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)

This article is contributed by **Nishant Singh**. 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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