Introduction

In this blog post, we will explore how to solve the LeetCode problem "2065" 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

There is an undirected graph with n nodes numbered from 0 to n - 1 (inclusive). You are given a 0-indexed integer array values where values[i] is the value of the ith node. You are also given a 0-indexed 2D integer array edges, where each edges[j] = [uj, vj, timej] indicates that there is an undirected edge between the nodes uj and vj, and it takes timej seconds to travel between the two nodes. Finally, you are given an integer maxTime. A valid path in the graph is any path that starts at node 0, ends at node 0, and takes at most maxTime seconds to complete. You may visit the same node multiple times. The quality of a valid path is the sum of the values of the unique nodes visited in the path (each node's value is added at most once to the sum). Return the maximum quality of a valid path. Note: There are at most four edges connected to each node. Example 1: Input: values = [0,32,10,43], edges = [[0,1,10],[1,2,15],[0,3,10]], maxTime = 49 Output: 75 Explanation: One possible path is 0 -> 1 -> 0 -> 3 -> 0. The total time taken is 10 + 10 + 10 + 10 = 40 <= 49. The nodes visited are 0, 1, and 3, giving a maximal path quality of 0 + 32 + 43 = 75. Example 2: Input: values = [5,10,15,20], edges = [[0,1,10],[1,2,10],[0,3,10]], maxTime = 30 Output: 25 Explanation: One possible path is 0 -> 3 -> 0. The total time taken is 10 + 10 = 20 <= 30. The nodes visited are 0 and 3, giving a maximal path quality of 5 + 20 = 25. Example 3: Input: values = [1,2,3,4], edges = [[0,1,10],[1,2,11],[2,3,12],[1,3,13]], maxTime = 50 Output: 7 Explanation: One possible path is 0 -> 1 -> 3 -> 1 -> 0. The total time taken is 10 + 13 + 13 + 10 = 46 <= 50. The nodes visited are 0, 1, and 3, giving a maximal path quality of 1 + 2 + 4 = 7. Constraints: n == values.length 1 <= n <= 1000 0 <= values[i] <= 108 0 <= edges.length <= 2000 edges[j].length == 3 0 <= uj < vj <= n - 1 10 <= timej, maxTime <= 100 All the pairs [uj, vj] are unique. There are at most four edges connected to each node. The graph may not be connected.

Explanation

Here's a breakdown of the approach, followed by the Python code:

• DFS with Backtracking: Explore all possible paths starting from node 0. Use Depth-First Search (DFS) to traverse the graph. Backtrack when a path exceeds maxTime or reaches a dead end.
• Track Visited Nodes and Time: Maintain a visited set to keep track of the unique nodes in the current path and a current_time variable to track the time taken so far.

• Maximize Quality: Update the maximum quality found so far whenever a valid path (starts and ends at 0, time <= maxTime) is found.

• Time Complexity: O(4^{maxTime/min_edge_time}), where min_edge_time is the minimum time of any edge. In the worst case, we explore all possible paths within the time limit. The problem statement says that the graph is sparse and each node has at most 4 neighbors. Also, maxTime <= 100 and all edge times are >= 10, so the exponent maxTime/min_edge_time is at most 10, which is a relatively small constant.

• Space Complexity: O(N + E + maxTime/min_edge_time), where N is the number of nodes and E is the number of edges. We need space for the adjacency list, the visited set, and the recursion stack.

Code

    def maximalPathQuality(values, edges, maxTime):
    n = len(values)
    adj = [[] for _ in range(n)]
    for u, v, time in edges:
        adj[u].append((v, time))
        adj[v].append((u, time))

    max_quality = 0
    visited = [0] * n

    def dfs(node, current_time, current_quality):
        nonlocal max_quality
        visited[node] += 1
        new_quality = current_quality
        if visited[node] == 1:
            new_quality += values[node]

        if node == 0:
            max_quality = max(max_quality, new_quality)

        for neighbor, time in adj[node]:
            if current_time + time <= maxTime:
                dfs(neighbor, current_time + time, new_quality)

        visited[node] -= 1  # Backtrack

    dfs(0, 0, 0)
    return max_quality