LeetCode-in-Java

743. Network Delay Time

Medium

You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target.

We will send a signal from a given node k. Return the time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.

Example 1:

Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2

Output: 2

Example 2:

Input: times = [[1,2,1]], n = 2, k = 1

Output: 1

Example 3:

Input: times = [[1,2,1]], n = 2, k = 2

Output: -1

Constraints:

• 1 <= k <= n <= 100
• 1 <= times.length <= 6000
• times[i].length == 3
• 1 <= ui, vi <= n
• ui != vi
• 0 <= wi <= 100
• All the pairs (ui, vi) are unique. (i.e., no multiple edges.)

Solution

import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;

public class Solution {
    public int networkDelayTime(int[][] times, int n, int k) {
        int[][] graph = new int[n + 1][n + 1];
        for (int[] g : graph) {
            Arrays.fill(g, -1);
        }
        for (int[] t : times) {
            graph[t[0]][t[1]] = t[2];
        }
        boolean[] visited = new boolean[n + 1];
        int[] dist = new int[n + 1];
        Arrays.fill(dist, Integer.MAX_VALUE);
        dist[k] = 0;
        Queue<Integer> spfa = new LinkedList<>();
        spfa.add(k);
        visited[k] = true;
        while (!spfa.isEmpty()) {
            int curr = spfa.poll();
            visited[curr] = false;
            for (int i = 1; i <= n; i++) {
                if (graph[curr][i] != -1 && dist[i] > dist[curr] + graph[curr][i]) {
                    dist[i] = dist[curr] + graph[curr][i];
                    if (!visited[i]) {
                        spfa.add(i);
                        visited[i] = true;
                    }
                }
            }
        }
        int result = 0;
        for (int i = 1; i <= n; i++) {
            result = Math.max(dist[i], result);
        }
        return result == Integer.MAX_VALUE ? -1 : result;
    }
}

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