# Connected Components in an undirected graph

Given an undirected graph, print all connected components line by line. For example consider the following graph.

We strongly recommend to minimize your browser and try this yourself first.

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

Finding connected components for an undirected graph is an easier task. We simple need to do either BFS or DFS starting from every unvisited vertex, and we get all strongly connected components. Below are steps based on DFS.

```1) Initialize all vertices as not visited.
2) Do following for every vertex 'v'.
(a) If 'v' is not visited before, call DFSUtil(v)
(b) Print new line character

DFSUtil(v)
1) Mark 'v' as visited.
2) Print 'v'
3) Do following for every adjacent 'u' of 'v'.
If 'u' is not visited, then recursively call DFSUtil(u)
```

Below is C++ implementation of above algorithm.

```// C++ program to print connected components in
// an undirected graph
#include<iostream>
#include <list>
using namespace std;

// Graph class represents a undirected graph
class Graph
{
int V;    // No. of vertices

// Pointer to an array containing adjacency lists

// A function used by DFS
void DFSUtil(int v, bool visited[]);
public:
Graph(int V);   // Constructor
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";
}
}
}

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
list<int>::iterator i;
if(!visited[*i])
DFSUtil(*i, visited);
}

Graph::Graph(int V)
{
this->V = V;
}

// method to add an undirected edge
{
}

// Drive program to test above
int main()
{
// Create a graph given in the above diagram
Graph g(5); // 5 vertices numbered from 0 to 4

cout << "Following are connected components \n";
g.connectedComponents();

return 0;
}
```

Output

```0 1
2 3 4```

Time complexity of above solution is O(V + E) as it does simple DFS for given graph.

# GATE CS Corner    Company Wise Coding Practice

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