Network Delay Time

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 minimum 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.)

Solutions

This problem can be solved using Dijkstra’s algorithm, which is a famous algorithm for finding the shortest paths from a single source in a graph with non-negative edge weights.

public class Solution {
    public int NetworkDelayTime(int[][] times, int n, int k) {
        var graph = new Dictionary<int, List<(int, int)>>();
        foreach (var time in times) {
            if (!graph.ContainsKey(time[0])) {
                graph[time[0]] = new List<(int, int)>();
            }
            graph[time[0]].Add((time[1], time[2]));
        }

        var dist = new Dictionary<int, int>();
        for (int i = 1; i <= n; ++i) {
            dist[i] = int.MaxValue;
        }
        dist[k] = 0;

        var heap = new SortedSet<(int, int)>() { (0, k) };
        while (heap.Count > 0) {
            var node = heap.Min;
            heap.Remove(node);
            int d = node.Item1, nodeIdx = node.Item2;
            if (d != dist[nodeIdx]) continue;
            if (graph.ContainsKey(nodeIdx)) {
                foreach (var pair in graph[nodeIdx]) {
                    int nei = pair.Item1, d2 = pair.Item2;
                    if (dist[nodeIdx] + d2 < dist[nei]) {
                        heap.Remove((dist[nei], nei));
                        dist[nei] = dist[nodeIdx] + d2;
                        heap.Add((dist[nei], nei));
                    }
                }
            }
        }

        int ans = 0;
        foreach (var cand in dist) {
            ans = Math.Max(ans, cand.Value);
        }
        return ans == int.MaxValue ? -1 : ans;
    }
}

The time complexity of this solution is O(E log E), where E is the number of edges (or the length of times), because in the worst case we need to iterate over each edge and update the queue. The space complexity is O(N + E), where N is the number of nodes, because we need to store the graph (O(E)) and the distance of each node (O(N)).