Below is a solution Network delay problem of leetcode. I have written a all test case success solution. But not able to analyse the time complexity. I believe its O(V^2 + E) where V is the number of nodes and E edges. In this solution though I am adding all adjacents of each node every time, but not processing them further if there exists a min distance for that node already. Leetcode question link https://leetcode.com/problems/network-delay-time
public int networkDelayTime(int[][] times, int n, int k) {
int[] distances = new int[n+1];
Arrays.fill(distances , -1);
if(n > 0){
List<List<int[]>> edges = new ArrayList<List<int[]>>();
for(int i = 0 ; i <= n ; i++){
edges.add(new ArrayList<int[]>());
}
for(int[] time : times){
edges.get(time[0]).add(new int[]{time[1] , time[2]});
}
Queue<Vertex> queue = new LinkedList<>();
queue.add(new Vertex(k , 0));
while(!queue.isEmpty()){
Vertex cx = queue.poll();
int index = cx.index;
int distance = cx.distance;
//process adjacents only if distance is updated
if(distances[index] == -1 || distances[index] > distance){
distances[index] = distance;
List<int[]> adjacents = edges.get(index);
for(int[] adjacent : adjacents){
queue.add(new Vertex(adjacent[0] , adjacent[1]+distance));
}
}
}
}
int sum = 0;
for(int i = 1 ; i <= n; i++){
int distance = distances[i];
if(distance == -1){
return -1;
}
sum = Math.max(sum , distance);
}
return sum;
}
public static class Vertex{
int index;
int distance;
public Vertex(int i , int d){
index = i;
distance = d;
}
}
