Maximum number of edges among all connected components of an undirected graph - GeeksforGeeks

12 Jul, 2021
Given integers ‘N’ and ‘K’ where, N is the number of vertices of an undirected graph and ‘K’ denotes the number of edges in the same graph (each edge is denoted by a pair of integers where i, j means that the vertex ‘i’ is directly connected to the vertex ‘j’ in the graph).The task is to find the maximum number of edges among all the connected components in the given graph.Examples:  
Input: N = 6, K = 4, Edges = {{1, 2}, {2, 3}, {3, 1}, {4, 5}} Output: 3 Here, graph has 3 components 1st component 1-2-3-1 : 3 edges 2nd component 4-5 : 1 edges 3rd component 6 : 0 edges max(3, 1, 0) = 3 edgesInput: N = 3, K = 2, Edges = {{1, 2}, {2, 3}} Output: 2  
 
Approach:  
Using Depth First Search, find the sum of the degrees of each of the edges in all the connected components separately.
Now, according to Handshaking Lemma, the total number of edges in a connected component of an undirected graph is equal to half of the total sum of the degrees of all of its vertices.
Print the maximum number of edges among all the connected components.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ program to find the connected component// with maximum number of edges#include <bits/stdc++.h>using namespace std; // DFS functionint dfs(int s, vector<int> adj[], vector<bool> visited,                                             int nodes){    // Adding all the edges connected to the vertex    int adjListSize = adj[s].size();    visited[s] = true;    for (long int i = 0; i < adj[s].size(); i++) {        if (visited[adj[s][i]] == false) {            adjListSize += dfs(adj[s][i], adj, visited, nodes);        }    }    return adjListSize;} int maxEdges(vector<int> adj[], int nodes){    int res = INT_MIN;    vector<bool> visited(nodes, false);    for (long int i = 1; i <= nodes; i++) {        if (visited[i] == false) {            int adjListSize = dfs(i, adj, visited, nodes);            res = max(res, adjListSize/2);        }         }    return res;} // Driver codeint main(){    int nodes = 3;    vector<int> adj[nodes+1];     // Edge from vertex 1 to vertex 2    adj[1].push_back(2);    adj[2].push_back(1);     // Edge from vertex 2 to vertex 3    adj[2].push_back(3);    adj[3].push_back(2);     cout << maxEdges(adj, nodes);     return 0;}
// Java program to find the connected component// with maximum number of edgesimport java.util.*; class GFG{     // DFS functionstatic int dfs(int s, Vector<Vector<Integer>> adj,boolean visited[],                        int nodes){    // Adding all the edges connected to the vertex    int adjListSize = adj.get(s).size();    visited[s] = true;    for (int i = 0; i < adj.get(s).size(); i++)    {        if (visited[adj.get(s).get(i)] == false)        {            adjListSize += dfs(adj.get(s).get(i), adj, visited, nodes);        }    }    return adjListSize;} static int maxEdges(Vector<Vector<Integer>> adj, int nodes){    int res = Integer.MIN_VALUE;    boolean visited[]=new boolean[nodes+1];    for (int i = 1; i <= nodes; i++)    {        if (visited[i] == false)        {            int adjListSize = dfs(i, adj, visited, nodes);            res = Math.max(res, adjListSize/2);        }    }    return res;} // Driver codepublic static void main(String args[]){    int nodes = 3;    Vector<Vector<Integer>> adj=new Vector<Vector<Integer>>();         for(int i = 0; i < nodes + 1; i++)    adj.add(new Vector<Integer>());     // Edge from vertex 1 to vertex 2    adj.get(1).add(2);    adj.get(2).add(1);     // Edge from vertex 2 to vertex 3    adj.get(2).add(3);    adj.get(3).add(2);          System.out.println(maxEdges(adj, nodes));}} // This code is contributed by Arnab Kundu
# Python3 program to find the connected# component with maximum number of edgesfrom sys import maxsize INT_MIN = -maxsize # DFS functiondef dfs(s: int, adj: list,        visited: list, nodes: int) -> int:     # Adding all the edges    # connected to the vertex    adjListSize = len(adj[s])    visited[s] = True         for i in range(len(adj[s])):         if visited[adj[s][i]] == False:            adjListSize += dfs(adj[s][i], adj,                               visited, nodes)                                    return adjListSize     def maxEdges(adj: list, nodes: int) -> int:    res = INT_MIN    visited = [False] * (nodes + 1)         for i in range(1, nodes + 1):         if visited[i] == False:            adjListSize = dfs(i, adj,                              visited, nodes)            res = max(res, adjListSize // 2)                 return res # Driver Codeif __name__ == "__main__":         nodes = 3    adj = [0] * (nodes + 1)         for i in range(nodes + 1):        adj[i] = []     # Edge from vertex 1 to vertex 2    adj[1].append(2)    adj[2].append(1)     # Edge from vertex 2 to vertex 3    adj[2].append(3)    adj[3].append(2)     print(maxEdges(adj, nodes)) # This code is contributed by sanjeev2552
// C# program to find the connected component// with maximum number of edgesusing System;using System.Collections.Generic;             class GFG{     // DFS functionstatic int dfs(int s, List<List<int>> adj,               bool []visited, int nodes){    // Adding all the edges connected to the vertex    int adjListSize = adj[s].Count;    visited[s] = true;    for (int i = 0; i < adj[s].Count; i++)    {        if (visited[adj[s][i]] == false)        {            adjListSize += dfs(adj[s][i], adj,                               visited, nodes);        }    }    return adjListSize;} static int maxEdges(List<List<int>> adj, int nodes){    int res = int.MinValue;    bool []visited = new bool[nodes + 1];    for (int i = 1; i <= nodes; i++)    {        if (visited[i] == false)        {            int adjListSize = dfs(i, adj, visited, nodes);            res = Math.Max(res, adjListSize / 2);        }    }    return res;} // Driver codepublic static void Main(String []args){    int nodes = 3;    List<List<int>> adj = new List<List<int>>();         for(int i = 0; i < nodes + 1; i++)    adj.Add(new List<int>());     // Edge from vertex 1 to vertex 2    adj[1].Add(2);    adj[2].Add(1);     // Edge from vertex 2 to vertex 3    adj[2].Add(3);    adj[3].Add(2);         Console.WriteLine(maxEdges(adj, nodes));}} // This code is contributed by PrinciRaj1992
<script> // JavaScript program to find the connected component// with maximum number of edges // DFS functionfunction dfs(s,adj,visited,nodes){    // Adding all the edges connected to the vertex    let adjListSize = adj[s].length;    visited[s] = true;    for (let i = 0; i < adj[s].length; i++)    {        if (visited[adj[s][i]] == false)        {            adjListSize += dfs(adj[s][i], adj, visited, nodes);        }    }    return adjListSize;} function maxEdges(adj,nodes){    let res = Number.MIN_VALUE;    let visited=new Array(nodes+1);    for(let i=0;i<nodes+1;i++)    {        visited[i]=false;    }    for (let i = 1; i <= nodes; i++)    {        if (visited[i] == false)        {            let adjListSize = dfs(i, adj, visited, nodes);            res = Math.max(res, adjListSize/2);        }    }    return res;} // Driver codelet nodes = 3;let adj=[]; for(let i = 0; i < nodes + 1; i++)    adj.push([]); // Edge from vertex 1 to vertex 2adj[1].push(2);adj[2].push(1); // Edge from vertex 2 to vertex 3adj[2].push(3);adj[3].push(2);  document.write(maxEdges(adj, nodes)+"<br>");  // This code is contributed by avanitrachhadiya2155 </script>
2
 
Time Complexity : O(nodes + edges) (Same as DFS)Note : We can also use BFS to solve this problem. We simply need to traverse connected components in an undirected graph. 
Palash Sethi
andrew1234
princiraj1992
sanjeev2552
avanitrachhadiya2155
DFS
graph-connectivity
Technical Scripter 2018
Graph
Technical Scripter
DFS
Graph
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