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

  • Difficulty Level : Hard
  • Last Updated : 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:
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 edges
Input: N = 3, K = 2, 
Edges = {{1, 2}, {2, 3}} 
Output:
 

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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++




// C++ program to find the connected component
// with maximum number of edges
#include <bits/stdc++.h>
using namespace std;
 
// DFS function
int 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 code
int 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




// Java program to find the connected component
// with maximum number of edges
import java.util.*;
 
class GFG
{
     
// DFS function
static 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 code
public 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




# Python3 program to find the connected
# component with maximum number of edges
from sys import maxsize
 
INT_MIN = -maxsize
 
# DFS function
def 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 Code
if __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#




// C# program to find the connected component
// with maximum number of edges
using System;
using System.Collections.Generic;            
 
class GFG
{
     
// DFS function
static 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 code
public 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

Javascript




<script>
 
// JavaScript program to find the connected component
// with maximum number of edges
 
// DFS function
function 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 code
let nodes = 3;
let adj=[];
 
for(let i = 0; i < nodes + 1; i++)
    adj.push([]);
 
// Edge from vertex 1 to vertex 2
adj[1].push(2);
adj[2].push(1);
 
// Edge from vertex 2 to vertex 3
adj[2].push(3);
adj[3].push(2);
 
 
document.write(maxEdges(adj, nodes)+"<br>");
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>
Output: 
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.
 




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