# Check if a graph is strongly connected | Set 1 (Kosaraju using DFS)

Given a directed graph, find out whether the graph is strongly connected or not. A directed graph is strongly connected if there is a path between any two pair of vertices. For example, following is a strongly connected graph.

It is easy for undirected graph, we can just do a BFS and DFS starting from any vertex. If BFS or DFS visits all vertices, then the given undirected graph is connected. This approach won’t work for a directed graph. For example, consider the following graph which is not strongly connected. If we start DFS (or BFS) from vertex 0, we can reach all vertices, but if we start from any other vertex, we cannot reach all vertices.

How to do for directed graph?

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple idea is to use a all pair shortest path algorithm like Floyd Warshall or find Transitive Closure of graph. Time complexity of this method would be O(v3).

We can also do DFS V times starting from every vertex. If any DFS, doesn’t visit all vertices, then graph is not strongly connected. This algorithm takes O(V*(V+E)) time which can be same as transitive closure for a dense graph.

A better idea can be Strongly Connected Components (SCC) algorithm. We can find all SCCs in O(V+E) time. If number of SCCs is one, then graph is strongly connected. The algorithm for SCC does extra work as it finds all SCCs.

Following is Kosaraju’s DFS based simple algorithm that does two DFS traversals of graph:
1) Initialize all vertices as not visited.

2) Do a DFS traversal of graph starting from any arbitrary vertex v. If DFS traversal doesn’t visit all vertices, then return false.

3) Reverse all arcs (or find transpose or reverse of graph)

4) Mark all vertices as not-visited in reversed graph.

5) Do a DFS traversal of reversed graph starting from same vertex v (Same as step 2). If DFS traversal doesn’t visit all vertices, then return false. Otherwise return true.

The idea is, if every node can be reached from a vertex v, and every node can reach v, then the graph is strongly connected. In step 2, we check if all vertices are reachable from v. In step 4, we check if all vertices can reach v (In reversed graph, if all vertices are reachable from v, then all vertices can reach v in original graph).

Following is the implementation of above algorithm.

## C++

```// C++ program to check if a given directed graph is strongly
// connected or not
#include <iostream>
#include <list>
#include <stack>
using namespace std;

class Graph
{
int V;    // No. of vertices
list<int> *adj;    // An array of adjacency lists

// A recursive function to print DFS starting from v
void DFSUtil(int v, bool visited[]);
public:
// Constructor and Destructor
Graph(int V) { this->V = V;  adj = new list<int>[V];}
~Graph() { delete [] adj; }

// Method to add an edge
void addEdge(int v, int w);

// The main function that returns true if the graph is strongly
// connected, otherwise false
bool isSC();

// Function that returns reverse (or transpose) of this graph
Graph getTranspose();
};

// A recursive function to print DFS starting from v
void Graph::DFSUtil(int v, bool visited[])
{
// Mark the current node as visited and print it
visited[v] = true;

// 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);
}

// Function that returns reverse (or transpose) of this graph
Graph Graph::getTranspose()
{
Graph g(V);
for (int v = 0; v < V; v++)
{
// Recur for all the vertices adjacent to this vertex
list<int>::iterator i;
for(i = adj[v].begin(); i != adj[v].end(); ++i)
{
g.adj[*i].push_back(v);
}
}
return g;
}

void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}

// The main function that returns true if graph is strongly connected
bool Graph::isSC()
{
// St1p 1: Mark all the vertices as not visited (For first DFS)
bool visited[V];
for (int i = 0; i < V; i++)
visited[i] = false;

// Step 2: Do DFS traversal starting from first vertex.
DFSUtil(0, visited);

// If DFS traversal doesn’t visit all vertices, then return false.
for (int i = 0; i < V; i++)
if (visited[i] == false)
return false;

// Step 3: Create a reversed graph
Graph gr = getTranspose();

// Step 4: Mark all the vertices as not visited (For second DFS)
for(int i = 0; i < V; i++)
visited[i] = false;

// Step 5: Do DFS for reversed graph starting from first vertex.
// Staring Vertex must be same starting point of first DFS
gr.DFSUtil(0, visited);

// If all vertices are not visited in second DFS, then
// return false
for (int i = 0; i < V; i++)
if (visited[i] == false)
return false;

return true;
}

// Driver program to test above functions
int main()
{
// Create graphs given in the above diagrams
Graph g1(5);
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.isSC()? cout << "Yes\n" : cout << "No\n";

Graph g2(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
g2.isSC()? cout << "Yes\n" : cout << "No\n";

return 0;
}
```

## Java

```// Java program to check if a given directed graph is strongly
// connected or not
import java.io.*;
import java.util.*;
import java.util.LinkedList;

// This class represents a directed graph using adjacency
// list representation
class Graph
{
private int V;   // No. of vertices
private LinkedList<Integer> adj[]; //Adjacency List

//Constructor
Graph(int v)
{
V = v;
adj = new LinkedList[v];
for (int i=0; i<v; ++i)
adj[i] = new LinkedList();
}

//Function to add an edge into the graph
void addEdge(int v,int w) {  adj[v].add(w); }

// A recursive function to print DFS starting from v
void DFSUtil(int v,Boolean visited[])
{
// Mark the current node as visited and print it
visited[v] = true;

int n;

// Recur for all the vertices adjacent to this vertex
Iterator<Integer> i = adj[v].iterator();
while (i.hasNext())
{
n = i.next();
if (!visited[n])
DFSUtil(n,visited);
}
}

// Function that returns transpose of this graph
Graph getTranspose()
{
Graph g = new Graph(V);
for (int v = 0; v < V; v++)
{
// Recur for all the vertices adjacent to this vertex
Iterator<Integer> i = adj[v].listIterator();
while (i.hasNext())
g.adj[i.next()].add(v);
}
return g;
}

// The main function that returns true if graph is strongly
// connected
Boolean isSC()
{
// Step 1: Mark all the vertices as not visited
// (For first DFS)
Boolean visited[] = new Boolean[V];
for (int i = 0; i < V; i++)
visited[i] = false;

// Step 2: Do DFS traversal starting from first vertex.
DFSUtil(0, visited);

// If DFS traversal doesn't visit all vertices, then
// return false.
for (int i = 0; i < V; i++)
if (visited[i] == false)
return false;

// Step 3: Create a reversed graph
Graph gr = getTranspose();

// Step 4: Mark all the vertices as not visited (For
// second DFS)
for (int i = 0; i < V; i++)
visited[i] = false;

// Step 5: Do DFS for reversed graph starting from
// first vertex. Staring Vertex must be same starting
// point of first DFS
gr.DFSUtil(0, visited);

// If all vertices are not visited in second DFS, then
// return false
for (int i = 0; i < V; i++)
if (visited[i] == false)
return false;

return true;
}

public static void main(String args[])
{
// Create graphs given in the above diagrams
Graph g1 = new Graph(5);
g1.addEdge(0, 1);
g1.addEdge(1, 2);
g1.addEdge(2, 3);
g1.addEdge(3, 0);
g1.addEdge(2, 4);
g1.addEdge(4, 2);
if (g1.isSC())
System.out.println("Yes");
else
System.out.println("No");

Graph g2 = new Graph(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
if (g2.isSC())
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Aakash Hasija
```

## Python

```# Python program to check if a given directed graph is strongly
# connected or not

from collections import defaultdict

#This class represents a directed graph using adjacency list representation
class Graph:

def __init__(self,vertices):
self.V= vertices #No. of vertices
self.graph = defaultdict(list) # default dictionary to store graph

# function to add an edge to graph
def addEdge(self,u,v):
self.graph[u].append(v)

#A function used by isSC() to perform DFS
def DFSUtil(self,v,visited):

# Mark the current node as visited
visited[v]= True

#Recur for all the vertices adjacent to this vertex
for i in self.graph[v]:
if visited[i]==False:
self.DFSUtil(i,visited)

# Function that returns reverse (or transpose) of this graph
def getTranspose(self):

g = Graph(self.V)

# Recur for all the vertices adjacent to this vertex
for i in self.graph:
for j in self.graph[i]:
g.addEdge(j,i)

return g

# The main function that returns true if graph is strongly connected
def isSC(self):

# Step 1: Mark all the vertices as not visited (For first DFS)
visited =[False]*(self.V)

# Step 2: Do DFS traversal starting from first vertex.
self.DFSUtil(0,visited)

# If DFS traversal doesnt visit all vertices, then return false
if any(i == False for i in visited):
return False

# Step 3: Create a reversed graph
gr = self.getTranspose()

# Step 4: Mark all the vertices as not visited (For second DFS)
visited =[False]*(self.V)

# Step 5: Do DFS for reversed graph starting from first vertex.
# Staring Vertex must be same starting point of first DFS
gr.DFSUtil(0,visited)

# If all vertices are not visited in second DFS, then
# return false
if any(i == False for i in visited):
return False

return True

# Create a graph given in the above diagram
g1 = Graph(5)
g1.addEdge(0, 1)
g1.addEdge(1, 2)
g1.addEdge(2, 3)
g1.addEdge(3, 0)
g1.addEdge(2, 4)
g1.addEdge(4, 2)
print "Yes" if g1.isSC() else "No"

g2 = Graph(4)
g2.addEdge(0, 1)
g2.addEdge(1, 2)
g2.addEdge(2, 3)
print "Yes" if g2.isSC() else "No"

#This code is contributed by Neelam Yadav
```

Output:
```Yes
No```

Time Complexity: Time complexity of above implementation is sane as Depth First Search which is O(V+E) if the graph is represented using adjacency list representation.

Can we improve further?
The above approach requires two traversals of graph. We can find whether a graph is strongly connected or not in one traversal using Tarjan’s Algorithm to find Strongly Connected Components.

Exercise:
Can we use BFS instead of DFS in above algorithm? See this.

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