Given an undirected acyclic graph having N nodes and M edges, the task is to find the size of the largest tree in the forest formed by the graph.
A forest is a collection of disjoint trees. In other words, we can also say that forest is a collection of an acyclic graph which is not connected.
Examples:
Input: N = 5, edges[][] = {{0, 1}, {0, 2}, {3, 4}}
Output: 3
Explanation:
There are 2 trees, each having size 3 and 2 respectively.0 / \ 1 2and
3 \ 4Hence the size of the largest tree is 3.
Input: N = 5, edges[][] = {{0, 1}, {0, 2}, {3, 4}, {0, 4}, {3, 5}}
Output: 6
Approach: The idea is to first count the number of reachable nodes from every forest. Therefore:
- Apply DFS on every node and obtain the size of the tree formed by this node and check if every connected node is visited from one source.
- If the size of the current tree is greater than the answer then update the answer to the current tree’s size.
- Again perform DFS traversal if some set of nodes are not yet visited.
- Finally, the max of all the answers when all the nodes are visited is the final answer.
Below is the implementation of the above approach:
C++
// C++ program to find the size// of the largest tree in the forest#include <bits/stdc++.h>using namespace std;// A utility function to add// an edge in an undirected graph.void addEdge(vector<int> adj[], int u, int v){ adj[u].push_back(v); adj[v].push_back(u);}// A utility function to perform DFS of a// graph recursively from a given vertex u// and returns the size of the tree formed by uint DFSUtil(int u, vector<int> adj[], vector<bool>& visited){ visited[u] = true; int sz = 1; // Iterating through all the nodes for (int i = 0; i < adj[u].size(); i++) if (visited[adj[u][i]] == false) // Perform DFS if the node is // not yet visited sz += DFSUtil( adj[u][i], adj, visited); return sz;}// Function to return the size of the// largest tree in the forest given as// the adjacency listint largestTree(vector<int> adj[], int V){ vector<bool> visited(V, false); int answer = 0; // Iterating through all the vertices for (int u = 0; u < V; u++) { if (visited[u] == false) { // Find the answer answer = max(answer, DFSUtil(u, adj, visited)); } } return answer;}// Driver codeint main(){ int V = 5; vector<int> adj[V]; addEdge(adj, 0, 1); addEdge(adj, 0, 2); addEdge(adj, 3, 4); cout << largestTree(adj, V); return 0;} |
Java
// Java program to find the size// of the largest tree in the forestimport java.util.*;class GFG{// A utility function to add// an edge in an undirected graph.static void addEdge(Vector<Integer> adj[], int u, int v){ adj[u].add(v); adj[v].add(u);}// A utility function to perform DFS of a// graph recursively from a given vertex u// and returns the size of the tree formed by ustatic int DFSUtil(int u, Vector<Integer> adj[], Vector<Boolean> visited){ visited.add(u, true); int sz = 1; // Iterating through all the nodes for (int i = 0; i < adj[u].size(); i++) if (visited.get(adj[u].get(i)) == false) // Perform DFS if the node is // not yet visited sz += DFSUtil(adj[u].get(i), adj, visited); return sz;}// Function to return the size of the// largest tree in the forest given as// the adjacency liststatic int largestTree(Vector<Integer> adj[], int V){ Vector<Boolean> visited = new Vector<>(); for(int i = 0; i < V; i++) { visited.add(false); } int answer = 0; // Iterating through all the vertices for (int u = 0; u < V; u++) { if (visited.get(u) == false) { // Find the answer answer = Math.max(answer, DFSUtil(u, adj, visited)); } } return answer;}// Driver codepublic static void main(String[] args){ int V = 5; Vector<Integer> adj[] = new Vector[V]; for (int i = 0; i < adj.length; i++) adj[i] = new Vector<Integer>(); addEdge(adj, 0, 1); addEdge(adj, 0, 2); addEdge(adj, 3, 4); System.out.print(largestTree(adj, V));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to find the size# of the largest tree in the forest # A utility function to add# an edge in an undirected graph.def addEdge(adj, u, v): adj[u].append(v) adj[v].append(u)# A utility function to perform DFS of a# graph recursively from a given vertex u# and returns the size of the tree formed by udef DFSUtil(u, adj, visited): visited[u] = True sz = 1 # Iterating through all the nodes for i in range(0, len(adj[u])): if (visited[adj[u][i]] == False): # Perform DFS if the node is # not yet visited sz += DFSUtil(adj[u][i], adj, visited) return sz# Function to return the size of the# largest tree in the forest given as# the adjacency listdef largestTree(adj, V): visited = [False for i in range(V)] answer = 0 # Iterating through all the vertices for u in range(V): if (visited[u] == False): # Find the answer answer = max(answer,DFSUtil( u, adj, visited)) return answer# Driver codeif __name__=="__main__": V = 5 adj = [[] for i in range(V)] addEdge(adj, 0, 1) addEdge(adj, 0, 2) addEdge(adj, 3, 4) print(largestTree(adj, V)) # This code is contributed by rutvik_56 |
C#
// C# program to find the size// of the largest tree in the forestusing System;using System.Collections.Generic;class GFG{// A utility function to add// an edge in an undirected graph.static void addEdge(List<int> []adj, int u, int v){ adj[u].Add(v); adj[v].Add(u);}// A utility function to perform DFS of a// graph recursively from a given vertex u// and returns the size of the tree formed by ustatic int DFSUtil(int u, List<int> []adj, List<Boolean> visited){ visited.Insert(u, true); int sz = 1; // Iterating through all the nodes for (int i = 0; i < adj[u].Count; i++) if (visited[adj[u][i]] == false) // Perform DFS if the node is // not yet visited sz += DFSUtil(adj[u][i], adj, visited); return sz;}// Function to return the size of the// largest tree in the forest given as// the adjacency liststatic int largestTree(List<int> []adj, int V){ List<Boolean> visited = new List<Boolean>(); for(int i = 0; i < V; i++) { visited.Add(false); } int answer = 0; // Iterating through all the vertices for (int u = 0; u < V; u++) { if (visited[u] == false) { // Find the answer answer = Math.Max(answer, DFSUtil(u, adj, visited)); } } return answer;}// Driver codepublic static void Main(String[] args){ int V = 5; List<int> []adj = new List<int>[V]; for (int i = 0; i < adj.Length; i++) adj[i] = new List<int>(); addEdge(adj, 0, 1); addEdge(adj, 0, 2); addEdge(adj, 3, 4); Console.Write(largestTree(adj, V));}}// This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to find the size // of the largest tree in the forest // A utility function to add // an edge in an undirected graph. function addEdge(adj, u, v) { adj[u].push(v); adj[v].push(u); } // A utility function to perform DFS of a // graph recursively from a given vertex u // and returns the size of the tree formed by u function DFSUtil(u, adj, visited) { visited[u] = true; let sz = 1; // Iterating through all the nodes for(let i = 0; i < adj[u].length; i++) { if (visited[adj[u][i]] == false) { // Perform DFS if the node is // not yet visited sz += DFSUtil(adj[u][i], adj, visited); } } return sz; } // Function to return the size of the // largest tree in the forest given as // the adjacency list function largestTree(adj, V) { let visited = new Array(V); visited.fill(false); let answer = 0; // Iterating through all the vertices for(let u = 0; u < V; u++) { if (visited[u] == false) { // Find the answer answer = Math.max(answer,DFSUtil(u, adj, visited)); } } return answer; } let V = 5; let adj = []; for(let i = 0; i < V; i++) { adj.push([]); } addEdge(adj, 0, 1); addEdge(adj, 0, 2); addEdge(adj, 3, 4); document.write(largestTree(adj, V)); // This code is contributed by divyesh072019.</script> |
Output:
3
Time Complexity: O(V + E), where V is the number of vertices and E is the number of edges.
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