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++

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// 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;
}

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Java

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// 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

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Output:

Sum of dependencies is 4

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

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