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Maximum number of edges among all connected components of an undirected graph
  • Difficulty Level : Hard
  • Last Updated : 18 May, 2020

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 edges

Input: 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++




// 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


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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