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: 3
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 vertex
int parent[1000000];

// Function to find the topmost
// parent of vertex a
int root(int a)
{
    // If current vertex is
    // the topmost vertex
    if (a == parent[a]) {
        return a;
    }

    // Otherwise, 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
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
void connectedComponents(int n)
{
    set<int> s;

    // Traverse all vertices
    for (int i = 0; i < n; i++) {

        // Insert all topmost
        // vertices obtained
        s.insert(root(parent[i]));
    }

    // Print count of connected components
    cout << s.size() << '\n';
}

// Function to print answer
void 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 Code
int 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;
}

Java

// Java program for the above approach
import java.util.*;
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;
    }

    // Otherwise, 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<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 answer
static 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 Code
public 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

Python3

# Python3 program for the above approach
from collections import defaultdict
# Given N
N = 8

# Given edges
edges = [[1, 0 ], [ 0, 2 ], [ 5, 3 ], [ 3, 4 ], [ 6, 7 ]]

# Stores the parent of each vertex
parent = list(range(N))

# Function to find the topmost
# parent of vertex x
def find(x):
  if x != parent[x]:
    parent[x] = find(parent[x])
  return parent[x]

def union(x,y):
  parent_x = find(x)
  parent_y = find(y)
  if parent_x != parent_y:
    parent[parent_y] = parent_x

for x,y in edges:
  union(x,y)

dict_pair = defaultdict(list)

for idx, val in enumerate(parent):
  dict_pair[find(val)].append(idx)

print(len(dict_pair.keys()))



# This code is contributed by Shivam Dwivedi

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;
        }

        // Otherwise, 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

Javascript

<script>

// Javascript program for the above approach

// Stores the parent of each vertex
var parent = Array(1000000);

// Function to find the topmost
// parent of vertex a
function root(a)
{
    // If current vertex is
    // the topmost vertex
    if (a == parent[a]) {
        return a;
    }

    // Otherwise, 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
function connect( a, b)
{
    // Connect edges
    a = root(a);
    b = root(b);

    if (a != b) {
        parent[b] = a;
    }
}

// Function to find unique top most parents
function connectedComponents( n)
{
    var s = new Set();

    // Traverse all vertices
    for (var i = 0; i < n; i++) {

        // Insert all topmost
        // vertices obtained
        s.add(parent[i]);
    }

    // Print count of connected components
    document.write( s.size + "<br>");
}

// Function to print answer
function printAnswer( N, edges)
{

    // Setting parent to itself
    for (var i = 0; i <= N; i++) {
        parent[i] = i;
    }

    // Traverse all edges
    for (var i = 0; i < edges.length; i++) {
        connect(edges[i][0], edges[i][1]);
    }

    // Print answer
    connectedComponents(N);
}

// Driver Code
// Given N
var N = 8;
// Given edges
var edges = [
    [ 1, 0 ], [ 0, 2 ], [ 5, 3 ], [ 3, 4 ], [ 6, 7 ]
];

// Function call
printAnswer(N, edges);


</script>

Output

3

Time Complexity: O(NLOGN+M)
Auxiliary Space: O(N+M)