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; ` `} ` |

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## 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 ` |

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Output:

Sum of dependencies is 4

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

This article is contributed by **Sahil Chhabra (akku)**. 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.

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