# Check if a directed graph is connected or not

Given a directed graph. The task is to check if the given graph is connected or not.

Examples:

Input:

Output: Yes

Input:

Output: No

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

Approach:

1. Take two bool arrays vis1 and vis2 of size N (number of nodes of a graph) and keep false in all indexes.
2. Start at a random vertex v of the graph G, and run a DFS(G, v).
3. Make all visited vertices v as vis1[v] = true.
4. Now reverse the direction of all the edges.
5. Start DFS at the vertex which was chosen at step 2.
6. Make all visited vertices v as vis2[v] = true.
7. If any vertex v has vis1[v] = false and vis2[v] = false then the graph is not connected.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std; #define N 100000    // To keep correct and reverse direction vector gr1[N], gr2[N];    bool vis1[N], vis2[N];    // Function to add edges void Add_edge(int u, int v) {     gr1[u].push_back(v);     gr2[v].push_back(u); }    // DFS function void dfs1(int x) {     vis1[x] = true;        for (auto i : gr1[x])         if (!vis1[i])             dfs1(i); }    // DFS function void dfs2(int x) {     vis2[x] = true;        for (auto i : gr2[x])         if (!vis2[i])             dfs2(i); }    bool Is_Connected(int n) {     // Call for correct direction     memset(vis1, false, sizeof vis1);     dfs1(1);        // Call for reverse direction     memset(vis2, false, sizeof vis2);     dfs2(1);        for (int i = 1; i <= n; i++) {            // If any vertex it not visited in any direction         // Then graph is not connected         if (!vis1[i] and !vis2[i])             return false;     }        // If graph is connected     return true; }    // Driver code int main() {     int n = 4;        // Add edges     Add_edge(1, 2);     Add_edge(1, 3);     Add_edge(2, 3);     Add_edge(3, 4);        // Function call     if (Is_Connected(n))         cout << "Yes";     else         cout << "No";        return 0; }

## Java

 // Java implementation of the approach import java.util.*;    class GFG  {     static int N = 100000;        // To keep correct and reverse direction     @SuppressWarnings("unchecked")     static Vector[] gr1 = new Vector[N];     @SuppressWarnings("unchecked")     static Vector[] gr2 = new Vector[N];        static boolean[] vis1 = new boolean[N];     static boolean[] vis2 = new boolean[N];        static {         for (int i = 0; i < N; i++)         {             gr1[i] = new Vector<>();             gr2[i] = new Vector<>();         }     }        // Function to add edges     static void Add_edge(int u, int v)     {         gr1[u].add(v);         gr2[v].add(u);     }        // DFS function     static void dfs1(int x)     {         vis1[x] = true;         for (int i : gr1[x])             if (!vis1[i])                 dfs1(i);     }        // DFS function     static void dfs2(int x)      {         vis2[x] = true;         for (int i : gr2[x])             if (!vis2[i])                 dfs2(i);     }        static boolean Is_connected(int n)     {            // Call for correct direction         Arrays.fill(vis1, false);         dfs1(1);            // Call for reverse direction         Arrays.fill(vis2, false);         dfs2(1);            for (int i = 1; i <= n; i++)         {                // If any vertex it not visited in any direction             // Then graph is not connected             if (!vis1[i] && !vis2[i])                 return false;         }            // If graph is connected         return true;     }        // Driver Code     public static void main(String[] args)     {         int n = 4;            // Add edges         Add_edge(1, 2);         Add_edge(1, 3);         Add_edge(2, 3);         Add_edge(3, 4);            // Function call         if (Is_connected(n))             System.out.println("Yes");         else             System.out.println("No");     } }    // This code is contributed by // sanjeev2552

## Python3

 # Python3 implementation of the approach  N = 100000    # To keep correct and reverse direction  gr1 = {}; gr2 = {};     vis1 = [0] * N; vis2 = [0] * N;     # Function to add edges  def Add_edge(u, v) :         if u not in gr1 :         gr1[u] = [];                if v not in gr2 :         gr2[v] = [];                gr1[u].append(v);     gr2[v].append(u);     # DFS function  def dfs1(x) :      vis1[x] = True;     if x not in gr1 :         gr1[x] = {};                for i in gr1[x] :         if (not vis1[i]) :             dfs1(i)     # DFS function  def dfs2(x) :         vis2[x] = True;         if x not in gr2 :         gr2[x] = {};                for i in gr2[x] :          if (not vis2[i]) :             dfs2(i);     def Is_Connected(n) :         global vis1;     global vis2;            # Call for correct direction     vis1 = [False] * len(vis1);     dfs1(1);            # Call for reverse direction     vis2 = [False] * len(vis2);     dfs2(1);            for i in range(1, n + 1) :                    # If any vertex it not visited in any direction         # Then graph is not connected         if (not vis1[i] and not vis2[i]) :             return False;                    # If graph is connected     return True;     # Driver code  if __name__ == "__main__" :         n = 4;         # Add edges      Add_edge(1, 2);      Add_edge(1, 3);      Add_edge(2, 3);      Add_edge(3, 4);         # Function call      if (Is_Connected(n)) :         print("Yes");      else :         print("No");     # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approach using System; using System.Collections.Generic;    class GFG  {     static int N = 100000;        // To keep correct and reverse direction     static List[] gr1 = new List[N];        static List[] gr2 = new List[N];        static bool[] vis1 = new bool[N];     static bool[] vis2 = new bool[N];        // Function to add edges     static void Add_edge(int u, int v)     {         gr1[u].Add(v);         gr2[v].Add(u);     }        // DFS function     static void dfs1(int x)     {         vis1[x] = true;         foreach (int i in gr1[x])             if (!vis1[i])                 dfs1(i);     }        // DFS function     static void dfs2(int x)      {         vis2[x] = true;         foreach (int i in gr2[x])             if (!vis2[i])                 dfs2(i);     }        static bool Is_connected(int n)     {            // Call for correct direction         for (int i = 0; i < n; i++)             vis1[i] = false;         dfs1(1);            // Call for reverse direction         for (int i = 0; i < n; i++)             vis2[i] = false;         dfs2(1);            for (int i = 1; i <= n; i++)         {                // If any vertex it not visited in any direction             // Then graph is not connected             if (!vis1[i] && !vis2[i])                 return false;         }            // If graph is connected         return true;     }        // Driver Code     public static void Main(String[] args)     {         int n = 4;         for (int i = 0; i < N; i++)         {             gr1[i] = new List();             gr2[i] = new List();         }                    // Add edges         Add_edge(1, 2);         Add_edge(1, 3);         Add_edge(2, 3);         Add_edge(3, 4);            // Function call         if (Is_connected(n))             Console.WriteLine("Yes");         else             Console.WriteLine("No");     } }    // This code is contributed by PrinciRaj1992

Output:

Yes

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.