Formatted question description: https://leetcode.ca/all/743.html
743. Network Delay Time (Medium)
There are N
network nodes, labelled 1
to N
.
Given times
, a list of travel times as directed edges times[i] = (u, v, w)
, where u
is the source node, v
is the target node, and w
is the time it takes for a signal to travel from source to target.
Now, we send a signal from a certain node K
. How long will it take for all nodes to receive the signal? If it is impossible, return -1
.
Example 1:
Input: times = [[2,1,1],[2,3,1],[3,4,1]], N = 4, K = 2 Output: 2
Note:
N
will be in the range[1, 100]
.K
will be in the range[1, N]
.- The length of
times
will be in the range[1, 6000]
. - All edges
times[i] = (u, v, w)
will have1 <= u, v <= N
and0 <= w <= 100
.
Related Topics:
Heap, Depth-first Search, Breadth-first Search, Graph
Solution 1. Dijkstra
// OJ: https://leetcode.com/problems/network-delay-time/
// Time: O(E + VlogV)
// Space: O(E)
class Solution {
typedef unordered_map<int, unordered_map<int, int>> Graph;
typedef pair<int, int> iPair;
vector<int> dijkstra(Graph graph, int N, int source) {
priority_queue<iPair, vector<iPair>, greater<iPair>> pq;
vector<int> dists(N, INT_MAX);
pq.emplace(0, source);
dists[source] = 0;
while (pq.size()) {
int u = pq.top().second;
pq.pop();
for (auto neighbor : graph[u]) {
int v = neighbor.first, weight = neighbor.second;
if (dists[v] > dists[u] + weight) {
dists[v] = dists[u] + weight;
pq.emplace(dists[v], v);
}
}
}
return dists;
}
public:
int networkDelayTime(vector<vector<int>>& times, int N, int K) {
Graph graph;
for (auto e : times) graph[e[0] - 1][e[1] - 1] = e[2];
auto dists = dijkstra(graph, N, K - 1);
int ans = 0;
for (int d : dists) {
if (d == INT_MAX) return -1;
ans = max(ans, d);
}
return ans;
}
};
Solution 2. Bellman-Ford
// OJ: https://leetcode.com/problems/network-delay-time/
// Time: O(VE)
// Space: O(V)
class Solution {
vector<int> bellmanFord(vector<vector<int>>& edges, int V, int src) {
vector<int> dist(V, INT_MAX);
dist[src - 1] = 0;
for (int i = 1; i < V; ++i) {
for (auto &e : edges) {
int u = e[0] - 1, v = e[1] - 1, w = e[2];
if (dist[u] == INT_MAX) continue;
dist[v] = min(dist[v], dist[u] + w);
}
}
return dist;
}
public:
int networkDelayTime(vector<vector<int>>& times, int N, int K) {
auto dist = bellmanFord(times, N, K);
int ans = *max_element(dist.begin(), dist.end());
return ans == INT_MAX ? -1 : ans;
}
};
Java
class Solution {
public int networkDelayTime(int[][] times, int N, int K) {
Map<Integer, List<int[]>> travelMap = new HashMap<Integer, List<int[]>>();
for (int[] time : times) {
int source = time[0] - 1, target = time[1] - 1, lapse = time[2];
List<int[]> travels = travelMap.getOrDefault(source, new ArrayList<int[]>());
travels.add(new int[]{target, lapse});
travelMap.put(source, travels);
}
int[] received = new int[N];
for (int i = 0; i < N; i++)
received[i] = Integer.MAX_VALUE;
received[K - 1] = 0;
Queue<int[]> queue = new LinkedList<int[]>();
queue.offer(new int[]{K - 1, 0});
while (!queue.isEmpty()) {
int[] nodeTime = queue.poll();
int node = nodeTime[0], time = nodeTime[1];
List<int[]> travels = travelMap.getOrDefault(node, new ArrayList<int[]>());
for (int[] travel : travels) {
int target = travel[0], lapse = travel[1];
int totalLapse = time + lapse;
if (totalLapse < received[target]) {
received[target] = totalLapse;
queue.offer(new int[]{target, totalLapse});
}
}
}
int maxTime = 0;
for (int time : received) {
if (time == Integer.MAX_VALUE)
return -1;
else
maxTime = Math.max(maxTime, time);
}
return maxTime;
}
}