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. 

Implementation:

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;
          @SuppressWarnings("unchecked")
        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

Python3

# Python3 program to find the sum
# of dependencies

# To add an edge
def addEdge(adj, u, v):

    adj[u].append(v)

# Find the sum of all dependencies
def findSum(adj, V):

    sum = 0

    # Just find the size at each
    # vector's index
    for u in range(V):
        sum += len(adj[u])

    return sum

# Driver code
if __name__=='__main__':

    V = 4
    adj = [[] for i in range(V)]

    addEdge(adj, 0, 2)
    addEdge(adj, 0, 3)
    addEdge(adj, 1, 3)
    addEdge(adj, 2, 3)

    print("Sum of dependencies is",
          findSum(adj, V))

# This code is contributed by rutvik_56

C#

// C# program to find the sum of dependencies
using System;
using System.Collections;

class GFG{

// To add an edge
static void addEdge(ArrayList []adj, int u,
                                     int v)
{
    adj[u].Add(v);
}

// Find the sum of all dependencies
static int findSum(ArrayList []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].Count;

    return sum;
}

// Driver code
public static void Main(string[] args)
{
    int V = 4;

    ArrayList []adj = new ArrayList[V];

    for(int i = 0; i < V; i++)
    {
        adj[i] = new ArrayList();
    }

    addEdge(adj, 0, 2);
    addEdge(adj, 0, 3);
    addEdge(adj, 1, 3);
    addEdge(adj, 2, 3);

    Console.Write("Sum of dependencies is " +
                  findSum(adj, V));
}
}

// This code is contributed by pratham76

Output

Sum of dependencies is 4

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

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