## Introduction🎄

Hi Ninja🥷! In this article, we will learn to solve the problem of checking if a given graph is strongly connected ( Kosaraju's BFS ). A strongly connected component of a graph is a part in which there is a path from every vertex to another vertex. It is only valid in the case of directed graphs.

We hope you will enjoy and have fun learning a hard-level DSA problem on graph data structure "Check if a given graph is strongly connected ( Kosaraju's BFS )".

### Problem Description🗒️

Given a directed graph, we have to find out whether the graph is strongly connected or not. A directed graph is strongly connected if there is a path from every vertex to another vertex.

There are many algorithms to solve this problem, but one of the most efficient ones is Kosaraju’s BFS-based algorithm.

## Approach 🤖

Following are the steps involved in Kosaraju’s BFS-based algorithm.

### Algorithm👽

- We do two BFS traversals of the given graph:
- First, we Initialize all vertices as not visited.
- Then we do a BFS traversal of a graph starting from any vertex, let's say vertex V.
- After BFS traversal, if we cannot visit all vertices, then return false.
- Second, We reverse all the edges present in the graph.
- We Initialize all vertices as not visited in the reversed graph.
- Then we repeat a BFS traversal of the reversed graph starting from the same vertex V.
- After BFS traversal, if we cannot visit all vertices, then return false.
- Otherwise, return true.

### Sample Examples😄

**Input:**

**Output:**

**Explanation:**

We say a directed graph is strongly connected if there is a path from every vertex to another vertex.

Here, If we start from vertex 1, we get BFS as 1 2 3 4 5

After reversing the given graph, we get,

Again, starting with vertex 1, we get BFS as 1 0

Vertex 0 in the original graph and 2 3 4 5 in the reverse graph remain unvisited.

So, this graph is not strongly connected.

**Input:**

**Output:**

Given graph is strongly connected.

**Explanation:**

If we start from vertex 2, we get BFS as 2 4 3 5 0 6 1

After reversing the given graph, we get,

Again, starting with vertex 2, we get BFS as 2 4 1 6 0 5 3

No vertex in the given and reversed graph remains unvisited.

So, this graph is strongly connected.

### C++ Code⌨️

```
// Program to check if the given directed graph
// is strongly connected or not.
#include <bits/stdc++.h>
using namespace std;
class Graph
{
int vertices; // To store no. of vertices
list<int> *adjList;
// Function to print DFS starting from v
void BFSUtil(int v, bool visited[]);
public:
Graph(int V) { this->vertices = V; adjList = new list<int>[V];}
~Graph() { delete [] adjList; }
// Function to add an edge
void addEdge(int v, int w);
// Function that returns if the
// graph is strongly connected or not.
bool isStronglyConnected();
// Function that returns reverse (or transpose)
// of this graph
Graph getTranspose();
};
// A recursive function to print DFS starting from v
void Graph::BFSUtil(int v, bool visited[])
{
// Creating a queue for BFS traversal
list<int> queue;
visited[v] = true;
queue.push_back(v);
list<int>::iterator i;
while (!queue.empty())
{
// Dequeue a vertex from queue
v = queue.front();
queue.pop_front();
for (i = adjList[v].begin(); i != adjList[v].end(); ++i)
{
if (!visited[*i])
{
visited[*i] = true;
queue.push_back(*i);
}
}
}
}
Graph Graph::getTranspose()
{
Graph g(vertices);
for (int v = 0; v < vertices; v++)
{
list<int>::iterator i;
for (i = adjList[v].begin(); i != adjList[v].end(); ++i)
g.adjList[*i].push_back(v);
}
return g;
}
void Graph::addEdge(int v, int w)
{
adjList[v].push_back(w);
}
// Function that returns true if graph
// is strongly connected
bool Graph::isStronglyConnected()
{
// Step 1: Mark all the vertices as not
// visited (For first BFS)
bool visited[vertices];
for (int i = 0; i < vertices; i++)
visited[i] = false;
// Do BFS traversal starting from the first vertex.
BFSUtil(0, visited);
// If BFS traversal doesn’t visit all vertices
// return false.
for (int i = 0; i < vertices; i++)
if (visited[i] == false)
return false;
//Create a reversed graph
Graph gr = getTranspose();
// Step 4: Mark all the vertices as not
// visited (For second BFS)
for(int i = 0; i < vertices; i++)
visited[i] = false;
gr.BFSUtil(0, visited);
// If all vertices are not visited in the second DFS
//return false
for (int i = 0; i < vertices; i++)
if (visited[i] == false)
return false;
return true;
}
// Driver program to test the above functions
int main()
{
Graph g1(7);
g1.addEdge(0, 1);
g1.addEdge(1, 2);
g1.addEdge(2, 3);
g1.addEdge(3, 0);
g1.addEdge(2, 4);
g1.addEdge(4, 2);
g1.addEdge(4, 5);
g1.addEdge(5, 6);
g1.addEdge(6, 4);
g1.isStronglyConnected()?cout<<"Yes":cout<<"No";
return 0;
}
```

Output:

`Yes`

#### Complexity Analysis

**Time Complexity: **O(V+E), where V and E are the numbers of vertices and edges, respectively. Time taken by this algorithm if the graph is represented using adjacency matrix representation is the same as BFS traversal.

**Space Complexity:** O(V), Where V is the number of vertices in the graph. We have to maintain a visited array and list of size V.