Introduction

In this blog post, we will explore how to solve the LeetCode problem "1575" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array of distinct positive integers locations where locations[i] represents the position of city i. You are also given integers start, finish and fuel representing the starting city, ending city, and the initial amount of fuel you have, respectively. At each step, if you are at city i, you can pick any city j such that j != i and 0 <= j < locations.length and move to city j. Moving from city i to city j reduces the amount of fuel you have by |locations[i] - locations[j]|. Please notice that |x| denotes the absolute value of x. Notice that fuel cannot become negative at any point in time, and that you are allowed to visit any city more than once (including start and finish). Return the count of all possible routes from start to finish. Since the answer may be too large, return it modulo 109 + 7. Example 1: Input: locations = [2,3,6,8,4], start = 1, finish = 3, fuel = 5 Output: 4 Explanation: The following are all possible routes, each uses 5 units of fuel: 1 -> 3 1 -> 2 -> 3 1 -> 4 -> 3 1 -> 4 -> 2 -> 3 Example 2: Input: locations = [4,3,1], start = 1, finish = 0, fuel = 6 Output: 5 Explanation: The following are all possible routes: 1 -> 0, used fuel = 1 1 -> 2 -> 0, used fuel = 5 1 -> 2 -> 1 -> 0, used fuel = 5 1 -> 0 -> 1 -> 0, used fuel = 3 1 -> 0 -> 1 -> 0 -> 1 -> 0, used fuel = 5 Example 3: Input: locations = [5,2,1], start = 0, finish = 2, fuel = 3 Output: 0 Explanation: It is impossible to get from 0 to 2 using only 3 units of fuel since the shortest route needs 4 units of fuel. Constraints: 2 <= locations.length <= 100 1 <= locations[i] <= 109 All integers in locations are distinct. 0 <= start, finish < locations.length 1 <= fuel <= 200

Explanation

Here's the solution:

• Dynamic Programming: Use a 2D DP table dp[city][fuel] to store the number of ways to reach city with fuel remaining.

• Base Case: If fuel is 0 and the current city is the destination, there's one way to reach the destination.

• Recursion with Memoization: Iterate through all other cities and recursively calculate the number of ways to reach the destination from those cities with reduced fuel.

• Runtime & Storage Complexity: O(n² fuel), where n is the number of locations and fuel is the maximum fuel allowed. This is because the dp table stores n fuel values, and calculating each value iterates through n cities. The space complexity is also O(n * fuel) due to the memoization table.

Code

    def countRoutes(locations, start, finish, fuel):
    n = len(locations)
    MOD = 10**9 + 7

    dp = {}  # Memoization table: (city, fuel) -> count

    def solve(city, remaining_fuel):
        if (city, remaining_fuel) in dp:
            return dp[(city, remaining_fuel)]

        if remaining_fuel < 0:
            return 0

        if remaining_fuel == 0 and city == finish:
            return 1 if city == finish else 0

        if remaining_fuel == 0 and city != finish:
            return 0

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

        for next_city in range(n):
            if next_city != city:
                cost = abs(locations[city] - locations[next_city])
                count = (count + solve(next_city, remaining_fuel - cost)) % MOD

        dp[(city, remaining_fuel)] = count
        return count

    return solve(start, fuel)