Minimum edges required to make a Directed Graph Strongly Connected

Given a Directed graph of N vertices and M edges, the task is to find the minimum number of edges required to make the given graph Strongly Connected.

Examples:

Input: N = 3, M = 3, source[] = {1, 2, 1}, destination[] = {2, 3, 3} 
Output: 1 
Explanation:
Adding a directed edge joining the pair of vertices {3, 1} makes the graph strongly connected. 
Hence, the minimum number of edges required is 1. 
Below is the illustration of the above example:  

Input: N = 5, M = 5, source[] = {1, 3, 1, 3, 4}, destination[] = {2, 2, 3, 4, 5} 
Output:  2 
Explanation:
Adding 2 directed edges to join the following pair of vertices makes the graph strongly connected: 

• {2, 1}
• {5, 2}

Hence, the minimum number of edges required is 2.

Approach:
For a Strongly Connected Graph, each vertex must have an in-degree and an out-degree of at least 1. Therefore, in order to make a graph strongly connected, each vertex must have an incoming edge and an outgoing edge. The maximum number of incoming edges and the outgoing edges required to make the graph strongly connected is the minimum edges required to make it strongly connected. 
Follow the steps below to solve the problem: 

• Find the count of in-degrees and out-degrees of each vertex of the graph, using DFS.
• If the in-degree or out-degree of a vertex is greater than 1, then consider it as only 1.
• Count the total in-degree and out-degree of the given graph.
• The minimum number of edges required to make the graph strongly connected is then given by max(N-totalIndegree, N-totalOutdegree).
• Print the count of minimum edges as the result.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Perform DFS to count the in-degree
// and out-degree of the graph
void dfs(int u, vector<int> adj[], int* vis, int* inDeg,
         int* outDeg)
{
    // Mark the source as visited
    vis[u] = 1;


    // Traversing adjacent nodes
    for (auto v : adj[u])
    {
        // Mark out-degree as 1
        outDeg[u] = 1;
        // Mark in-degree as 1
        inDeg[v] = 1;

        // If not visited
        if (vis[v] == 0)
        {
            // DFS Traversal on
            // adjacent vertex
            dfs(v, adj, vis, inDeg, outDeg);
        }
    }
}

// Function to return minimum number
// of edges required to make the graph
// strongly connected
int findMinimumEdges(int source[], int N, int M, int dest[])
{
    // For Adjacency List
    vector<int> adj[N + 1];

    // Create the Adjacency List
    for (int i = 0; i < M; i++)
    {
        adj[ source[i] ].push_back(dest[i]);
    }

    // Initialize the in-degree array
    int inDeg[N + 1] = { 0 };

    // Initialize the out-degree array
    int outDeg[N + 1] = { 0 };

    // Initialize the visited array
    int vis[N + 1] = { 0 };

    // Perform DFS to count in-degrees
    // and out-degreess
    dfs(1, adj, vis, inDeg, outDeg);

    // To store the result
    int minEdges = 0;

    // To store total count of in-degree
    // and out-degree
    int totalIndegree = 0;
    int totalOutdegree = 0;

    // Find total in-degree
    // and out-degree
    for (int i = 1; i <= N; i++)
    {
        if (inDeg[i] == 1)
            totalIndegree++;
        if (outDeg[i] == 1)
            totalOutdegree++;
    }

    // Calculate the minimum
    // edges required
    minEdges = max(N - totalIndegree, N - totalOutdegree);

    // Return the minimum edges
    return minEdges;
}

// Driver Code
int main()
{
    int N = 5, M = 5;

    int source[] = { 1, 3, 1, 3, 4 };
    int destination[] = { 2, 2, 3, 4, 5 };

    // Function call
    cout << findMinimumEdges(source, N, M, destination);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{

// Perform DFS to count the
// in-degree and out-degree
// of the graph
static void dfs(int u, Vector<Integer> adj[],
                int[] vis, int[] inDeg,
         int[] outDeg)
{
  // Mark the source
  // as visited
  vis[u] = 1;

  // Traversing adjacent nodes
  for (int v : adj[u])
  {
    // Mark out-degree as 1
      outDeg[u] = 1;
    // Mark in-degree as 1
      inDeg[v] = 1;

    // If not visited
    if (vis[v] == 0)
    {

      // DFS Traversal on
      // adjacent vertex
      dfs(v, adj, vis,
          inDeg, outDeg);
    }
  }
}

// Function to return minimum
// number of edges required
// to make the graph strongly
// connected
static int findMinimumEdges(int source[],
                            int N, int M,
                            int dest[])
{
  // For Adjacency List
  @SuppressWarnings("unchecked")
  Vector<Integer> []adj =
         new Vector[N + 1];

  for (int i = 0; i < adj.length; i++)
    adj[i] = new Vector<Integer>();

  // Create the Adjacency List
  for (int i = 0; i < M; i++)
  {
    adj].add(dest[i]);
  }

  // Initialize the in-degree array
  int inDeg[] = new int[N + 1];

  // Initialize the out-degree array
  int outDeg[] = new int[N + 1];

  // Initialize the visited array
  int vis[] = new int[N + 1];

  // Perform DFS to count
  // in-degrees and out-degreess
  dfs(1, adj, vis, inDeg, outDeg);

  // To store the result
  int minEdges = 0;

  // To store total count of
  // in-degree and out-degree
  int totalIndegree = 0;
  int totalOutdegree = 0;

  // Find total in-degree
  // and out-degree
  for (int i = 1; i <= N; i++)
  {
    if (inDeg[i] == 1)
      totalIndegree++;
    if (outDeg[i] == 1)
      totalOutdegree++;
  }

  // Calculate the minimum
  // edges required
  minEdges = Math.max(N - totalIndegree,
                      N - totalOutdegree);

  // Return the minimum edges
  return minEdges;
}

// Driver Code
public static void main(String[] args)
{
  int N = 5, M = 5;
  int source[] = {1, 3, 1, 3, 4};
  int destination[] = {2, 2, 3, 4, 5};

  // Function call
  System.out.print(findMinimumEdges(source,
                                    N, M,
                                    destination));
}
}

Python3

# Python3 program to implement
# the above approach

# Perform DFS to count the in-degree
# and out-degree of the graph
def dfs(u, adj, vis,inDeg, outDeg):

    # Mark the source as visited
    vis[u] = 1;

    # Traversing adjacent nodes
    for v in adj[u]:
        # Mark out-degree as 1
        outDeg[u] = 1;
        # Mark in-degree as 1
        inDeg[u] = 1;

        # If not visited
        if (vis[v] == 0):

            # DFS Traversal on
            # adjacent vertex
            dfs(v, adj, vis,
                inDeg, outDeg)

# Function to return minimum
# number of edges required
# to make the graph strongly
# connected
def findMinimumEdges(source, N,
                     M, dest):

    # For Adjacency List
    adj = [[] for i in range(N + 1)]

    # Create the Adjacency List
    for i in range(M):
        adj].append(dest[i]);


    # Initialize the in-degree array
    inDeg = [0 for i in range(N + 1)]

    # Initialize the out-degree array
    outDeg = [0 for i in range(N + 1)]

    # Initialize the visited array
    vis = [0 for i in range(N + 1)]

    # Perform DFS to count in-degrees
    # and out-degreess
    dfs(1, adj, vis, inDeg, outDeg);

    # To store the result
    minEdges = 0;

    # To store total count of
    # in-degree and out-degree
    totalIndegree = 0;
    totalOutdegree = 0;

    # Find total in-degree
    # and out-degree
    for i in range(1, N):
        if (inDeg[i] == 1):
            totalIndegree += 1;
        if (outDeg[i] == 1):
            totalOutdegree += 1;

    # Calculate the minimum
    # edges required
    minEdges = max(N - totalIndegree,
                   N - totalOutdegree);

    # Return the minimum edges
    return minEdges;

# Driver code
if __name__ == "__main__":

    N = 5
    M = 5

    source = [1, 3, 1, 3, 4]
    destination = [2, 2, 3, 4, 5]

    # Function call
    print(findMinimumEdges(source, N,
                           M, destination))

# This code is contributed by rutvik_56

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;

class GFG{

// Perform DFS to count the
// in-degree and out-degree
// of the graph
static void dfs(int u, List<int> []adj,
                int[] vis, int[] inDeg,
                int[] outDeg)
{

    // Mark the source
    // as visited
    vis[u] = 1;

    // Traversing adjacent nodes
    foreach (int v in adj[u])
    {
          // Mark out-degree as 1
        outDeg[u] = 1;
        // Mark in-degree as 1
        inDeg[v] = 1;

        // If not visited
        if (vis[v] == 0)
        {
            // DFS Traversal on
            // adjacent vertex
            dfs(v, adj, vis,
                inDeg, outDeg);
        }
    }
}

// Function to return minimum
// number of edges required
// to make the graph strongly
// connected
static int findMinimumEdges(int []source,
                            int N, int M,
                            int []dest)
{

    // For Adjacency List
    List<int> []adj = new List<int>[N + 1];

    for(int i = 0; i < adj.Length; i++)
        adj[i] = new List<int>();

    // Create the Adjacency List
    for(int i = 0; i < M; i++)
    {
        adj].Add(dest[i]);
    }

    // Initialize the in-degree array
    int []inDeg = new int[N + 1];

    // Initialize the out-degree array
    int []outDeg = new int[N + 1];

    // Initialize the visited array
    int []vis = new int[N + 1];

    // Perform DFS to count
    // in-degrees and out-degreess
    dfs(1, adj, vis, inDeg, outDeg);

    // To store the result
    int minEdges = 0;

    // To store total count of
    // in-degree and out-degree
    int totalIndegree = 0;
    int totalOutdegree = 0;

    // Find total in-degree
    // and out-degree
    for (int i = 1; i <= N; i++)
    {
        if (inDeg[i] == 1)
            totalIndegree++;
        if (outDeg[i] == 1)
            totalOutdegree++;
    }

    // Calculate the minimum
    // edges required
    minEdges = Math.Max(N - totalIndegree,
                        N - totalOutdegree);

    // Return the minimum edges
    return minEdges;
}

// Driver Code
public static void Main(String[] args)
{
    int N = 5, M = 5;
    int []source = { 1, 3, 1, 3, 4 };
    int []destination = { 2, 2, 3, 4, 5 };

    // Function call
    Console.Write(findMinimumEdges(source,
                                   N, M,
                                   destination));
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>
// Javascript program to implement
// the above approach

// Perform DFS to count the
// in-degree and out-degree
// of the graph
function dfs(u, adj, vis, inDeg, outDeg)
{

    // Mark the source
    // as visited
    vis[u] = 1;

    // Traversing adjacent nodes
    for(var v of adj[u])
    {
          // Mark out-degree as 1
        outDeg[u] = 1;
        // Mark in-degree as 1
        inDeg[v] = 1;

        // If not visited
        if (vis[v] == 0)
        {
            // DFS Traversal on
            // adjacent vertex
            dfs(v, adj, vis,
                inDeg, outDeg);
        }
    }
}

// Function to return minimum
// number of edges required
// to make the graph strongly
// connected
function findMinimumEdges(source, N, M, dest)
{

    // For Adjacency List
    var adj = Array.from(Array(N+1), ()=>Array());

    // Create the Adjacency List
    for(var i = 0; i < M; i++)
    {
        adj].push(dest[i]);
    }

    // Initialize the in-degree array
    var inDeg = Array(N+1).fill(0);

    // Initialize the out-degree array
    var outDeg = Array(N+1).fill(0);

    // Initialize the visited array
    var vis = Array(N+1).fill(0);

    // Perform DFS to count
    // in-degrees and out-degreess
    dfs(1, adj, vis, inDeg, outDeg);

    // To store the result
    var minEdges = 0;

    // To store total count of
    // in-degree and out-degree
    var totalIndegree = 0;
    var totalOutdegree = 0;

    // Find total in-degree
    // and out-degree
    for (var i = 1; i <= N; i++)
    {
        if (inDeg[i] == 1)
            totalIndegree++;
        if (outDeg[i] == 1)
            totalOutdegree++;
    }

    // Calculate the minimum
    // edges required
    minEdges = Math.max(N - totalIndegree,
                        N - totalOutdegree);

    // Return the minimum edges
    return minEdges;
}

// Driver Code
var N = 5, M = 5;
var source = [1, 3, 1, 3, 4];
var destination = [2, 2, 3, 4, 5];

// Function call
document.write(findMinimumEdges(source,
                               N, M,
                               destination));


</script>

Output

2

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