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Sum of dependencies in a graph

Given a directed and connected graph with n nodes. If there is an edge from u to v then u depends on v. Our task was to find out the sum of dependencies for every node.

Example:
For the graph in diagram,
A depends on C and D i.e. 2
B depends on C i.e. 1
D depends on C i.e. 1
And C depends on none.
Hence answer -> 0 + 1 + 1 + 2 = 4

Asked in : Flipkart Interview



Idea is to check adjacency list and find how many edges are there from each vertex and return the total number of edges.

C++

// C++ program to find the sum of dependencies
#include <bits/stdc++.h>
using namespace std;
  
// To add an edge
void addEdge(vector <int> adj[], int u,int v)
{
    adj[u].push_back(v);
}
  
// find the sum of all dependencies
int findSum(vector<int> adj[], int V)
{
    int sum = 0;
  
    // just find the size at each vector's index
    for (int u = 0; u < V; u++)
        sum += adj[u].size();
  
    return sum;
}
  
// Driver code
int main()
{
    int V = 4;
    vector<int >adj[V];
    addEdge(adj, 0, 2);
    addEdge(adj, 0, 3);
    addEdge(adj, 1, 3);
    addEdge(adj, 2, 3);
  
    cout << "Sum of dependencies is "
         << findSum(adj, V);
    return 0;
}

Java

// Java program to find the sum of dependencies
  
import java.util.Vector;
  
class Test
{
    // To add an edge
    static void addEdge(Vector <Integer> adj[], int u,int v)
    {
        adj[u].addElement((v));
    }
      
    // find the sum of all dependencies
    static int findSum(Vector<Integer> adj[], int V)
    {
        int sum = 0;
       
        // just find the size at each vector's index
        for (int u = 0; u < V; u++)
            sum += adj[u].size();
       
        return sum;
    }
      
    // Driver method
    public static void main(String[] args) 
    {
        int V = 4;
        Vector<Integer> adj[] = new Vector[V];
          
        for (int i = 0; i < adj.length; i++) {
            adj[i] = new Vector<>();
        }
          
        addEdge(adj, 0, 2);
        addEdge(adj, 0, 3);
        addEdge(adj, 1, 3);
        addEdge(adj, 2, 3);
       
        System.out.println("Sum of dependencies is " +
                            findSum(adj, V));
    }
}
// This code is contributed by Gaurav Miglani


Output:

Sum of dependencies is 4

Time complexity : O(V) where V is number of vertices in graph.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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