Welcome to Subscribe On Youtube

743. Network Delay Time

Description

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

  • class Solution {
    private static final int N = 110;
    private static final int INF = 0x3f3f;

    public int networkDelayTime(int[][] times, int n, int k) {
        int[][] g = new int[N][N];
        for (int i = 0; i < N; ++i) {
            Arrays.fill(g[i], INF);
        }
        for (int[] e : times) {
            g[e[0]][e[1]] = e[2];
        }
        int[] dist = new int[N];
        Arrays.fill(dist, INF);
        dist[k] = 0;
        boolean[] vis = new boolean[N];
        for (int i = 0; i < n; ++i) {
            int t = -1;
            for (int j = 1; j <= n; ++j) {
                if (!vis[j] && (t == -1 || dist[t] > dist[j])) {
                    t = j;
                }
            }
            vis[t] = true;
            for (int j = 1; j <= n; ++j) {
                dist[j] = Math.min(dist[j], dist[t] + g[t][j]);
            }
        }
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            ans = Math.max(ans, dist[i]);
        }
        return ans == INF ? -1 : ans;
    }
}

  • class Solution {
public:
    const int inf = 0x3f3f;

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<vector<int>> g(n, vector<int>(n, inf));
        for (auto& t : times) g[t[0] - 1][t[1] - 1] = t[2];
        vector<bool> vis(n);
        vector<int> dist(n, inf);
        dist[k - 1] = 0;
        for (int i = 0; i < n; ++i) {
            int t = -1;
            for (int j = 0; j < n; ++j) {
                if (!vis[j] && (t == -1 || dist[t] > dist[j])) {
                    t = j;
                }
            }
            vis[t] = true;
            for (int j = 0; j < n; ++j) {
                dist[j] = min(dist[j], dist[t] + g[t][j]);
            }
        }
        int ans = *max_element(dist.begin(), dist.end());
        return ans == inf ? -1 : ans;
    }
};

  • class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        INF = 0x3F3F
        g = defaultdict(list)
        for u, v, w in times:
            g[u - 1].append((v - 1, w))
        dist = [INF] * n
        dist[k - 1] = 0
        q = [(0, k - 1)]
        while q:
            _, u = heappop(q)
            for v, w in g[u]:
                if dist[v] > dist[u] + w:
                    dist[v] = dist[u] + w
                    heappush(q, (dist[v], v))
        ans = max(dist)
        return -1 if ans == INF else ans


  • const Inf = 0x3f3f3f3f

type pair struct {
	first  int
	second int
}

var _ heap.Interface = (*pairs)(nil)

type pairs []pair

func (a pairs) Len() int { return len(a) }
func (a pairs) Less(i int, j int) bool {
	return a[i].first < a[j].first || a[i].first == a[j].first && a[i].second < a[j].second
}
func (a pairs) Swap(i int, j int) { a[i], a[j] = a[j], a[i] }
func (a *pairs) Push(x any)       { *a = append(*a, x.(pair)) }
func (a *pairs) Pop() any         { l := len(*a); t := (*a)[l-1]; *a = (*a)[:l-1]; return t }

func networkDelayTime(times [][]int, n int, k int) int {
	graph := make([]pairs, n)
	for _, time := range times {
		from, to, time := time[0]-1, time[1]-1, time[2]
		graph[from] = append(graph[from], pair{to, time})
	}

	dis := make([]int, n)
	for i := range dis {
		dis[i] = Inf
	}
	dis[k-1] = 0

	vis := make([]bool, n)
	h := make(pairs, 0)
	heap.Push(&h, pair{0, k - 1})
	for len(h) > 0 {
		from := heap.Pop(&h).(pair).second
		if vis[from] {
			continue
		}
		vis[from] = true
		for _, e := range graph[from] {
			to, d := e.first, dis[from]+e.second
			if d < dis[to] {
				dis[to] = d
				heap.Push(&h, pair{d, to})
			}
		}
	}
	ans := slices.Max(dis)
	if ans == Inf {
		return -1
	}
	return ans
}

All Problems

All Solutions