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++
// 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; } |
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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 |
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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 |
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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 |
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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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