Given an adjacency list representation undirected graph. Write a function to count the number of edges in the undirected graph.

Expected time complexity : O(V)

Examples:

Input : Adjacency list representation of below graph. Output : 9

Idea is based on Handshaking Lemma. Handshaking lemma is about undirected graph. In every finite undirected graph number of vertices with odd degree is always even. The handshaking lemma is a consequence of the degree sum formula (also sometimes called the handshaking lemma)

So we traverse all vertices, compute sum of sizes of their adjacency lists, and finally returns sum/2. Below c++ implementation of above idea

// C++ program to count number of edge in // undirected graph #include<bits/stdc++.h> using namespace std; // Adjacency list representation of graph class Graph { int V ; list < int > *adj; public : Graph( int V ) { this->V = V ; adj = new list<int>[V]; } void addEdge ( int u, int v ) ; int countEdges () ; }; // add edge to graph void Graph :: addEdge ( int u, int v ) { adj[u].push_back(v); adj[v].push_back(u); } // Returns count of edge in undirected graph int Graph :: countEdges() { int sum = 0; //traverse all vertex for (int i = 0 ; i < V ; i++) // add all edge that are linked to the // current vertex sum += adj[i].size(); // The count of edge is always even because in // undirected graph every edge is connected // twice between two vertices return sum/2; } // driver program to check above function int main() { int V = 9 ; Graph g(V); // making above uhown graph g.addEdge(0, 1 ); g.addEdge(0, 7 ); g.addEdge(1, 2 ); g.addEdge(1, 7 ); g.addEdge(2, 3 ); g.addEdge(2, 8 ); g.addEdge(2, 5 ); g.addEdge(3, 4 ); g.addEdge(3, 5 ); g.addEdge(4, 5 ); g.addEdge(5, 6 ); g.addEdge(6, 7 ); g.addEdge(6, 8 ); g.addEdge(7, 8 ); cout << g.countEdges() << endl; return 0; }

Output:

14

Time Complexity : O(V)

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