Given a directed graph G with N vertices and M edges. The task is to find the length of the longest directed path in Graph.
Note: Length of a directed path is the number of edges in it.
Examples:
Input: N = 4, M = 5
Output: 3
The directed path 1->3->2->4
Input: N = 5, M = 8
Output: 3
Simple Approach: A naive approach is to calculate the length of the longest path from every node using DFS.
The time complexity of this approach is O(N2).
Efficient Approach: An efficient approach is to use Dynamic Programming and DFS together to find the longest path in the Graph.
Let dp[i] be the length of the longest path starting from the node i. Initially all positions of dp will be 0. We can call the DFS function from every node and traverse for all its children. The recursive formula will be:
dp[node] = max(dp[node], 1 + max(dp[child1], dp[child2], dp[child3]..))
At the end check for the maximum value in dp[] array, which will be the longest path in the DAG.
Below is the implementation of the above approach:
C++
// C++ program to find the longest// path in the DAG#include <bits/stdc++.h>using namespace std;// Function to traverse the DAG// and apply Dynamic Programming// to find the longest pathvoid dfs(int node, vector<int> adj[], int dp[], bool vis[]){ // Mark as visited vis[node] = true; // Traverse for all its children for (int i = 0; i < adj[node].size(); i++) { // If not visited if (!vis[adj[node][i]]) dfs(adj[node][i], adj, dp, vis); // Store the max of the paths dp[node] = max(dp[node], 1 + dp[adj[node][i]]); }}// Function to add an edgevoid addEdge(vector<int> adj[], int u, int v){ adj[u].push_back(v);}// Function that returns the longest pathint findLongestPath(vector<int> adj[], int n){ // Dp array int dp[n + 1]; memset(dp, 0, sizeof dp); // Visited array to know if the node // has been visited previously or not bool vis[n + 1]; memset(vis, false, sizeof vis); // Call DFS for every unvisited vertex for (int i = 1; i <= n; i++) { if (!vis[i]) dfs(i, adj, dp, vis); } int ans = 0; // Traverse and find the maximum of all dp[i] for (int i = 1; i <= n; i++) { ans = max(ans, dp[i]); } return ans;}// Driver Codeint main(){ int n = 5; vector<int> adj[n + 1]; // Example-1 addEdge(adj, 1, 2); addEdge(adj, 1, 3); addEdge(adj, 3, 2); addEdge(adj, 2, 4); addEdge(adj, 3, 4); cout << findLongestPath(adj, n); return 0;} |
Java
// Java program to find the longest// path in the DAGimport java.util.ArrayList;// graph classclass Graph{ int vertices; ArrayList<Integer> edge[]; Graph(int vertices) { this.vertices = vertices; edge = new ArrayList[vertices+1]; for (int i = 0; i <= vertices; i++) { edge[i] = new ArrayList<>(); } } void addEdge(int a,int b) { edge[a].add(b); } void dfs(int node, ArrayList<Integer> adj[], int dp[], boolean visited[]) { // Mark as visited visited[node] = true; // Traverse for all its children for (int i = 0; i < adj[node].size(); i++) { // If not visited if (!visited[adj[node].get(i)]) dfs(adj[node].get(i), adj, dp, visited); // Store the max of the paths dp[node] = Math.max(dp[node], 1 + dp[adj[node].get(i)]); } } // Function that returns the longest path int findLongestPath( int n) { ArrayList<Integer> adj[] = edge; // Dp array int[] dp = new int[n+1]; // Visited array to know if the node // has been visited previously or not boolean[] visited = new boolean[n + 1]; // Call DFS for every unvisited vertex for (int i = 1; i <= n; i++) { if (!visited[i]) dfs(i, adj, dp, visited); } int ans = 0; // Traverse and find the maximum of all dp[i] for (int i = 1; i <= n; i++) { ans = Math.max(ans, dp[i]); } return ans; }}public class Main{ // Driver code public static void main(String[] args) { int n = 5; Graph graph = new Graph(n); // Example-1 graph.addEdge( 1, 2); graph.addEdge( 1, 3); graph.addEdge( 3, 2); graph.addEdge( 2, 4); graph.addEdge( 3, 4); graph.findLongestPath(n); System.out.println( graph.findLongestPath( n)); }}// This code is contributed by SumanSaurav |
Python3
# Python3 program to find the # longest path in the DAG # Function to traverse the DAG # and apply Dynamic Programming # to find the longest path def dfs(node, adj, dp, vis): # Mark as visited vis[node] = True # Traverse for all its children for i in range(0, len(adj[node])): # If not visited if not vis[adj[node][i]]: dfs(adj[node][i], adj, dp, vis) # Store the max of the paths dp[node] = max(dp[node], 1 + dp[adj[node][i]]) # Function to add an edge def addEdge(adj, u, v): adj[u].append(v) # Function that returns the longest path def findLongestPath(adj, n): # Dp array dp = [0] * (n + 1) # Visited array to know if the node # has been visited previously or not vis = [False] * (n + 1) # Call DFS for every unvisited vertex for i in range(1, n + 1): if not vis[i]: dfs(i, adj, dp, vis) ans = 0 # Traverse and find the maximum of all dp[i] for i in range(1, n + 1): ans = max(ans, dp[i]) return ans # Driver Code if __name__ == "__main__": n = 5 adj = [[] for i in range(n + 1)] # Example-1 addEdge(adj, 1, 2) addEdge(adj, 1, 3) addEdge(adj, 3, 2) addEdge(adj, 2, 4) addEdge(adj, 3, 4) print(findLongestPath(adj, n)) # This code is contributed by Rituraj Jain |
C#
// C# program to find the longest// path in the DAGusing System;using System.Collections.Generic;// graph classclass Graph{ public int vertices; public List<int> []edge; public Graph(int vertices) { this.vertices = vertices; edge = new List<int>[vertices + 1]; for (int i = 0; i <= vertices; i++) { edge[i] = new List<int>(); } } public void addEdge(int a, int b) { edge[a].Add(b); } public void dfs(int node, List<int> []adj, int []dp, Boolean []visited) { // Mark as visited visited[node] = true; // Traverse for all its children for (int i = 0; i < adj[node].Count; i++) { // If not visited if (!visited[adj[node][i]]) dfs(adj[node][i], adj, dp, visited); // Store the max of the paths dp[node] = Math.Max(dp[node], 1 + dp[adj[node][i]]); } } // Function that returns the longest path public int findLongestPath( int n) { List<int> []adj = edge; // Dp array int[] dp = new int[n + 1]; // Visited array to know if the node // has been visited previously or not Boolean[] visited = new Boolean[n + 1]; // Call DFS for every unvisited vertex for (int i = 1; i <= n; i++) { if (!visited[i]) dfs(i, adj, dp, visited); } int ans = 0; // Traverse and find the maximum of all dp[i] for (int i = 1; i <= n; i++) { ans = Math.Max(ans, dp[i]); } return ans; }}class GFG{ // Driver code public static void Main(String[] args) { int n = 5; Graph graph = new Graph(n); // Example-1 graph.addEdge( 1, 2); graph.addEdge( 1, 3); graph.addEdge( 3, 2); graph.addEdge( 2, 4); graph.addEdge( 3, 4); graph.findLongestPath(n); Console.WriteLine(graph.findLongestPath(n)); }}// This code is contributed by Princi Singh |
Javascript
<script>// JavaScript program to find the longest// path in the DAG// Function to traverse the DAG// and apply Dynamic Programming// to find the longest pathfunction dfs(node, adj, dp, vis){ // Mark as visited vis[node] = true; // Traverse for all its children for (var i = 0; i < adj[node].length; i++) { // If not visited if (!vis[adj[node][i]]) dfs(adj[node][i], adj, dp, vis); // Store the max of the paths dp[node] = Math.max(dp[node], 1 + dp[adj[node][i]]); }}// Function to add an edgefunction addEdge(adj, u, v){ adj[u].push(v);}// Function that returns the longest pathfunction findLongestPath(adj, n){ // Dp array var dp = Array(n+1).fill(0); // Visited array to know if the node // has been visited previously or not var vis = Array(n+1).fill(false); // Call DFS for every unvisited vertex for (var i = 1; i <= n; i++) { if (!vis[i]) dfs(i, adj, dp, vis); } var ans = 0; // Traverse and find the maximum of all dp[i] for (var i = 1; i <= n; i++) { ans = Math.max(ans, dp[i]); } return ans;}// Driver Codevar n = 5;var adj = Array.from(Array(n+1), ()=>Array());// Example-1addEdge(adj, 1, 2);addEdge(adj, 1, 3);addEdge(adj, 3, 2);addEdge(adj, 2, 4);addEdge(adj, 3, 4);document.write( findLongestPath(adj, n));</script> |
Output:
3
Time Complexity: O(N+M)
Auxiliary Space: O(N)
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