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Minimum edges required to make a Directed Graph Strongly Connected

Last Updated : 16 Apr, 2024
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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++ 
#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 from all unvisited vertices
    for (int i = 1; i <= N; ++i)
    {
        if (vis[i] == 0)
        {
            dfs(i, 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 
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[source[i]].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 from all unvisited vertices
        for (int i = 1; i <= N; ++i) {
            if (vis[i] == 0) {
                dfs(i, 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 
# 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[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
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[source[i]].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 from all unvisited vertices
    for i in range(1, N + 1):
        if vis[i] == 0:
            dfs(i, 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 + 1):
        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))
C# 
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[source[i]].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 from all unvisited vertices
        for (int i = 1; i <= N; ++i) {
            if (vis[i] == 0) {
                dfs(i, 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));
    }
}
Javascript 
<script>
// 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[source[i]].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 from all unvisited vertices
    for (var i = 1; i <= N; i++) 
    {
        if (vis[i] == 0) 
        {
            dfs(i, 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)



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