping stones across a river. Each stone (array element) has a number written on it that indicates the maximum number of stones you can skip. You start on the first stone, and your goal is to make it to the other side of the river (last index).

Detailed Steps

  1. Initialization: Stand on the first stone (current_position = 0) and see how far you can jump from there (farthest_reachable = nums[0]).

  2. Traverse Array: Move through the array from left to right, one stone at a time.

    • Can’t Reach: Before jumping, always check if you can actually reach the stone you are considering. If current_position > farthest_reachable, it means you’re stuck. Return false.

    • Extend Reach: At each stone, you check the farthest stone you can reach from there by calculating max(farthest_reachable, current_position + nums[current_position]). This is like extending your leg as far as it goes to see if you can reach a further stone.

    • Move: Once you’ve checked your reach, move on to the next stone (current_position += 1).

  3. Check Success: At the end of your journey, if farthest_reachable >= last index, you’ve successfully crossed the river. Return true.

Effect of Problem’s Parameters

  • Array Length: A longer array increases computational time linearly but doesn’t affect the logic.

  • Element Value: The larger the elements, the easier it generally is to reach the end. However, a single zero at the wrong place can make the problem unsolvable.

Example Cases

  • Example 1: nums = [2, 3, 1, 1, 4]

    1. Initialize current_position = 0 and farthest_reachable = 2
    2. current_position = 1, farthest_reachable = max(2, 1+3) = 4
    3. current_position = 2, farthest_reachable = max(4, 2+1) = 4
    4. current_position = 3, farthest_reachable = max(4, 3+1) = 4
    5. current_position = 4 (last index), farthest_reachable >= 4, Return true

  • Example 2: nums = [3, 2, 1, 0, 4]

    1. Initialize current_position = 0 and farthest_reachable = 3
    2. current_position = 1, farthest_reachable = max(3, 1+2) = 3
    3. current_position = 2, farthest_reachable = max(3, 2+1) = 3
    4. current_position = 3, can’t move further (3 > 3), Return false

With this step-by-step process, you can reliably determine if it’s possible to reach the last index of the given array.

Identify Invariant

The invariant in this problem is that at any point during traversal through the array, farthest_reachable indicates the maximum index you can reach from the starting position up to the current position. It is maintained and updated as you move through the array. This invariant holds true at the start, during each iteration, and at the end of the process.

The concept of farthest_reachable encapsulates your progress and options at any point. It is updated at every step using the formula farthest_reachable = max(farthest_reachable, current_position + nums[current_position]), ensuring that it always represents the farthest index you can reach so far. This is essential for determining whether you can reach the end of the array or not.

Identify Loop Invariant

The loop invariant in this problem is that at the beginning of each iteration, the variable farthest_reachable contains the maximum index reachable from index 0 to the current current_position in the array. This loop invariant holds before the loop starts, during each iteration, and after the loop terminates.

Here’s how the loop invariant is maintained:

  1. Initialization: Before the loop starts, farthest_reachable is set to the first element’s value (nums[0]), which is the farthest index we can reach from the starting position. At this point, current_position is 0.

  2. Maintenance: Inside the loop, farthest_reachable is updated using the formula farthest_reachable = max(farthest_reachable, current_position + nums[current_position]). This guarantees that farthest_reachable always contains the maximum index that can be reached so far.

  3. Termination: When the loop terminates, if farthest_reachable is greater than or equal to the last index of the array, the function returns true, confirming that the last index is reachable. Otherwise, it returns false.

The loop invariant ensures that at each step, we are correctly updating our reachability information, helping us to make an accurate final determination.

In this specific problem, the invariant and the loop invariant effectively represent the same condition: that the variable farthest_reachable holds the maximum index that can be reached from the starting index (0) up to the current_position at any given point in time.

However, it’s worth distinguishing between the two terms:

  • Invariant: An invariant is a condition that holds true in a broader sense throughout the system or algorithm, not just within a loop.

  • Loop Invariant: This is a specialized form of an invariant that specifically applies to loops. It’s a condition that holds true before the loop starts, during each iteration, and after the loop ends.

In this case, the broader algorithmic structure consists mainly of the loop that iterates through the array, so the invariant and loop invariant coincide. But in more complex algorithms, these could very well be different.

In this specific problem, the invariant and the loop invariant effectively represent the same condition: that the variable farthest_reachable holds the maximum index that can be reached from the starting index (0) up to the current_position at any given point in time.

However, it’s worth distinguishing between the two terms:

  • Invariant: An invariant is a condition that holds true in a broader sense throughout the system or algorithm, not just within a loop.

  • Loop Invariant: This is a specialized form of an invariant that specifically applies to loops. It’s a condition that holds true before the loop starts, during each iteration, and after the loop ends.

In this case, the broader algorithmic structure consists mainly of the loop that iterates through the array, so the invariant and loop invariant coincide. But in more complex algorithms, these could very well be different.

Let’s structure the content for this problem to verify that the loop invariant holds at all stages of the algorithm.

Initialization

Prior to the first iteration of the loop, we initialize farthest_reachable with the value nums[0]. At this point, current_position is 0. The loop invariant here states that farthest_reachable contains the maximum index reachable from index 0 up to the current_position. This holds true as, initially, you can only reach up to nums[0] from the starting index.

Maintenance

To see that each iteration maintains the loop invariant, consider the lines where we update farthest_reachable. These lines perform the operation farthest_reachable = max(farthest_reachable, current_position + nums[current_position]). This ensures that farthest_reachable contains the maximum index that can be reached so far, maintaining the loop invariant.

For instance, if farthest_reachable is 5 and at current_position = 3 we have nums[3] = 4, then the new farthest_reachable will be max(5, 3 + 4) = max(5, 7) = 7. This change upholds the loop invariant that farthest_reachable represents the farthest index we can reach from the start to the current position.

Termination

At termination, the loop invariant provides a useful property to show correctness. If the loop exits and farthest_reachable is greater than or equal to the last index of the array, then we can confidently say that it is possible to reach the end of the array from the starting index. This is because the loop invariant has been maintained throughout the algorithm, and it vouches for the reachability from start to end.

If farthest_reachable is less than the last index at termination, the loop invariant also supports the conclusion that it’s not possible to reach the end of the array, affirming the algorithm’s correctness in this scenario as well.

Thought Process

Certainly, let’s dissect the problem step-by-step, focusing on cues and direction given in the problem statement, and then code the final solution.

Thought Process

  1. Understand the Problem: You start at the first index and can jump up to nums[i] steps from the i-th index. You need to determine if you can reach the end of the array.

  2. Identify Cues:

    • Each element in the array represents your maximum jump length at that position.
    • You need to find out if it’s possible to reach the last index.

  3. Generate Insights:

    • You don’t have to take the maximum jump at each step. Sometimes a smaller jump is beneficial.
    • You should focus on how far you can get at each point, not just where you can get from the current position.
    • The objective is to see if you can reach the end, not to make the fewest jumps.

  4. Identify Key Variables:

    • current_position to keep track of where you are in the array.
    • farthest_reachable to keep track of the farthest index you can reach from the start to the current position.

  5. Algorithm Approach: Greedy

    • At each index, update farthest_reachable.
    • If farthest_reachable is greater than or equal to the last index, return true.
    • If current_position surpasses farthest_reachable, you’re stuck; return false.

Code

Here’s how to implement the solution based on the above insights and thought process:

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def can_jump(nums):
    farthest_reachable = 0
    for i in range(len(nums)):
        if i > farthest_reachable:
            return False
        farthest_reachable = max(farthest_reachable, i + nums[i])
    return farthest_reachable >= len(nums) - 1

# Test the function
print(can_jump([2, 3, 1, 1, 4]))  # Should return True
print(can_jump([3, 2, 1, 0, 4]))  # Should return False

By following these steps and maintaining the loop invariant, the final algorithm correctly solves the problem in an efficient manner.

Establishing Preconditions and Postconditions

1. Parameters:

  • Inputs: The method takes one input, nums, which is a list of integers.
  • Types: The parameter is a list where each element is an integer.
  • Context: Each integer in the list nums represents the maximum jump length from that position.

2. Preconditions:

  • State: The program doesn’t have a predefined state before the method is called.
  • Constraints:
    • The length of the list nums should be at least 1 and at most 10^4.
    • Each integer in nums should be between 0 and 10^5.

3. Method Functionality:

  • Expectation: The method should return a boolean value—True if you can reach the last index of the array, otherwise False.
  • Interaction: It iterates through the input list nums, keeping track of the farthest reachable index.

4. Postconditions:

  • State: The state of the program remains unchanged as the method does not have any side effects.
  • Return Value:
    • Returns True if reaching the last index is possible.
    • Returns False otherwise.

5. Error Handling:

  • Response: The method does not explicitly handle errors but assumes that the input meets the constraints specified in the problem statement.
  • Exception: No exceptions are thrown.

Problem Decomposition

1. Problem Understanding:

  • In My Own Words: Given a list of numbers, each number represents the maximum steps you can jump forward from that position. You start at the first index. The problem is to find out if you can reach the last index by jumping.
  • Key Components:
    1. The list of integers (nums)
    2. The maximum jump length at each index
    3. Starting at the first index
    4. Need to reach the last index

2. Initial Breakdown:

  • Major Parts:
    1. Initialization: Start at the first index and track how far you can jump.
    2. Iteration: Loop through the list, updating the farthest reachable index.
    3. Termination: Determine if the last index is reachable.

3. Subproblem Refinement:

  • For Each Subproblem:
    1. Initialization:
      • Set the initial position to 0.
      • Set the farthest reachable index to 0.
    2. Iteration:
      • Update the farthest reachable index based on the current index and its maximum jump.
    3. Termination:
      • Check if the last index is within the farthest reachable index.

4. Task Identification:

  • Similar Tasks:
    • Updating the farthest reachable index is a repeated operation.

5. Task Abstraction:

  • Clear and Reusable Tasks:
    • Initialization is abstracted enough to only require the starting point and the farthest reachable index.
    • Iteration is also well-abstracted, as it only needs the current index and maximum jump to update the farthest reachable index.

6. Method Naming:

  • Descriptive Names:
    1. Initialization: initialize_position()
    2. Iteration: update_farthest_reachable()
    3. Termination: check_reachability()

7. Subproblem Interactions:

  • Order and Dependencies:
    1. Initialization must happen first.
    2. Iteration follows and is based on the initialized values.
    3. Termination is the last step, relying on the values updated during iteration.

From Brute Force to Optimal Solution

Brute Force Solution:

The brute-force approach to solving the “Can you reach the last index?” problem involves trying out all possible jumps at each index and checking if any of the paths reach the end. Here’s some simple Python3 code for the brute-force solution:

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def can_jump_from_position(pos, nums):
    if pos == len(nums) - 1:
        return True
    
    max_jump = min(pos + nums[pos], len(nums) - 1)
    
    for next_pos in range(pos + 1, max_jump + 1):
        if can_jump_from_position(next_pos, nums):
            return True
            
    return False

def can_jump(nums):
    return can_jump_from_position(0, nums)

Inefficiencies:

  1. Time Complexity: This has a time complexity of (O(2^N)), where (N) is the length of the nums list. This is because, in the worst case, we are visiting every index multiple times.
  2. Space Complexity: It takes (O(N)) space due to the recursion stack in the worst case.

Optimizing the Solution:

Step 1: Memoization

The first step to optimizing this is to use dynamic programming to store the results of subproblems so that we don’t solve them multiple times.

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def can_jump_from_position(pos, nums, memo):
    if pos == len(nums) - 1:
        return True
    
    if memo[pos] != -1:
        return True if memo[pos] == 1 else False
    
    max_jump = min(pos + nums[pos], len(nums) - 1)
    
    for next_pos in range(pos + 1, max_jump + 1):
        if can_jump_from_position(next_pos, nums, memo):
            memo[pos] = 1
            return True
            
    memo[pos] = 0
    return False

def can_jump(nums):
    memo = [-1] * len(nums)
    return can_jump_from_position(0, nums, memo)

Impact on Time and Space Complexity:

  • Time Complexity: (O(N^2))
  • Space Complexity: (O(N))

Step 2: Greedy Approach

We can observe that the farthest we can reach from index i is i + nums[i]. So, we can simply iterate through the list and keep track of the farthest index we can reach.

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def can_jump(nums):
    farthest = 0
    for i in range(len(nums)):
        if i > farthest:
            return False
        farthest = max(farthest, i + nums[i])
    return True

Impact on Time and Space Complexity:

  • Time Complexity: (O(N))
  • Space Complexity: (O(1))

Summary:

We started with a brute-force approach with an unacceptable time complexity of (O(2^N)). We first optimized it to (O(N^2)) using memoization, and then finally to (O(N)) using a greedy approach. The space complexity also improved from (O(N)) to (O(1)).

Code Explanation and Design Decisions

1. Initial Parameters

  • nums: This is an array of integers where each integer at index i represents the maximum jump length from that index. It is the main input and contains the elements that dictate the flow and feasibility of reaching the last index.

2. Primary Loop or Iteration

The primary loop iterates over each index in the nums array, from the first to the last. Each iteration checks how far you can jump from the current index, and whether you can reach the last index.

  • Significance: The loop helps us calculate the farthest we can reach from the current index. By doing so iteratively for each index, we can check if we can reach the end of the array.

3. Conditions or Branches within the Loop

  • if i > farthest: This condition checks if the current index is beyond the farthest index we can reach. If it is, then it’s impossible to reach the end from here.

  • Reasoning: If you’re at an index that is further than the farthest reachable index, then it’s logically impossible to proceed further.

4. Updates or Modifications to Parameters

  • farthest = max(farthest, i + nums[i]): This updates the maximum index you can reach from the current position.

  • Why Necessary: The objective is to reach the last index. By keeping track of the farthest index you can reach at each step, you’re constantly updating the upper limit of your reach. This reflects whether reaching the last index is feasible.

5. Invariant

The farthest variable is an invariant that keeps track of the maximum index that can be reached so far.

  • How it Helps: By maintaining this invariant, we guarantee that as long as the farthest index is greater than or equal to the last index, the end of the array can be reached. It helps to quickly eliminate scenarios where reaching the end is not possible.

6. Significance of the Final Output

The final output is a boolean value (True or False) indicating whether it’s possible to reach the last index from the first index based on the jump values in nums.

  • What it Represents: A True value means reaching the last index is feasible, satisfying the problem’s constraints. A False means it’s impossible to reach the end based on the current state of nums.

The intent of the code is to efficiently determine reachability to the last index by constantly updating and checking the farthest reachable index. This is done while adhering to the problem’s constraints and objectives.

Coding Constructs

1. High-Level Strategies

The high-level strategies used by this code are iteration and state tracking. The code iterates through the array while maintaining a variable (farthest) that keeps track of the farthest index that can be reached.

2. Explanation to a Non-Programmer

Imagine you’re on a series of stepping stones across a river. Each stone has a number written on it. That number tells you how far you can jump forward. This code figures out if you can make it to the last stone or if you’ll end up falling into the river at some point.

3. Logical Elements

  • Iteration: Looping through each element of the array.
  • Conditional Statements: Checking if certain conditions are met.
  • State Update: Modifying the farthest variable based on current conditions.
  • Final Evaluation: Assessing whether reaching the end of the array is possible.

4. Algorithmic Approach in Plain English

Start at the first stepping stone and keep track of the farthest stone you can reach. As you go from one stone to the next, update this ‘farthest reachable stone’ based on the number written on your current stone. If you ever find yourself on a stone that’s beyond this ‘farthest reachable stone’, you know you’ll fall into the river. Otherwise, you can make it to the last stone.

5. Key Steps or Operations

  • Initialize farthest: Set it to zero initially.
  • Iterate through nums: For each number (i.e., stepping stone), calculate how far you could jump from that stone.
  • Update farthest: Use the current stone’s number to update the ‘farthest reachable stone’ if it lets you reach further.
  • Check Conditions: If you’re on a stone that is beyond the ‘farthest reachable stone’, stop and declare that reaching the end is impossible.

6. Algorithmic Patterns

  • Greedy Algorithm: The code is always updating the farthest based on the best option available at the current stone, without worrying about future stones.
  • Traversal: Iterating through the array to examine each element.
  • State Tracking: Keeping track of important information (farthest) as we traverse the array.

Language Agnostic Coding Drills

  1. Variable Definitions: Understanding how to define variables and assign them values.

  2. Lists and List Indexing: Lists are a basic data structure in Python that hold an ordered collection of items, which can be of any type. Indexing is used to access the individual items of the list.

  3. Loops: Understanding how to use a loop (for loop in this case) to iterate over a sequence of elements.

  4. Conditional Statements: Using if statements to perform different actions based on certain conditions.

  5. Functions: Declaring and invoking functions. In this case, a method canJump is defined within a class Solution.

  6. Boolean Operations: Understanding how to use boolean values (True and False) in logic statements.

  7. Classes: Defining classes and methods in Python.

Targeted Drills in Python

  1. Variable Definitions:
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x = 10
print(x)  # Output: 10
  1. Lists and List Indexing:
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numbers = [1, 2, 3, 4, 5]
print(numbers[1])  # Output: 2
  1. Loops:
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for i in range(5):
    print(i)  # Output: 0 1 2 3 4
  1. Conditional Statements:
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x = 10
if x > 0:
    print("Positive")  # Output: Positive
else:
    print("Non-positive")
  1. Functions:
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def greet():
    return "Hello, World!"

print(greet())  # Output: Hello, World!
  1. Boolean Operations:
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x = True
y = False
print(x and y)  # Output: False
  1. Classes:
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class Test:
    def say_hello(self):
        return "Hello, World!"

test = Test()
print(test.say_hello())  # Output: Hello, World!

Q&A

Similar Problems

Here are 10 problems that use similar underlying concepts:

  1. Jump Game II

    • Similarity: Requires you to find the minimum number of jumps to reach the end, using a similar greedy approach.

  2. Climbing Stairs

    • Similarity: Also a traversal problem where you have limited steps to reach a goal, although simpler than Jump Game.

  3. Maximal Square

    • Similarity: Involves traversal and state tracking to find the largest square of 1’s in a 2D grid.

  4. Gas Station

    • Similarity: Another circular traversal problem where you need to find if you can complete the loop based on the given conditions.

  5. Coin Change

    • Similarity: Uses a similar greedy approach to make a certain value with the least amount of coins.

  6. Maximum Subarray

    • Similarity: Involves traversing an array while maintaining state (max sum so far) and updating it based on new elements.

  7. Trapping Rain Water

    • Similarity: Traverses an array to find trapped water, requiring state tracking of highest bars on both sides.

  8. House Robber

    • Similarity: You have to decide at each step (house) whether robbing it will maximize your loot, somewhat similar to deciding whether jumping is optimal.

  9. Longest Increasing Subsequence

    • Similarity: Involves traversing an array and tracking state (longest subsequence so far) to make optimal decisions.

  10. Best Time to Buy and Sell Stock

    • Similarity: As you traverse the array, you keep track of the minimum price so far to decide the maximum profit you can make, similar to how farthest is updated in Jump Game.

Each of these problems involves traversal, state tracking, and making optimal decisions based on current and past information, similar to what’s required in Jump Game.