LeetCode 1575: Count All Possible Routes Solution

Master LeetCode problem 1575 (Count All Possible Routes), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1575. Count All Possible Routes

Hard

Array Dynamic Programming Memoization

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We define a 3D dp array where dp[i][j][k] represents the number of ways to reach city j with k fuel units starting from city i. We initialize the dp array with base cases where dp[i][finish][0] is set to 1 for all i such that locations[i] == finish.

Next, we iterate over all possible cities, fuel units, and steps remaining. For each step, we calculate the number of ways to reach the current city with the current fuel units by considering all possible transitions from the previous cities with the remaining fuel units. We update the dp array accordingly.

Finally, the answer will be the sum of dp[start][finish][0] to dp[start][finish][fuel] modulo 10^9 + 7. :

Solution Code

class Solution {
    public int countRoutes(int[] locations, int start, int finish, int fuel) {
        int n = locations.length;
        int MOD = 1000000007;
        
        int[][][] dp = new int[n][n][fuel + 1];
        for (int i = 0; i < n; i++) {
            if (i != finish) {
                dp[i][finish][0] = 1;
            }
        }
        
        for (int f = 0; f <= fuel; f++) {
            for (int s = 0; s < n; s++) {
                for (int i = 0; i < n; i++) {
                    for (int j = 0; j < n; j++) {
                        if (i != j && f >= Math.abs(locations[i] - locations[j])) {
                            dp[i][s][f] = (dp[i][s][f] + dp[j][s][f - Math.abs(locations[i] - locations[j])]) % MOD;
                        }
                    }
                }
            }
        }
        
        int result = 0;
        for (int f = 0; f <= fuel; f++) {
            result = (result + dp[start][start][f]) % MOD;
        }
        
        return result;
    }
}

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