LeetCode 2065: Maximum Path Quality of a Graph Solution

Master LeetCode problem 2065 (Maximum Path Quality of a Graph), 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.

2065. Maximum Path Quality of a Graph

Problem Explanation

Explanation

To solve this problem, we can use a dynamic programming approach. We will iterate through all possible paths starting and ending at node 0 within the given time constraint. At each step, we will calculate the maximum quality of the path by considering the values of the nodes visited so far.

We will maintain a 3D dp array where dp[i][j][k] represents the maximum quality of a path starting at node 0, ending at node i, and taking exactly j seconds to travel, with k representing the bitmask of nodes visited in the path.

Finally, we will iterate through all possible paths within the time constraint and return the maximum quality found.

Time Complexity

The time complexity of this approach is O(n * maxTime * 2^n), where n is the number of nodes and maxTime is the maximum time constraint.

Space Complexity

The space complexity is O(n * maxTime * 2^n) for the dp array.

Solution Code

class Solution {
    public int maximumPathQuality(int[] values, int[][] edges, int maxTime) {
        int n = values.length;
        int[][][] dp = new int[n][maxTime + 1][1 << n];

        for (int t = 1; t <= maxTime; t++) {
            for (int[] edge : edges) {
                int u = edge[0], v = edge[1], time = edge[2];
                for (int mask = 0; mask < (1 << n); mask++) {
                    dp[v][t][mask | (1 << v)] = Math.max(dp[v][t][mask | (1 << v)], dp[u][t - time][mask] + values[v]);
                    dp[u][t][mask | (1 << u)] = Math.max(dp[u][t][mask | (1 << u)], dp[v][t - time][mask] + values[u]);
                }
            }
        }

        int maxQuality = 0;
        for (int t = 1; t <= maxTime; t++) {
            for (int i = 0; i < n; i++) {
                maxQuality = Math.max(maxQuality, dp[i][t][(1 << n) - 1]);
            }
        }

        return maxQuality;
    }
}

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