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 |
Sum of dependencies is 4
Time complexity: O(V) where V is number of vertices in graph.
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