# Count number of edges in an undirected graph

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 implementation of above idea

## C++

`// 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; ` `} ` |

*chevron_right*

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## Python3

`# Python3 program to count number of ` `# edge in undirected graph ` ` ` `# Adjacency list representation of graph ` `class` `Graph: ` ` ` `def` `__init__(` `self` `, V): ` ` ` `self` `.V ` `=` `V ` ` ` `self` `.adj ` `=` `[[] ` `for` `i ` `in` `range` `(V)] ` ` ` ` ` `# add edge to graph ` ` ` `def` `addEdge (` `self` `, u, v ): ` ` ` `self` `.adj[u].append(v) ` ` ` `self` `.adj[v].append(u) ` ` ` ` ` `# Returns count of edge in undirected graph ` ` ` `def` `countEdges(` `self` `): ` ` ` `Sum` `=` `0` ` ` ` ` `# traverse all vertex ` ` ` `for` `i ` `in` `range` `(` `self` `.V): ` ` ` ` ` `# add all edge that are linked ` ` ` `# to the current vertex ` ` ` `Sum` `+` `=` `len` `(` `self` `.adj[i]) ` ` ` ` ` `# The count of edge is always even ` ` ` `# because in undirected graph every edge ` ` ` `# is connected twice between two vertices ` ` ` `return` `Sum` `/` `/` `2` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `V ` `=` `9` ` ` `g ` `=` `Graph(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` `) ` ` ` ` ` `print` `(g.countEdges()) ` ` ` `# This code is contributed by PranchalK ` |

*chevron_right*

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**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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