Count number of edges in an undirected graph
Last Updated :
06 Feb, 2023
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++
#include<bits/stdc++.h>
using namespace std;
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 () ;
};
void Graph :: addEdge ( int u, int v )
{
adj[u].push_back(v);
adj[v].push_back(u);
}
int Graph :: countEdges()
{
int sum = 0;
for ( int i = 0 ; i < V ; i++)
sum += adj[i].size();
return sum/2;
}
int main()
{
int V = 9 ;
Graph g(V);
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;
}
|
Java
import java.io.*;
import java.util.*;
class Graph
{
int V;
Vector<Integer>[] adj;
Graph( int V)
{
this .V = V;
this .adj = new Vector[V];
for ( int i = 0 ; i < V; i++)
adj[i] = new Vector<Integer>();
}
void addEdge( int u, int v)
{
adj[u].add(v);
adj[v].add(u);
}
int countEdges()
{
int sum = 0 ;
for ( int i = 0 ; i < V; i++)
sum += adj[i].size();
return sum / 2 ;
}
}
class GFG
{
public static void main(String[] args) throws IOException
{
int V = 9 ;
Graph g = new Graph(V);
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 );
System.out.println(g.countEdges());
}
}
|
Python3
class Graph:
def __init__( self , V):
self .V = V
self .adj = [[] for i in range (V)]
def addEdge ( self , u, v ):
self .adj[u].append(v)
self .adj[v].append(u)
def countEdges( self ):
Sum = 0
for i in range ( self .V):
Sum + = len ( self .adj[i])
return Sum / / 2
if __name__ = = '__main__' :
V = 9
g = Graph(V)
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())
|
C#
using System;
using System.Collections.Generic;
class Graph
{
public int V;
public List< int >[] adj;
public Graph( int V)
{
this .V = V;
this .adj = new List< int >[V];
for ( int i = 0; i < V; i++)
adj[i] = new List< int >();
}
public void addEdge( int u, int v)
{
adj[u].Add(v);
adj[v].Add(u);
}
public int countEdges()
{
int sum = 0;
for ( int i = 0; i < V; i++)
sum += adj[i].Count;
return sum / 2;
}
}
class GFG
{
public static void Main(String[] args)
{
int V = 9;
Graph g = new Graph(V);
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);
Console.WriteLine(g.countEdges());
}
}
|
Javascript
class Graph {
constructor(V) {
this .V = V;
this .adj = new Array(V);
for (let i = 0; i < V; i++) {
this .adj[i] = [];
}
}
addEdge(u, v) {
this .adj[u].push(v);
this .adj[v].push(u);
}
countEdges() {
let sum = 0;
for (let i = 0; i < this .V; i++) {
sum += this .adj[i].length;
}
return sum / 2;
}
}
function main() {
let V = 9;
let g = new Graph(V);
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);
console.log(g.countEdges());
}
main();
|
Output:
14
Time Complexity: O(V)
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