Problem 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.
Solution
This can be solved using Dijkstra’s algorithm (shortest path to all nodes)
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class Solution {
public int networkDelayTime(int[][] times, int n, int k) {
//create adjacency list (directed weighted graph)
//edge -> times[0] to times[1] with weight as times[2]
List<Node>[] g = new ArrayList[n+1];
for(int i=0; i<g.length; i++) g[i] = new ArrayList<>();
for(int i=0; i<times.length; i++){
int u = times[i][0];
int v = times[i][1];
int w = times[i][2];
g[u].add(new Node(v, w));
}
//initilize delay array to store the time taken to reach any node from the source node k
int[] delay = new int[n+1];
//the time taken for source node will be 0
//for others, set it to +infinity
Arrays.fill(delay, Integer.MAX_VALUE);
delay[k] = 0;
//we need to get the minimum weighted edge which are possible from the currently visited nodes
//for this we can use minHeap sorted by weights (time in this case)
PriorityQueue<Pair> pq = new PriorityQueue<>((p1, p2) -> p1.time - p2.time);
//add the source node with time 0 to min Heap
pq.add(new Pair(0, k));
//while queue is not empty
while(!pq.isEmpty()){
//get the node with min delay
Pair p = pq.poll();
//get its neighbours
List<Node> neighbours = g[p.node];
//for all neighbours, update the time needed for signal to reach
for(int i=0; i<neighbours.size(); i++){
Node current = neighbours.get(i);
int tempDelay = current.weight + p.time;
//if the current calculated delay is lesser, update it in delay array
//and add that node to minHeap so that it can be used to reach other nodes
if(tempDelay < delay[current.node]){
delay[current.node] = tempDelay;
pq.add(new Pair(tempDelay, current.node));
}
//if the tempDelay was more, that indicates that the signal had already reached before and we can continue
}
}
//the min time needed will be the max delay in delay array
int minTimeNeeded = Integer.MIN_VALUE;
for(int i=1; i<delay.length; i++){
if(delay[i] == Integer.MAX_VALUE) return -1;
minTimeNeeded = Math.max(minTimeNeeded, delay[i]);
}
return minTimeNeeded;
}
}
class Node{
int node;
int weight;
Node(int n, int w){
node = n;
weight = w;
}
}
class Pair{
int time;
int node;
Pair(int t, int n){
time = t;
node = n;
}
}