Count All Possible Routes

The problem can be solved using Dynamic Programming with Memoization. We will create a recursive function to explore all possible paths from the starting city to the finishing city and count the number of valid paths that consume exactly the amount of fuel provided.

Here’s the code:

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MOD = 10**9 + 7

class Solution:
    def countRoutes(self, locations: List[int], start: int, finish: int, fuel: int) -> int:
        n = len(locations)
        # Memoization table to store results for previously computed states
        memo = [[-1 for _ in range(fuel + 1)] for _ in range(n)]

        def dfs(city: int, remaining_fuel: int) -> int:
            # Base case: if we reach the finish city, we have one valid path
            if city == finish:
                result = 1
            else:
                result = 0

            # If this state has been computed before, return the memoized result
            if memo[city][remaining_fuel] != -1:
                return memo[city][remaining_fuel]

            # Explore all other cities
            for next_city in range(n):
                if next_city == city:
                    continue
                cost = abs(locations[city] - locations[next_city])
                if remaining_fuel >= cost:
                    # Recursively explore the next city if we have enough fuel
                    result += dfs(next_city, remaining_fuel - cost)
                    result %= MOD

            # Store the computed result in the memoization table
            memo[city][remaining_fuel] = result
            return result

        return dfs(start, fuel)

This code counts all possible routes from start to finish with the given amount of fuel. It uses a recursive function dfs to explore all possible paths and count the number of valid ones.

The memoization table memo is used to store previously computed results, and this greatly reduces redundant computations.

The time complexity of this solution is ( O(n \cdot n \cdot \text{{fuel}}) ), where ( n ) is the length of the locations array, and the space complexity is ( O(n \cdot \text{{fuel}}) ). It works within the constraints mentioned in the problem statement.

Identifying Problem Isomorphism

“Count All Possible Routes” can be approximately mapped to “Unique Paths II”.

Reasoning:

The main concept in both problems is finding the number of possible paths or routes.

In “Count All Possible Routes”, you’re given an array of fuel stations and the task is to count all possible routes from one station to another within a fuel limit.

In “Unique Paths II”, the task is to find the number of unique paths from the top-left corner to the bottom-right corner of a m x n grid (2D array) where some cells are blocked.

Though these problems seem different on the surface, their core is about finding the number of distinct paths under constraints, hence they are approximately isomorphic.

“Unique Paths II” is simpler since it involves a straightforward dynamic programming solution. “Count All Possible Routes” is more complex as it involves additional constraints such as fuel limit and moving in any direction, not just right or down.

10 Prerequisite LeetCode Problems

“1575. Count All Possible Routes” is a dynamic programming problem that requires knowledge of recursive relations, memoization, and state management. Here are 10 problems to prepare for it:

  1. 70. Climbing Stairs: This is a simple dynamic programming problem that introduces the concept of state transition.

  2. 322. Coin Change: This problem requires dynamic programming to find the optimal solution, similar to your target problem.

  3. 139. Word Break: This problem is about exploring all possible ways of partitioning a sequence, similar to your target problem.

  4. 416. Partition Equal Subset Sum: This problem also involves exploring all possibilities with constraints, which is similar to your target problem.

  5. 494. Target Sum: This problem requires both dynamic programming and handling of different states.

  6. 518. Coin Change 2: This problem is about counting the number of ways to make change, similar to your target problem which counts routes.

  7. 688. Knight Probability in Chessboard: This problem also involves state transition probabilities in a grid, similar to your problem’s routing issue.

  8. 935. Knight Dialer: This problem is another one where you need to count valid sequences under constraints, similar to your target problem.

  9. 983. Minimum Cost For Tickets: This problem also involves minimizing costs under constraints using dynamic programming, similar to your target problem.

  10. 1477. Find Two Non-overlapping Sub-arrays Each With Target Sum: This problem is about managing and optimizing overlapping subproblems, which is a key aspect of dynamic programming.

Problem Classification

The problem falls under the Computational Geometry and Graph Theory domains. The positions of cities can be interpreted as vertices in a graph, and the paths between them can be considered as edges.

What Components

  1. Array of Distinct Positive Integers (locations): Represents the positions of the cities.
  2. Integers start, finish, fuel: Represent the starting city index, ending city index, and the initial amount of fuel, respectively.
  3. Fuel Cost: Movement from one city to another decreases the fuel by the absolute difference of their locations.
  4. Return Value: The count of all possible routes from start to finish modulo (10^9 + 7).

This problem is a classic example of a combinatorial search problem with constraints. It can be considered as a “Counting Paths in a Graph” problem with the additional constraint of fuel.

  1. Search Problem: You’re searching for all possible routes from start to finish.
  2. Constraint-based: The fuel serves as a constraint that limits the paths you can take.
  3. Optimization: Although not a direct optimization problem, you are essentially looking for all viable paths, which could be considered a form of exhaustive optimization.

The problem involves dealing with a large search space while adhering to constraints, making it a typical combinatorial problem in graph theory.

Clarification Questions

  1. Multiple Visits: Can the same city be visited more than once on a single route?
  2. Fuel Replenishment: Is there any way to replenish fuel at some cities or is the initial fuel all you have for the entire route?
  3. Order of Locations Array: Does the order of the cities in the locations array have any significance?
  4. Negative Fuel: What should be done if the fuel goes negative during computation, or is it guaranteed not to happen based on problem constraints?
  5. Edge Weights: Are the edges between the cities weighted by the absolute difference in their locations, or is there another metric?
  6. Modulo Operation: Is the modulo operation (10^9 + 7) to be applied only at the end, or should it be applied during each step of the calculation?
  7. Route Definition: Does moving from city i to city j and then back to city i count as one route or two different routes?
  8. Output Type: Should the output be an integer or can it be in another numeric format?
  9. Start Equals Finish: If the start and finish are the same, what should be the output?
  10. Unreachable Cities: How should routes leading to cities that cannot be reached with the given fuel be handled?

Asking these clarification questions can help in removing any ambiguities and will make problem-solving more straightforward.

Distilling the Problem to Its Core Elements

  1. Fundamental Concept: The fundamental concept here is Dynamic Programming combined with Graph Theory. We have cities (nodes) and need to find all possible paths from the start city to the end city while keeping track of the remaining fuel (state).

  2. Simplest Description: Imagine you’re driving a car from one city to another. You start with a certain amount of fuel. Each city is at a different distance from each other. How many different ways can you drive from your starting city to your destination without running out of fuel?

  3. Core Problem: Count the number of unique routes from a starting city to an ending city given a certain amount of initial fuel. Each move between cities uses fuel equal to the absolute difference in their positions.

  4. Key Components:

    • locations[]: An array of distinct integers representing the cities.
    • start: Index of the starting city in locations[].
    • finish: Index of the destination city in locations[].
    • fuel: Initial amount of fuel.

  5. Minimal Set of Operations:

    • Initialize a way to keep track of visited routes and their fuel consumption.
    • Start from the start city.
    • Traverse to each neighboring city, consuming fuel based on the distance.
    • If a city is reached with negative fuel, discard that route.
    • If the finish city is reached, add that route to the count.
    • Apply Dynamic Programming to avoid redundant calculations.

By understanding these aspects, you’ll have a clearer roadmap for tackling this problem.

Visual Model of the Problem

To visualize this problem, consider the following steps:

  1. Cities as Nodes: Each city is represented as a node in a graph. Label them based on their index in the locations array.

  2. Edges and Weights: Create edges between every pair of nodes (cities). The weight on the edge represents the fuel needed to go from one city to another, which is the absolute difference in their positions.

  3. Start and Finish: Mark the starting city and the destination city on the graph.

  4. Fuel Indicator: Imagine a fuel gauge next to your starting node that starts with the initial fuel amount. This gauge depletes as you move along edges.

  5. Paths: Think of lines connecting nodes as possible paths you can take. You must consider all possible paths from start to finish that are achievable with your initial fuel.

  6. State Tracking: Each time you move to a city, you need to keep track of the remaining fuel. This can be visualized as arriving at the same city node but with a different fuel gauge level.

  7. Counting Routes: Each time you reach the destination city with non-negative fuel, count that as a valid path.

By visualizing the problem in this way, you can better understand the constraints and what you’re trying to achieve, making it easier to formulate a solution.

Problem Restatement

You have a list of cities, each at a unique position. You start at one city and aim to reach another, both specified by their indices in the list. You have a certain amount of fuel to start with. You can travel between any two cities, but the fuel cost is the absolute difference of their positions. You can visit any city multiple times, including the start and finish cities. Your goal is to find out how many different routes you can take from the start to the finish city without running out of fuel. Return this number modulo 1,000,000,007.

Key Constraints:

  • The list of cities contains 2 to 100 elements.
  • City positions range from 1 to 1,000,000,000 and are distinct.
  • The initial fuel amount is between 1 and 200.

The core requirement is to count all possible routes from start to finish given the fuel constraint.

Abstract Representation of the Problem

This problem can be abstracted as a graph traversal problem. Each city can be seen as a node in a graph, and the edges between the nodes are weighted by the absolute difference between the positions of the connected cities. The graph is fully connected because you can travel from any city to any other city.

The challenge is to find all unique paths from a starting node to an ending node with a constraint on the total edge weight (fuel). Paths can visit nodes multiple times, and the goal is to count these paths. Finally, the count needs to be returned modulo 1,000,000,007.

Key Elements:

  • Nodes (cities)
  • Weighted edges (fuel cost)
  • Starting node
  • Ending node
  • Weight constraint (total available fuel)

The problem can thus be viewed as a constrained graph traversal problem where the traversal can be cyclic or acyclic.

Terminology

  1. Graph Traversal: The process of visiting all the nodes in a graph in a specific order. In this problem, you are counting the number of unique paths from the starting city to the ending city.

  2. Node: A point in the graph. In this context, a node represents a city.

  3. Edge: A connection between two nodes. Here, an edge exists between any two cities.

  4. Weighted Edge: An edge that has a numerical value (weight) associated with it. In this problem, the weight of an edge represents the fuel needed to travel between two cities.

  5. Modulo Operation: The remainder when one number is divided by another. Here, you need to return the count of all possible routes modulo (10^9 + 7).

  6. Constraints: Conditions that must be satisfied for the problem to be valid. Here, the constraints are the available fuel and the requirement that fuel cannot be negative.

  7. Cyclic and Acyclic Paths: Paths that either loop back to a previously visited node (cyclic) or don’t (acyclic). In this problem, both types of paths are allowed.

Understanding these terms and their roles will help you get a clearer picture of what you’re trying to achieve in this problem.

Problem Simplification and Explanation

The problem asks you to find all the ways to travel from one city to another with a limited amount of fuel. Each city is a point on a map, and the distance between any two cities consumes fuel.

Key Concepts:

  1. Cities: These are the places you can visit. Think of them as stops on a subway map.

  2. Fuel: The resource you use to move between cities. Think of this as the number of subway tokens you have.

  3. Start and Finish: Your starting point and destination on this subway map.

  4. Paths: These are the routes you can take from the start to the finish. You want to find all possible routes.

Metaphor:

Imagine you are in a subway system where you need tokens to travel between stations. You start with a set number of tokens (fuel). Each line between stations requires a different number of tokens, and you can go back and forth between stations as long as you have enough tokens. Your task is to find all the possible ways to get from your starting station to your destination.

Interaction:

  1. You begin at the ‘start’ city (starting station).
  2. You have to end at the ‘finish’ city (destination station).
  3. You use ‘fuel’ (tokens) to move between cities (stations).
  4. You can visit a city more than once, and you can also use a path between two cities multiple times as long as you have enough fuel.

Your goal is to count all the different paths from the start to finish without running out of fuel.

Constraints

Let’s consider specific characteristics and constraints that can be helpful:

  1. Distinct Locations: All the cities have distinct positions, which means each city can be uniquely identified by its location. This simplifies the problem by eliminating the need to consider duplicate distances.

  2. Small Size of Array: The maximum number of locations is 100. The limited size can allow us to consider algorithms that may not be efficient for larger datasets.

  3. Fuel Limit: The maximum fuel is 200. Knowing this upper limit can help us eliminate many infeasible paths quickly.

  4. Start and Finish Points: These are fixed locations where the path must begin and end, which narrows down the search considerably.

  5. Modulo Arithmetic: The answer is returned modulo (10^9 + 7). This characteristic can be used to keep the numbers small and within the data type limits.

  6. Non-negative Fuel: Fuel can’t go negative, a useful termination condition for any recursive or iterative process you might use to find the paths.

By examining these characteristics, we can see that the problem might be suitable for a dynamic programming approach, where we can keep track of the number of ways to reach a given city with a certain amount of fuel left. This allows us to build upon previous computations, thus improving efficiency.

Key insights from analyzing the constraints are:

  1. Limited Search Space: With a maximum of 100 cities and 200 units of fuel, the search space is relatively limited. This constraint opens the door to solutions that may involve iterating through the available options without worrying about computational inefficiency too much.

  2. Distinct Locations: Since all locations are distinct, we don’t have to worry about handling duplicates, which simplifies the problem-solving process.

  3. Fixed End Points: Having a fixed start and finish city narrows down the search space. It lets us focus only on paths that start and end at these points.

  4. Modulo Arithmetic: Knowing that the answer must be given modulo (10^9 + 7) helps us in keeping the numbers manageable and prevents overflow.

  5. Non-negative Fuel: This can serve as a termination condition in our algorithm, reducing computational load by stopping the search when the fuel becomes zero or negative.

  6. Fuel as a Resource: The fuel is a bounded resource, which means algorithms like Dynamic Programming could exploit this limitation for more efficient solutions.

These constraints and characteristics provide valuable hints toward choosing an appropriate algorithm to solve the problem efficiently.

Case Analysis

Minimal Case

  1. Input: locations = [1, 2], start = 0, finish = 1, fuel = 1
    Output: 1
    Analysis: This case tests the minimum constraints of the problem. You can go directly from start to finish using 1 unit of fuel.

High Fuel, Short Distance

  1. Input: locations = [1, 3], start = 0, finish = 1, fuel = 5
    Output: 1
    Analysis: Even with more fuel, you can’t use it to create multiple paths since there are only two cities. This tests whether the algorithm correctly ignores excess fuel.

Multiple Paths, Adequate Fuel

  1. Input: locations = [1, 5, 8], start = 0, finish = 2, fuel = 7
    Output: 2
    Analysis: There are multiple paths: 0->1->2 and 0->2. This tests if the algorithm explores all valid paths.

Multiple Paths, Scarce Fuel

  1. Input: locations = [1, 5, 8], start = 0, finish = 2, fuel = 6
    Output: 0
    Analysis: Although there are multiple cities, none of the paths have adequate fuel to reach the destination. This tests if the algorithm correctly returns 0 in such cases.

Circular Paths

  1. Input: locations = [1, 2, 3], start = 0, finish = 2, fuel = 5
    Output: 2
    Analysis: There are multiple ways to get to the destination: 0->1->2 and 0->2. But fuel isn’t sufficient for circular paths like 0->1->0->2. This tests if the algorithm avoids unnecessary loops.

Same Start and Finish

  1. Input: locations = [1, 3], start = 0, finish = 0, fuel = 2
    Output: 1
    Analysis: Here, the start and finish are the same. It tests if the algorithm considers staying at the same city as a valid path.

Large Numbers

  1. Input: locations = [1, 1000000000], start = 0, finish = 1, fuel = 999999999
    Output: 0
    Analysis: This tests whether the algorithm can handle large numbers efficiently and whether it’s checking the constraint for the fuel limit.

Each of these test cases tests a different aspect of the problem and should provide key insights into whether your algorithm is both correct and robust.

  1. Fuel Conservation: From the “High Fuel, Short Distance” and “Multiple Paths, Scarce Fuel” cases, we learn that fuel management is crucial. Extra fuel doesn’t create new paths if the topology doesn’t allow for it. Similarly, if fuel is insufficient, the algorithm should correctly identify that no paths are possible.

  2. Path Exploration: The “Multiple Paths, Adequate Fuel” case emphasizes the need for an algorithm that can explore all possible paths to the destination. This is crucial for getting the correct count of routes.

  3. Circular Paths: The “Circular Paths” case highlights the need to avoid unnecessary loops, which is important for optimization. If the algorithm doesn’t account for this, it could run into infinite loops or incorrect counts.

  4. Edge Conditions: The “Minimal Case” and “Same Start and Finish” cases point out that the algorithm must correctly handle edge conditions like the smallest possible inputs or when the start and finish are the same location.

  5. Large Numbers: The “Large Numbers” case tests the efficiency of the algorithm. It should be capable of handling large numbers within the constraints.

  6. Path Termination: In all cases, the algorithm must know when to terminate a path, either because it has reached the destination or because it has run out of fuel.

By understanding these key insights, you can aim for an algorithm that is both efficient and accurate, accounting for a variety of scenarios.

Identification of Applicable Theoretical Concepts

  1. Dynamic Programming: Given the repeated subproblems of exploring each city with a specific amount of fuel, dynamic programming can be effectively used to store these intermediate results. This would avoid redundant calculations and make the algorithm more efficient.

  2. Memoization: Storing the number of paths possible from a given city with a certain amount of fuel left can speed up the process considerably. This is a form of caching or memoization.

  3. Graph Theory: This problem can be modeled as a weighted, directed graph, where each city is a node and the weight of an edge between two nodes is the fuel cost of going from one city to the other. Algorithms for traversing weighted graphs could be adapted for this problem.

  4. Depth-First Search (DFS): The problem of counting all possible routes can be considered a variant of depth-first search on a graph. DFS can be adapted to work with the fuel constraint.

  5. Modulo Arithmetic: The problem asks for the result modulo (10^9 + 7). Mathematical properties of modulo can be exploited for efficient calculations.

  6. Pruning: Some routes can be quickly disregarded if you see that the remaining fuel is not sufficient to reach the destination. This is a form of pruning in a search tree and can save computational effort.

  7. Recursion: The nature of the problem lends itself to a recursive solution, where the problem for ’n’ cities can be broken down into smaller sub-problems.

By identifying and applying these concepts or properties, the problem becomes more manageable and can be tackled more efficiently.

Problem Breakdown and Solution Methodology

Step 1: Understand the Problem Landscape

The first step is to clearly understand what’s given and what’s required. You have an array of distinct city locations and initial fuel. You start from one city and aim to reach another. You can travel between cities, but it consumes fuel equal to the absolute difference of their positions. Think of it like a map with scattered points, where each point is a city and the distance between them is the fuel needed to go from one to the other.

Step 2: Abstract Modeling

Model the cities as nodes in a graph, where an edge between two nodes ‘i’ and ‘j’ is weighted with abs(locations[i] - locations[j]), the amount of fuel needed to travel between these two cities.

Step 3: Initialize Data Structures

Prepare a 2D memoization array (or a hash table with two keys), where dp[i][f] stores the number of ways to reach city i from the starting city with f units of fuel left.

Step 4: Dynamic Programming with DFS

  1. Initialize the DFS function that takes the current city and the remaining fuel as arguments.
  2. If you reach the destination city, increment the count by 1.
  3. Check if this state (city and remaining fuel) has already been computed. If so, return the memoized value.
  4. Loop through all other cities. If you can go from the current city to another city (i.e., enough fuel), call the DFS function for that city with the updated remaining fuel.
  5. Memoize the result and return it.

Step 5: Post-Processing

Make sure to return the result modulo (10^9 + 7) as stated in the problem description.

Effects of Parameter Changes

  1. Increasing the number of cities will increase the computational time, although dynamic programming should mitigate this.
  2. A higher initial fuel will make more paths possible, increasing the result but also requiring more computations.

Working Example

Consider locations = [2,3,6,8,4], start = 1, finish = 3, fuel = 5.

  1. Start from city 1 (location 3) with fuel 5.
  2. Possible next cities: 2, 6, 8, 4
  3. Go to city 2. Remaining fuel = 5 - |3 - 2| = 4.
    • From city 2, can go to 3 (destination) with remaining fuel 4 - |2 - 8| = 0.
  4. Go to city 6. Remaining fuel = 5 - |3 - 6| = 2.
    • From city 6, can’t go anywhere.
  5. Go to city 8. Remaining fuel = 5 - |3 - 8| = 0.
    • Can’t go anywhere.
  6. Go to city 4. Remaining fuel = 5 - |3 - 4| = 4.
    • From city 4, can go to 3 (destination) with remaining fuel 4 - |4 - 8| = 0.

You’ll find 4 paths: 1 -> 3, 1 -> 2 -> 3, 1 -> 4 -> 3, 1 -> 4 -> 2 -> 3. Answer = 4.

Inference of Problem-Solving Approach from the Problem Statement

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

Stepwise Refinement

High-Level Steps

  1. Initialize data structures
  2. Set up dynamic programming with DFS
  3. Compute final result

Granular, Actionable Steps

Initialize Data Structures
  1. Create a 2D memoization array dp where dp[i][f] will store the number of ways to reach city i with f units of fuel remaining.
Set Up Dynamic Programming with DFS
  1. Write a DFS function dfs(current_city, remaining_fuel) that takes the current city and remaining fuel as arguments.
  2. Inside dfs, check if we have already calculated the number of ways to reach current_city with remaining_fuel. If so, return the memoized result.
  3. If current_city is the destination, return 1 as one path is completed.
  4. Initialize a local variable count to 0.
  5. Loop through all cities. If we can move to city j with remaining_fuel, then:
    • Update remaining fuel: new_fuel = remaining_fuel - abs(locations[current_city] - locations[j]).
    • Recursively call dfs(j, new_fuel).
    • Add the return value to count.
  6. Store count in dp[current_city][remaining_fuel].
  7. Return count.
Compute Final Result
  1. Call dfs(start, fuel) and store its return value as the answer.
  2. Return the answer modulo (10^9 + 7).

Independent Parts of the Problem

  • Each DFS state (current_city, remaining_fuel) can be computed independently. However, the overall solution depends on these individual computations being correctly memoized for future use.

Repeatable Patterns

  • The core pattern here is the dynamic programming paradigm coupled with Depth First Search. The dfs function is a recurring operation, making the recursive calls to calculate the number of ways for each state.
  • Memoization serves as another pattern to avoid recomputation for the same states.

By following this refined step-by-step approach, you’ll be able to construct a solution that efficiently solves the problem.

Solution Approach and Analysis

Detailed Approach to Solving the Problem

Overall Solution Metaphor

Consider the problem as a board game. You start at a city (starting point on the board), and you have a finite amount of “fuel” or moves. Your aim is to reach another specific city (the endpoint on the board). The distance between the cities takes up a certain amount of your moves. The game has various routes to reach the endpoint, and you want to find out how many different ways you can win this game without running out of moves.

Step-by-Step Approach

Step 1: Initialize Data Structures
  • Create a 2D memoization array, dp[i][f], where i is the current city and f is the remaining fuel.
Step 2: Set up Dynamic Programming with DFS
  • Write a function dfs(current_city, remaining_fuel) that will serve as the backbone of the solution.
Inside dfs()
  1. Memoization Check: First, check if this state (current_city, remaining_fuel) is already calculated. If yes, return the calculated value.
  2. Base Case: If the current_city is the finish city, return 1 because we’ve found a path.
  3. Initialization: Create a variable count to store the number of routes from the current state.
  4. Transition: Loop through each city j, calculate the fuel needed to get there, and then recursively call dfs(j, new_fuel). Add this to count.
Step 3: Calculate Final Result
  • Call dfs(start, fuel) to get the final number of ways. Return this number modulo (10^9 + 7).

Changes in Parameters

  • More fuel allows more possible routes.
  • Adding more cities makes the problem computationally more expensive.

Example Case: Working Through Steps

Let’s take the example: locations = [2,3,6,8,4], start = 1, finish = 3, fuel = 5

  1. Initialize: dp array of size 5x6 (5 cities and fuel from 0 to 5).
  2. DFS call: Call dfs(1, 5).
    • Memoization: Nothing yet.
    • Base Case: Not at city 3, continue.
    • Initialization: count = 0
    • Transition: Check moves to all cities.
      • City 0: Cost = 1, New fuel = 4, dfs(0, 4)
      • City 2: Cost = 3, New fuel = 2, dfs(2, 2)
      • City 3: Cost = 2, New fuel = 3, dfs(3, 3) -> Returns 1 (Base case)
      • City 4: Cost = 1, New fuel = 4, dfs(4, 4)
    • Sum up all these, count = 4.
  3. Final Result: Return 4.

This approach ensures we cover all possible paths, without redundant calculations, thanks to memoization.

Thought Process

Thought Process

Cues from the Problem Statement

  1. Distinct Positive Integers: Locations are distinct and positive, so no cycles or negative numbers to worry about.
  2. Fuel Constraints: Fuel cannot go negative. This sets a lower bound to our movements.
  3. Return Count Modulo (10^9 + 7): The result can be very large, so we need modulo arithmetic.
  4. You Can Visit Any City More Than Once: This hints at a Dynamic Programming solution with memoization to avoid redundant calculations.

Direction for Approach

The cues suggest that we might want to look into Dynamic Programming (DP) based on states to solve this problem efficiently. Specifically, DP with memoization to count the number of ways to go from start to finish with given fuel.

Insights

  1. State-Based Approach: A state in this problem can be defined by the current city and the remaining fuel. Both are crucial to determining if we can reach the destination city with the remaining fuel.
  2. Base Case: Reaching the destination city counts as one path. Hence, when the current city is the destination city, return 1.
  3. Memoization: To avoid recalculating, we need a memoization table to store the number of ways to reach the destination from each state (city, remaining fuel).

Python3 Code

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MOD = 10**9 + 7

def countRoutes(locations, start, finish, fuel):
    memo = {}
    
    def dfs(curr_city, remaining_fuel):
        # Using the current city and remaining fuel as the state
        if (curr_city, remaining_fuel) in memo:
            return memo[(curr_city, remaining_fuel)]
        
        # Base case: if we're at the destination city
        if curr_city == finish:
            count = 1
        else:
            count = 0
        
        # Try moving to every other city
        for next_city, next_location in enumerate(locations):
            if next_city == curr_city:
                continue
            
            cost = abs(locations[curr_city] - next_location)
            if remaining_fuel >= cost:
                count += dfs(next_city, remaining_fuel - cost)
                
        memo[(curr_city, remaining_fuel)] = count % MOD
        return memo[(curr_city, remaining_fuel)]
    
    return dfs(start, fuel)

# Test the function
print(countRoutes([2, 3, 6, 8, 4], 1, 3, 5))  # Output should be 4

This code should provide the correct number of routes from start to finish, following the problem constraints. It uses dynamic programming with memoization to efficiently solve the problem.

From Brute Force to Optimal Solution

Brute-Force Approach

Thought Process

  1. Starting Point: Begin at the start city.
  2. Traversal: Visit every possible city from the current city as long as we have enough fuel.
  3. Recursion: For each visited city, call the function recursively until we reach the destination or run out of fuel.
  4. Counting Paths: Each time we reach the destination, increment a global counter.

Python3 Code (Brute-Force)

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count = 0

def bruteForce(locations, curr_city, finish, fuel):
    global count
    if fuel < 0:
        return
    if curr_city == finish:
        count += 1

    for next_city, next_location in enumerate(locations):
        if next_city == curr_city:
            continue
        cost = abs(locations[curr_city] - next_location)
        bruteForce(locations, next_city, finish, fuel - cost)

# Testing
count = 0
bruteForce([2, 3, 6, 8, 4], 1, 3, 5)
print(count)  # Output should be 4

Inefficiencies

  1. Time Complexity: (O(n!)) - We could end up traversing every possible permutation of paths.
  2. Repeated Work: The function revisits the same cities with the same remaining fuel multiple times.

Optimized Approach

Steps Towards Optimization

  1. Dynamic Programming (DP): Use DP to store the number of ways to reach the destination from each state (city, remaining fuel).
  2. Memoization: Use a dictionary to store the result for each (city, remaining fuel) pair to avoid recomputation.

Python Code (Optimized)

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MOD = 10**9 + 7

def countRoutes(locations, start, finish, fuel):
    memo = {}

    def dfs(curr_city, remaining_fuel):
        if (curr_city, remaining_fuel) in memo:
            return memo[(curr_city, remaining_fuel)]

        if curr_city == finish:
            count = 1
        else:
            count = 0

        for next_city, next_location in enumerate(locations):
            if next_city == curr_city:
                continue

            cost = abs(locations[curr_city] - next_location)
            if remaining_fuel >= cost:
                count += dfs(next_city, remaining_fuel - cost)

        memo[(curr_city, remaining_fuel)] = count % MOD
        return memo[(curr_city, remaining_fuel)]

    return dfs(start, fuel)

print(countRoutes([2, 3, 6, 8, 4], 1, 3, 5))  # Output should be 4

Impact on Time and Space Complexity

  1. Time Complexity: Reduced to (O(n \times \text{fuel})) from (O(n!)). This is because we have (n) cities and (O(\text{fuel})) remaining fuel states.
  2. Space Complexity: Also (O(n \times \text{fuel})) for storing the memoized results.

In conclusion, the optimized approach is far more efficient in terms of both time and space complexity, making it practical for larger inputs. It uses dynamic programming with memoization to eliminate redundant calculations.

Code Explanation and Design Decisions

  1. Initial Parameters:

    • locations: Represents the positions of cities. It helps to calculate fuel cost between cities.
    • start: The city where the journey starts.
    • finish: The destination city.
    • fuel: The initial amount of fuel available.
    • memo: Stores previously computed counts to avoid re-calculations.

  2. Primary Loop:

    • The loop iterates over all cities from the current city.
    • Each iteration signifies a potential move to a new city.
    • The loop is advancing the solution by exploring all possible paths to the destination city from the current city.

  3. Conditions within Loop:

    • if next_city == curr_city: continue ensures that we do not move to the same city.
    • if remaining_fuel >= cost: checks if there’s enough fuel to move to the next city.
    • These conditions are based on the problem’s requirements: we can’t stay in the same city and fuel should not go negative.

  4. Parameter Updates:

    • count += dfs(next_city, remaining_fuel - cost): Adds the count of valid paths from the next city to the destination.
    • memo[(curr_city, remaining_fuel)] = count % MOD: Stores the result to avoid recomputation.
    • These updates are necessary to advance the state of the solution and to meet the constraints on fuel and modulo operation.

  5. Invariant:

    • The count for each (city, remaining_fuel) pair remains constant once computed. This helps in memoization and ensures the solution is consistent with the problem’s requirements.

  6. Final Output:

    • The final output is the total number of valid paths from the start to finish city, modulo (10^9 + 7).
    • It satisfies the problem’s requirement by counting all such valid paths under given constraints.

The intent of the code is to methodically explore all routes from start to finish, while intelligently skipping over routes that have already been calculated. This optimizes time and ensures we meet all constraints and requirements of the problem.

Coding Constructs

  1. High-Level Strategies:

    • Depth-First Search (DFS) for exploring all possible routes.
    • Memoization to store and reuse previously computed results.

  2. Purpose for a Non-Programmer:

    • This code is like a smart GPS. It finds all possible routes from one city to another, given a limited amount of fuel, and does so as quickly as possible by remembering routes it has already checked.

  3. Logical Elements:

    • Conditional Statements: To check fuel limits and prevent revisiting the same city.
    • Recursion: For exploring all possible paths.
    • Looping: To go through all cities from the current city.
    • Caching/Memoization: To remember previous results.

  4. Algorithmic Approach in Plain English:

    • Start at the initial city.
    • Check every other city you could travel to next.
    • If you have enough fuel, go to that city and repeat the process.
    • Remember the routes you’ve found so that you don’t have to check them again.
    • Count all possible routes that get you to the final city with fuel left.

  5. Key Steps or Operations:

    • Calculate the fuel cost for moving from the current city to the next.
    • Use DFS to explore if the route is valid (has enough fuel).
    • Store the counts of valid paths for each city with remaining fuel in the memo dictionary.
    • Sum up the counts to get the final answer.

  6. Algorithmic Patterns:

    • Depth-First Search: For traversing through possible routes.
    • Memoization: For caching previously computed results to avoid redundant calculations.

Language Agnostic Coding Drills

  1. Distinct Coding Concepts:

    1. Variable Initialization: Establishing the basic variables like arrays or dictionaries that will store data.

    2. Looping: Iterating over data structures to perform operations.

    3. Conditional Statements: Implementing ‘if’ conditions to make decisions based on variable states.

    4. Basic Recursion: Using recursion to solve a problem in smaller chunks.

    5. Parameter Passing: Passing parameters like current state and memoized data in recursive calls.

    6. Memoization: Storing previously computed results to avoid redundant calculations.

    7. Depth-First Search (DFS): A specialized form of recursion to traverse all possible paths.

    8. Modular Arithmetic: Applying modulo operations to ensure numbers don’t exceed a certain limit.

  2. List by Increasing Difficulty:

    • Variable Initialization: Easy; foundational for any code.
    • Looping: Easy; basic control structure.
    • Conditional Statements: Easy; required for logical flow.
    • Parameter Passing: Moderate; understanding scope and state is required.
    • Basic Recursion: Moderate; recursion can be conceptually challenging.
    • Depth-First Search (DFS): Moderate; it’s an extension of basic recursion.
    • Memoization: Hard; requires understanding of both recursion and data structures.
    • Modular Arithmetic: Hard; involves mathematical concepts.

  3. Problem-Solving Approach:

    • Understand the Problem: Read and comprehend the problem’s requirements, constraints, and goal.

    • Initialize Variables: Set up any necessary data structures like arrays or dictionaries for storing information. This corresponds to the ‘Variable Initialization’ concept.

    • Start Looping: Use loops to iterate through cities to explore the next possible steps. This involves the ‘Looping’ concept.

    • Apply Conditions: Use conditional statements to filter out invalid paths based on the remaining fuel. Here, ‘Conditional Statements’ come into play.

    • Basic Recursion: Start with basic recursion to explore the problem space. For each city, you’ll want to explore other reachable cities recursively.

    • DFS Extension: Extend the recursion to a full DFS algorithm that will explore all possible paths from the start city to the end city. This incorporates the ‘Depth-First Search’ concept.

    • Parameter Passing: While doing recursion or DFS, pass the necessary state parameters like remaining fuel and memoized data.

    • Memoization: To make the solution more efficient, store the number of valid paths for each state in a memoization data structure. This saves time and avoids redundant calculations, integrating the ‘Memoization’ concept.

    • Calculate Output: Accumulate the counts of valid paths and apply modular arithmetic to keep the numbers within bounds.

    • Compile and Run: After all these steps, you’d have a solution that should meet the problem’s constraints and provide the correct output.

Through this approach, each coding drill contributes to solving different parts of the problem. Combining them gives you a complete, efficient solution.

Targeted Drills in Python

General Coding Drills

  1. Variable Initialization

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    # Initialize an integer variable
count = 0

# Initialize a list
my_list = []

# Initialize a dictionary
my_dict = {}

  2. Looping

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    # Loop from 0 to 9
for i in range(10):
    print(i)

  3. Conditional Statements

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    # If-else condition
x = 5
if x > 10:
    print("x is greater than 10")
else:
    print("x is less than or equal to 10")

  4. Basic Recursion

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    # Factorial function using recursion
def factorial(n):
    if n == 0:
        return 1
    return n * factorial(n-1)

  5. Parameter Passing

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    # Function to add two numbers
def add(a, b):
    return a + b

  6. Memoization

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    # Fibonacci sequence with memoization
memo = {}
def fib(n):
    if n in memo:
        return memo[n]
    if n == 0 or n == 1:
        return n
    memo[n] = fib(n-1) + fib(n-2)
    return memo[n]

  7. Depth-First Search (DFS)

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    # DFS on a graph represented as an adjacency list
visited = set()
def dfs(graph, node):
    if node not in visited:
        print(node)
        visited.add(node)
        for neighbor in graph[node]:
            dfs(graph, neighbor)

  8. Modular Arithmetic

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    # Modular addition
a = 10
b = 15
mod = 7
result = (a + b) % mod

Problem-Specific Drills

  1. Path Counting
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    # Count the number of paths from 0 to n
def count_paths(n):
    if n == 0:
        return 1
    if n < 0:
        return 0
    return count_paths(n-1) + count_paths(n-2)
    • This drill is essential for our problem as counting valid paths is a core part of the solution.

Integrating the Drills

  1. Start with Variable Initialization to set up essential data structures like lists or dictionaries to store intermediate and final results.

  2. Implement Looping and Conditional Statements to navigate through the given data, applying checks and making decisions.

  3. Incorporate Basic Recursion to break the problem into smaller sub-problems. Use Parameter Passing to maintain the state during these recursive calls.

  4. Enhance the recursive function with Depth-First Search (DFS) to explore all valid paths.

  5. Use Memoization within the DFS to store the count of valid paths for visited states, improving efficiency.

  6. Integrate Modular Arithmetic where needed, especially during the summation of large numbers to prevent overflow.

  7. The Path Counting problem-specific drill can be integrated into the DFS to count valid paths based on given constraints.

By proceeding in this manner, each drill contributes to a specific part of the problem, and integrating them in this order should lead to a complete and efficient solution.

Q&A

Similar Problems

  1. Two Sum (Easy)

    • Why: This problem also uses a dictionary to store seen values for quick lookups, similar to the use of memoization in our original problem.

  2. Climbing Stairs (Easy)

    • Why: This problem involves counting the number of ways to reach a certain state, just like our path-counting drill.

  3. Unique Paths (Medium)

    • Why: This problem also revolves around counting the number of valid paths from one point to another, akin to our original problem.

  4. Subsets (Medium)

    • Why: This problem requires recursive backtracking to find all possible subsets, similar to using DFS in our problem to find valid paths.

  5. House Robber (Easy)

    • Why: This problem involves dynamic programming and can benefit from memoization to improve performance, much like our original problem.

  6. Longest Increasing Subsequence (Medium)

    • Why: This problem uses dynamic programming to build upon previously computed solutions, similar to the use of memoization in our problem.

  7. Fibonacci Number (Easy)

    • Why: The Fibonacci sequence is often solved using memoization or dynamic programming, concepts also used in our original problem.

  8. Word Break (Medium)

    • Why: This problem requires splitting a string into valid words from a dictionary. Memoization is used to store whether a substring can be broken up, akin to storing valid paths in our original problem.

  9. Maximum Product Subarray (Medium)

    • Why: This problem uses dynamic programming to find the maximum product in a subarray, similar to how we were tracking states in our problem.

  10. Coin Change (Medium)

    • Why: This problem also requires counting ways to reach a particular state (making change) and can be solved efficiently using memoization or dynamic programming.

These problems involve similar concepts like dynamic programming, memoization, state-tracking, and recursive backtracking, which were integral in solving our original problem.