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# 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:
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:
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 using namespace std;    // Perform DFS to count the in-degree // and out-degree of the graph void dfs(int u, vector 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 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; }

## 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

## Javascript



Output

2

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

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