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Connected Components in an Undirected Graph

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Given an undirected graph, the task is to print all the connected components line by line. 

Examples:

Input: Consider the following graph

Example of an undirected graph

Example of an undirected graph

Output:
0 1 2
3 4
Explanation: There are 2 different connected components.
They are {0, 1, 2} and {3, 4}.

Recommended Practice

We have discussed algorithms for finding strongly connected components in directed graphs in following posts. 
Kosaraju’s algorithm for strongly connected components
Tarjan’s Algorithm to find Strongly Connected Components

Connected Components for undirected graph using DFS:

Finding connected components for an undirected graph is an easier task. The idea is to 

Do either BFS or DFS starting from every unvisited vertex, and we get all strongly connected components. 

Follow the steps mentioned below to implement the idea using DFS:

  • Initialize all vertices as not visited.
  • Do the following for every vertex v:
    • If v is not visited before, call the DFS. and print the newline character to print each component in a new line
      • Mark v as visited and print v.
      • For every adjacent u of v, If u is not visited, then recursively call the DFS.

Below is the implementation of above algorithm.

C++

// C++ program to print connected components in
// an undirected graph
#include <bits/stdc++.h>
using namespace std;
 
// Graph class represents a undirected graph
// using adjacency list representation
class Graph {
    int V; // No. of vertices
 
    // Pointer to an array containing adjacency lists
    list<int>* adj;
 
    // A function used by DFS
    void DFSUtil(int v, bool visited[]);
 
public:
    Graph(int V); // Constructor
    ~Graph();
    void addEdge(int v, int w);
    void connectedComponents();
};
 
// Method to print connected components in an
// undirected graph
void Graph::connectedComponents()
{
    // Mark all the vertices as not visited
    bool* visited = new bool[V];
    for (int v = 0; v < V; v++)
        visited[v] = false;
 
    for (int v = 0; v < V; v++) {
        if (visited[v] == false) {
            // print all reachable vertices
            // from v
            DFSUtil(v, visited);
 
            cout << "\n";
        }
    }
    delete[] visited;
}
 
void Graph::DFSUtil(int v, bool visited[])
{
    // Mark the current node as visited and print it
    visited[v] = true;
    cout << v << " ";
 
    // Recur for all the vertices
    // adjacent to this vertex
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
        if (!visited[*i])
            DFSUtil(*i, visited);
}
 
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}
 
Graph::~Graph() { delete[] adj; }
 
// method to add an undirected edge
void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w);
    adj[w].push_back(v);
}
 
// Driver code
int main()
{
    // Create a graph given in the above diagram
    Graph g(5); // 5 vertices numbered from 0 to 4
    g.addEdge(1, 0);
    g.addEdge(2, 1);
    g.addEdge(3, 4);
 
    cout << "Following are connected components \n";
    g.connectedComponents();
 
    return 0;
}

                    

Java

// Java program to print connected components in
// an undirected graph
import java.util.ArrayList;
class Graph {
    // A user define class to represent a graph.
    // A graph is an array of adjacency lists.
    // Size of array will be V (number of vertices
    // in graph)
    int V;
    ArrayList<ArrayList<Integer> > adjListArray;
 
    // constructor
    Graph(int V)
    {
        this.V = V;
        // define the size of array as
        // number of vertices
        adjListArray = new ArrayList<>();
 
        // Create a new list for each vertex
        // such that adjacent nodes can be stored
 
        for (int i = 0; i < V; i++) {
            adjListArray.add(i, new ArrayList<>());
        }
    }
 
    // Adds an edge to an undirected graph
    void addEdge(int src, int dest)
    {
        // Add an edge from src to dest.
        adjListArray.get(src).add(dest);
 
        // Since graph is undirected, add an edge from dest
        // to src also
        adjListArray.get(dest).add(src);
    }
 
    void DFSUtil(int v, boolean[] visited)
    {
        // Mark the current node as visited and print it
        visited[v] = true;
        System.out.print(v + " ");
        // Recur for all the vertices
        // adjacent to this vertex
        for (int x : adjListArray.get(v)) {
            if (!visited[x])
                DFSUtil(x, visited);
        }
    }
    void connectedComponents()
    {
        // Mark all the vertices as not visited
        boolean[] visited = new boolean[V];
        for (int v = 0; v < V; ++v) {
            if (!visited[v]) {
                // print all reachable vertices
                // from v
                DFSUtil(v, visited);
                System.out.println();
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        // Create a graph given in the above diagram
        Graph g = new Graph(5);
 
        g.addEdge(1, 0);
        g.addEdge(2, 1);
        g.addEdge(3, 4);
        System.out.println(
            "Following are connected components");
        g.connectedComponents();
    }
}

                    

Python3

# Python program to print connected
# components in an undirected graph
 
 
class Graph:
 
    # init function to declare class variables
    def __init__(self, V):
        self.V = V
        self.adj = [[] for i in range(V)]
 
    def DFSUtil(self, temp, v, visited):
 
        # Mark the current vertex as visited
        visited[v] = True
 
        # Store the vertex to list
        temp.append(v)
 
        # Repeat for all vertices adjacent
        # to this vertex v
        for i in self.adj[v]:
            if visited[i] == False:
 
                # Update the list
                temp = self.DFSUtil(temp, i, visited)
        return temp
 
    # method to add an undirected edge
    def addEdge(self, v, w):
        self.adj[v].append(w)
        self.adj[w].append(v)
 
    # Method to retrieve connected components
    # in an undirected graph
    def connectedComponents(self):
        visited = []
        cc = []
        for i in range(self.V):
            visited.append(False)
        for v in range(self.V):
            if visited[v] == False:
                temp = []
                cc.append(self.DFSUtil(temp, v, visited))
        return cc
 
 
# Driver Code
if __name__ == "__main__":
 
    # Create a graph given in the above diagram
    # 5 vertices numbered from 0 to 4
    g = Graph(5)
    g.addEdge(1, 0)
    g.addEdge(2, 1)
    g.addEdge(3, 4)
    cc = g.connectedComponents()
    print("Following are connected components")
    print(cc)
 
# This code is contributed by Abhishek Valsan

                    

C#

using System;
using System.Collections.Generic;
 
class Graph
{
   
  // A user defined class to represent a graph.
  // A graph is an array of adjacency lists.
  // Size of array will be V (number of vertices
  // in graph)
  int V;
  List<List<int>> adjListArray;
 
  // constructor
  public Graph(int V)
  {
    this.V = V;
     
    // define the size of array as
    // number of vertices
    adjListArray = new List<List<int>>();
 
    // Create a new list for each vertex
    // such that adjacent nodes can be stored
    for (int i = 0; i < V; i++)
    {
      adjListArray.Add(new List<int>());
    }
  }
 
  // Adds an edge to an undirected graph
  public void addEdge(int src, int dest)
  {
    // Add an edge from src to dest.
    adjListArray[src].Add(dest);
 
    // Since graph is undirected, add an edge from dest
    // to src also
    adjListArray[dest].Add(src);
  }
 
  void DFSUtil(int v, bool[] visited)
  {
    // Mark the current node as visited and print it
    visited[v] = true;
    Console.Write(v + " ");
    // Recur for all the vertices
    // adjacent to this vertex
    foreach (int x in adjListArray[v])
    {
      if (!visited[x])
        DFSUtil(x, visited);
    }
  }
 
  public void connectedComponents()
  {
    // Mark all the vertices as not visited
    bool[] visited = new bool[V];
    for (int v = 0; v < V; ++v)
    {
      if (!visited[v])
      {
        // print all reachable vertices
        // from v
        DFSUtil(v, visited);
        Console.WriteLine();
      }
    }
  }
 
  // Driver code
  public static void Main(string[] args)
  {
     
    // Create a graph given in the above diagram
    Graph g = new Graph(5);
 
    g.addEdge(1, 0);
    g.addEdge(2, 1);
    g.addEdge(3, 4);
    Console.WriteLine("Following are connected components:");
    g.connectedComponents();
  }
}
 
// This code is contributed by lokeshpotta20.

                    

Javascript

<script>
// Javascript program to print connected components in
// an undirected graph
 
// A user define class to represent a graph.
    // A graph is an array of adjacency lists.
    // Size of array will be V (number of vertices
    // in graph)
let V;
let adjListArray=[];
 
// constructor
function Graph(v)
{   
    V = v
     
        // define the size of array as
        // number of vertices
  
        // Create a new list for each vertex
        // such that adjacent nodes can be stored
  
        for (let i = 0; i < V; i++) {
            adjListArray.push([]);
        }
}
 
// Adds an edge to an undirected graph
function addEdge(src,dest)
{
    // Add an edge from src to dest.
        adjListArray[src].push(dest);
  
        // Since graph is undirected, add an edge from dest
        // to src also
        adjListArray[dest].push(src);
}
 
function DFSUtil(v,visited)
{
    // Mark the current node as visited and print it
        visited[v] = true;
        document.write(v + " ");
         
        // Recur for all the vertices
        // adjacent to this vertex
        for (let x = 0; x < adjListArray[v].length; x++)
        {
            if (!visited[adjListArray[v][x]])
                DFSUtil(adjListArray[v][x], visited);
        }
}
 
function connectedComponents()
{
    // Mark all the vertices as not visited
        let visited = new Array(V);
        for(let i = 0; i < V; i++)
        {
            visited[i] = false;
        }
        for (let v = 0; v < V; ++v)
        {
            if (!visited[v])
            {
                // print all reachable vertices
                // from v
                DFSUtil(v, visited);
                document.write("<br>");
            }
        }
}
 
// Driver code
Graph(5);
 
addEdge(1, 0);
addEdge(2, 1);
addEdge(3, 4);
document.write(
"Following are connected components<br>");
connectedComponents();
 
// This code is contributed by rag2127
</script>

                    

Output
Following are connected components 
0 1 2 
3 4 

Time Complexity: O(V + E) where V is the number of vertices and E is the number of edges.
Auxiliary Space: O(V)

Connected Component for undirected graph using Disjoint Set Union:

The idea to solve the problem using DSU (Disjoint Set Union) is

Initially declare all the nodes as individual subsets and then visit them. When a new unvisited node is encountered, unite it with the under. In this manner, a single component will be visited in each traversal.

Follow the below steps to implement the idea:

  • Declare an array arr[] of size V where V is the total number of nodes.
  • For every index i of array arr[], the value denotes who the parent of ith vertex is.
  • Initialise every node as the parent of itself and then while adding them together, change their parents accordingly.
  • Traverse the nodes from 0 to V:
    • For each node that is the parent of itself start the DSU.
    • Print the nodes of that disjoint set as they belong to one component.

Below is the implementation of the above approach.

C++

#include <bits/stdc++.h>
using namespace std;
 
int merge(int* parent, int x)
{
    if (parent[x] == x)
        return x;
    return merge(parent, parent[x]);
}
 
int connectedcomponents(int n, vector<vector<int> >& edges)
{
    int parent[n];
    for (int i = 0; i < n; i++) {
        parent[i] = i;
    }
    for (auto x : edges) {
        parent[merge(parent, x[0])] = merge(parent, x[1]);
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
        ans += (parent[i] == i);
    }
    for (int i = 0; i < n; i++) {
        parent[i] = merge(parent, parent[i]);
    }
    map<int, list<int> > m;
    for (int i = 0; i < n; i++) {
        m[parent[i]].push_back(i);
    }
    for (auto it = m.begin(); it != m.end(); it++) {
        list<int> l = it->second;
        for (auto x : l) {
            cout << x << " ";
        }
        cout << endl;
    }
    return ans;
}
 
int main()
{
    int n = 5;
    vector<int> e1 = { 0, 1 };
    vector<int> e2 = { 2, 1 };
    vector<int> e3 = { 3, 4 };
    vector<vector<int> > e;
    e.push_back(e1);
    e.push_back(e2);
    e.push_back(e3);
 
    cout << "Following are connected components:\n";
    int a = connectedcomponents(n, e);
    return 0;
}

                    

Java

import java.util.*;
 
class ConnectedComponents {
    public static int merge(int[] parent, int x) {
        if (parent[x] == x)
            return x;
        return merge(parent, parent[x]);
    }
 
    public static int connectedComponents(int n, List<List<Integer>> edges) {
        int[] parent = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
 
        for (List<Integer> x : edges) {
            parent[merge(parent, x.get(0))] = merge(parent, x.get(1));
        }
 
        int ans = 0;
        for (int i = 0; i < n; i++) {
            if (parent[i] == i) ans++;
        }
 
        for (int i = 0; i < n; i++) {
            parent[i] = merge(parent, parent[i]);
        }
 
        Map<Integer, List<Integer>> m = new HashMap<>();
        for (int i = 0; i < n; i++) {
            m.computeIfAbsent(parent[i], k -> new ArrayList<>()).add(i);
        }
 
        for (Map.Entry<Integer, List<Integer>> it : m.entrySet()) {
            List<Integer> l = it.getValue();
            for (int x : l) {
                System.out.print(x + " ");
            }
            System.out.println();
        }
        return ans;
    }
 
    public static void main(String[] args) {
        int n = 5;
        List<List<Integer>> edges = new ArrayList<>();
        edges.add(Arrays.asList(0, 1));
        edges.add(Arrays.asList(2, 1));
        edges.add(Arrays.asList(3, 4));
 
        System.out.println("Following are connected components:");
        int ans = connectedComponents(n, edges);
    }
}

                    

C#

using System;
using System.Collections.Generic;
 
class ConnectedComponents {
public static int Merge(int[] parent, int x) {
if (parent[x] == x)
return x;
return Merge(parent, parent[x]);
}
 
 
public static int CountConnectedComponents(int n, List<List<int>> edges) {
    int[] parent = new int[n];
    for (int i = 0; i < n; i++) {
        parent[i] = i;
    }
 
    foreach (List<int> x in edges) {
        parent[Merge(parent, x[0])] = Merge(parent, x[1]);
    }
 
    int ans = 0;
    for (int i = 0; i < n; i++) {
        if (parent[i] == i) ans++;
    }
 
    for (int i = 0; i < n; i++) {
        parent[i] = Merge(parent, parent[i]);
    }
 
    Dictionary<int, List<int>> m = new Dictionary<int, List<int>>();
    for (int i = 0; i < n; i++) {
        if (!m.ContainsKey(parent[i])) {
            m[parent[i]] = new List<int>();
        }
        m[parent[i]].Add(i);
    }
 
    foreach (KeyValuePair<int, List<int>> it in m) {
        List<int> l = it.Value;
        foreach (int x in l) {
            Console.Write(x + " ");
        }
        Console.WriteLine();
    }
    return ans;
}
 
public static void Main() {
    int n = 5;
    List<List<int>> edges = new List<List<int>>();
    edges.Add(new List<int> { 0, 1 });
    edges.Add(new List<int> { 2, 1 });
    edges.Add(new List<int> { 3, 4 });
 
    Console.WriteLine("Following are connected components:");
    int ans = CountConnectedComponents(n, edges);
}
}

                    

Python3

# Python equivalent of above Java code
 
# importing utilities
import collections
 
# class to find connected components
class ConnectedComponents:
  # function to merge two components
  def merge(self, parent, x):
    if parent[x] == x:
      return x
    return self.merge(parent, parent[x])
 
  # function to find connected components
  def connectedComponents(self, n, edges):
    # list to store parents of each node
    parent = [i for i in range(n)]
 
    # loop to set parent of each node
    for x in edges:
      parent[self.merge(parent, x[0])] = self.merge(parent, x[1])
       
    # count to store number of connected components
    ans = 0
    # loop to count number of connected components
    for i in range(n):
      if parent[i] == i:
        ans += 1
     
    # loop to merge all components
    for i in range(n):
      parent[i] = self.merge(parent, parent[i])
 
    # map to store parent and its connected components
    m = collections.defaultdict(list)
    for i in range(n):
      m[parent[i]].append(i)
     
    # loop to print connected components
    for it in m.items():
      l = it[1]
      print(" ".join([str(x) for x in l]))
     
    return ans
 
 
# driver code
if __name__ == "__main__":
  n = 5
  edges = [[0, 1], [2, 1], [3, 4]]
 
  # print connected components
  print("Following are connected components:")
  ans = ConnectedComponents().connectedComponents(n, edges)

                    

Javascript

// JavaScript program to find connected components in a graph
 
// Function to find the parent of a node using path compression
function merge(parent, x) {
    if (parent[x] === x) {
        return x;
    }
    return merge(parent, parent[x]);
}
 
// Function to find the number of connected components in the graph
function connectedcomponents(n, edges) {
    // Initialize parent array for each node
    let parent = [];
    for (let i = 0; i < n; i++) {
        parent[i] = i;
    } // Union operation for all edges
    for (let x of edges) {
        parent[merge(parent, x[0])] = merge(parent, x[1]);
    }
 
    // Count the number of nodes with self as parent, which are the roots of connected components
    let ans = 0;
    for (let i = 0; i < n; i++) {
        ans += parent[i] === i;
    }
 
    // Find the parent of each node again, and group nodes with the same parent
    for (let i = 0; i < n; i++) {
        parent[i] = merge(parent, parent[i]);
    }
    let m = new Map();
    for (let i = 0; i < n; i++) {
        if (!m.has(parent[i])) {
            m.set(parent[i], []);
        }
        m.get(parent[i]).push(i);
    }
 
    // Print the nodes in each connected component
    console.log("Following are connected components:");
    for (let [key, value] of m) {
        console.log(value.join(" "));
    }
 
    // Return the number of connected components
    return ans;
}
 
// Sample input
let n = 5;
let e1 = [0, 1];
let e2 = [2, 1];
let e3 = [3, 4];
let e = [e1, e2, e3];
 
// Find connected components and print them
let a = connectedcomponents(n, e);

                    

Output
Following are connected components:
0 1 2 
3 4 

Time Complexity: O(V)
Auxiliary Space: O(V)



Last Updated : 07 Mar, 2023
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