# Solution For Network Delay Time

Problem Statement:

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] = (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.

We will send a signal from a given node k. Return the 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:

We are given a weighted directed graph where the weight represents the time taken to reach the destination node from the source node and we want to find the shortest path from a starting node to all other nodes in the graph. We can use Dijkstra’s algorithm to solve this problem.

Firstly, we create an adjacency list of the given graph. Then, we initialize the distance of the starting node to be zero and all the other nodes as infinity. We create a min-heap to maintain the minimum distance from the starting node to any node. We push the starting node into the min-heap with distance as 0.

While the min-heap is not empty, we extract the node with the minimum distance value from the min-heap, say u. We iterate through all its neighbours v, and for each neighbour, we calculate its distance through u, let’s call it dist. If the distance obtained is less than the current distance of v, then we update the distance of v and push the updated distance and v into the min-heap.

Finally, we return the maximum distance obtained from the starting node to all other nodes. If any node’s distance value remains infinity, it means that it is not reachable from the starting node and hence we return -1.

Here is the Python solution:

“`

import heapq

from collections import defaultdict

class Solution:

def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:

# Create adjacency list to represent the graph

graph = defaultdict(list)

for u, v, w in times:

graph[u].append((v, w))

```
# Initialize distance of all nodes from source
dist = [float('inf')] * (n+1)
dist[k] = 0
# Create a min-heap to maintain minimum distance
min_heap = [(0, k)]
visited = set()
while min_heap:
d, u = heapq.heappop(min_heap)
if u in visited:
continue
visited.add(u)
for v, w in graph[u]:
if dist[u] + w < dist[v]:
dist[v] = dist[u] + w
heapq.heappush(min_heap, (dist[v], v))
max_dist = max(dist[1:])
return max_dist if max_dist < float('inf') else -1
```

“`

Time Complexity:

The time complexity of Dijkstra’s algorithm is O(E + V log V), where E is the number of edges and V is the number of vertices, since we are using a min-heap to maintain minimum distance.

Space Complexity:

The space complexity is O(V + E) for the adjacency list and min-heap.

In conclusion, we can use Dijkstra’s algorithm to solve the network delay time problem on Leetcode. It is a standard graph algorithm to find the shortest path from a starting node to all other nodes in a weighted directed graph.

## Step by Step Implementation For Network Delay Time

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 have 1 <= u, v <= N and 0 <= w <= 100.

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 have 1 <= u, v <= N and 0 <= w <= 100.

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 have 1 <= u, v <= N and 0 <= w <= 100.

There are a few things to consider when solving this problem: 1) We need to find the shortest path from the source vertex to all other vertices. 2) We need to take into account the weight of each edge when finding the shortest path. We can use Dijkstra's algorithm to find the shortest path from the source vertex to all other vertices. This algorithm works by keeping track of the shortest distance from the source vertex to all other vertices. It does this by starting at the source vertex and considering all of its neighbors. For each neighbor, it calculates the distance from the source vertex to that neighbor. If this distance is shorter than the current shortest distance for that neighbor, then the shortest distance is updated. The algorithm then moves on to the next vertex and repeats the process. Once the algorithm has finished, the shortest distance from the source vertex to all other vertices will be stored in an array. We can then iterate through this array and find the longest distance, which will be the answer to the problem.

using System; using System.Collections.Generic; using System.Linq; public class Solution { public int NetworkDelayTime(int[,] times, int N, int K) { //Create a list of Nodes Listnodes = new List (); //Create a list of Edges List edges = new List (); //Initialize the list of Nodes for(int i = 0; i < N; i++){ nodes.Add(new Node(i)); } //Initialize the list of Edges for(int i = 0; i < times.GetLength(0); i++){ edges.Add(new Edge(nodes[times[i,0] - 1], nodes[times[i,1] - 1], times[i,2])); } //Sort the list of edges in ascending order of weight edges = edges.OrderBy(e => e.weight).ToList(); //Initialize the shortest distance of all nodes as infinity foreach(Node node in nodes){ node.distance = int.MaxValue; } //Initialize the source node's shortest distance as 0 nodes[K - 1].distance = 0; //Create a set to keep track of visited nodes HashSet visited = new HashSet (); //Variable to store the maximum distance int maxDistance = 0; //Loop until all nodes are visited while(visited.Count != N){ //Find the node with the shortest distance that has not been visited yet Node current = nodes.Where(n => !visited.Contains(n)).OrderBy(n => n.distance).First(); //Update the distance of all adjacent nodes foreach(Edge edge in edges.Where(e => e.source == current)){ if(edge.destination.distance > current.distance + edge.weight){ edge.destination.distance = current.distance + edge.weight; } } //Add the current node to the visited set visited.Add(current); //Update the maximum distance if(current.distance > maxDistance){ maxDistance = current.distance; } } //If any node's shortest distance is still infinity, then there is no solution if(nodes.Any(n => n.distance == int.MaxValue)){ return -1; } //Return the maximum distance return maxDistance; } } //Class to represent a Node public class Node{ public int id; public int distance; public Node(int id){ this.id = id; } } //Class to represent an Edge public class Edge{ public Node source; public Node destination; public int weight; public Edge(Node source, Node destination, int weight){ this.source = source; this.destination = destination; this.weight = weight; } }