Number of connected components of a graph ( using Disjoint Set Union )
Given an undirected graph G with vertices numbered in the range [0, N] and an array Edges[][] consisting of M edges, the task is to find the total number of connected components in the graph using Disjoint Set Union algorithm.
Examples:
Input: N = 4, Edges[][] = {{1, 0}, {2, 3}, {3, 4}}
Output: 2
Explanation: There are only 2 connected components as shown below:
Input: N = 4, Edges[][] = {{1, 0}, {0, 2}, {3, 5}, {3, 4}, {6, 7}}
Output: 2
Explanation: There are only 3 connected components as shown below:
Approach: The problem can be solved using Disjoint Set Union algorithm. Follow the steps below to solve the problem:
- In DSU algorithm, there are two main functions, i.e. connect() and root() function.
- connect(): Connects an edge.
- root(): Recursively determine the topmost parent of a given edge.
- For each edge {a, b}, check if a is connected to b or not. If found to be false, connect them by appending their top parents.
- After completing the above step for every edge, print the total number of the distinct top-most parents for each vertex.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Stores the parent of each vertexint parent[1000000];// Function to find the topmost// parent of vertex aint root(int a){ // If current vertex is // the topmost vertex if (a == parent[a]) { return a; } // Otherwsie, set topmost vertex of // its parent as its topmost vertex return parent[a] = root(parent[a]);}// Function to connect the component// having vertex a with the component// having vertex bvoid connect(int a, int b){ // Connect edges a = root(a); b = root(b); if (a != b) { parent[b] = a; }}// Function to find unique top most parentsvoid connectedComponents(int n){ set<int> s; // Traverse all vertices for (int i = 0; i < n; i++) { // Insert all topmost // vertices obtained s.insert(parent[i]); } // Print count of connected components cout << s.size() << '\n';}// Function to print answervoid printAnswer(int N, vector<vector<int> > edges){ // Setting parent to itself for (int i = 0; i <= N; i++) { parent[i] = i; } // Traverse all edges for (int i = 0; i < edges.size(); i++) { connect(edges[i][0], edges[i][1]); } // Print answer connectedComponents(N);}// Driver Codeint main(){ // Given N int N = 8; // Given edges vector<vector<int> > edges = { { 1, 0 }, { 0, 2 }, { 5, 3 }, { 3, 4 }, { 6, 7 } }; // Function call printAnswer(N, edges); return 0;} |
chevron_right
filter_none
Java
// Java program for the above approachimport java.util.*;class GFG{// Stores the parent of each vertexstatic int []parent = new int[1000000];// Function to find the topmost// parent of vertex astatic int root(int a){ // If current vertex is // the topmost vertex if (a == parent[a]) { return a; } // Otherwsie, set topmost vertex of // its parent as its topmost vertex return parent[a] = root(parent[a]);}// Function to connect the component// having vertex a with the component// having vertex bstatic void connect(int a, int b){ // Connect edges a = root(a); b = root(b); if (a != b) { parent[b] = a; }}// Function to find unique top most parentsstatic void connectedComponents(int n){ HashSet<Integer> s = new HashSet<Integer>(); // Traverse all vertices for (int i = 0; i < n; i++) { // Insert all topmost // vertices obtained s.add(parent[i]); } // Print count of connected components System.out.println(s.size());}// Function to print answerstatic void printAnswer(int N,int [][] edges){ // Setting parent to itself for (int i = 0; i <= N; i++) { parent[i] = i; } // Traverse all edges for (int i = 0; i < edges.length; i++) { connect(edges[i][0], edges[i][1]); } // Print answer connectedComponents(N);}// Driver Codepublic static void main(String[] args){ // Given N int N = 8; // Given edges int [][]edges = {{ 1, 0 }, { 0, 2 }, { 5, 3 }, { 3, 4 }, { 6, 7 }}; // Function call printAnswer(N, edges);}}// This code is contributed by 29AjayKumar |
chevron_right
filter_none
Python3
# Python3 program for the above approach# Stores the parent of each vertexparent = [0]*(1000000) # Function to find the topmost# parent of vertex adef root(a) : # If current vertex is # the topmost vertex if (a == parent[a]) : return a # Otherwsie, set topmost vertex of # its parent as its topmost vertex parent[a] = root(parent[a]) return parent[a] # Function to connect the component# having vertex a with the component# having vertex bdef connect(a, b) : # Connect edges a = root(a) b = root(b) if (a != b) : parent[b] = a # Function to find unique top most parentsdef connectedComponents(n) : s = set() # Traverse all vertices for i in range(n) : # Insert all topmost # vertices obtained s.add(parent[i]) # Print count of connected components print(len(s)) # Function to print answerdef printAnswer(N, edges) : # Setting parent to itself for i in range(N + 1) : parent[i] = i # Traverse all edges for i in range(len(edges)) : connect(edges[i][0], edges[i][1]) # Print answer connectedComponents(N) # Given NN = 8# Given edgesedges = [[ 1, 0 ], [ 0, 2 ], [ 5, 3 ], [ 3, 4 ], [ 6, 7 ]]# Function callprintAnswer(N, edges)# This code is contributed by divyesh072019 |
chevron_right
filter_none
C#
// C# program for the above approach using System;using System.Collections.Generic;class GFG { // Stores the parent of each vertex static int[] parent = new int[1000000]; // Function to find the topmost // parent of vertex a static int root(int a) { // If current vertex is // the topmost vertex if (a == parent[a]) { return a; } // Otherwsie, set topmost vertex of // its parent as its topmost vertex return parent[a] = root(parent[a]); } // Function to connect the component // having vertex a with the component // having vertex b static void connect(int a, int b) { // Connect edges a = root(a); b = root(b); if (a != b) { parent[b] = a; } } // Function to find unique top most parents static void connectedComponents(int n) { HashSet<int> s = new HashSet<int>(); // Traverse all vertices for (int i = 0; i < n; i++) { // Insert all topmost // vertices obtained s.Add(parent[i]); } // Print count of connected components Console.WriteLine(s.Count); } // Function to print answer static void printAnswer(int N, List<List<int> > edges) { // Setting parent to itself for (int i = 0; i <= N; i++) { parent[i] = i; } // Traverse all edges for (int i = 0; i < edges.Count; i++) { connect(edges[i][0], edges[i][1]); } // Print answer connectedComponents(N); } // Driver code static void Main() { // Given N int N = 8; // Given edges List<List<int>> edges = new List<List<int>>(); edges.Add(new List<int> { 1, 0 }); edges.Add(new List<int> { 0, 2 }); edges.Add(new List<int> { 5, 3 }); edges.Add(new List<int> { 3, 4 }); edges.Add(new List<int> { 6, 7 }); // Function call printAnswer(N, edges); }}// This code is contributed by divyeshrabadiya07 |
chevron_right
filter_none
Output:
3
Time Complexity: O(N+M)
Auxiliary Space: O(N+M)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
