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Number of Simple Graph with N Vertices and M Edges

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  • Difficulty Level : Easy
  • Last Updated : 23 Jun, 2022

Given two integers N and M, the task is to count the number of simple undirected graphs that can be drawn with N vertices and M edges. A simple graph is a graph that does not contain multiple edges and self-loops.

Examples: 

Input: N = 3, M = 1 
Output:
The 3 graphs are {1-2, 3}, {2-3, 1}, {1-3, 2}.

Input: N = 5, M = 1 
Output: 10 

Approach: The N vertices are numbered from 1 to N. As there are no self-loops or multiple edges, the edge must be present between two different vertices. So the number of ways we can choose two different vertices is NC2 which is equal to (N * (N – 1)) / 2. Assume it P
Now M edges must be used with these pairs of vertices, so the number of ways to choose M pairs of vertices between P pairs will be PCM
If P < M then the answer will be 0 as the extra edges can not be left alone.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the value of
// Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
 
    if (k > n)
        return 0;
 
    int res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate the value of
    // [n * (n - 1) *---* (n - k + 1)] / [k * (k - 1) * ... * 1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// Driver Code
int main()
{
    int N = 5, M = 1;
 
    int P = (N * (N - 1)) / 2;
 
    cout << binomialCoeff(P, M);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
    // Function to return the value of
    // Binomial Coefficient C(n, k)
    static int binomialCoeff(int n, int k)
    {
 
        if (k > n)
            return 0;
 
        int res = 1;
 
        // Since C(n, k) = C(n, n-k)
        if (k > n - k)
            k = n - k;
 
        // Calculate the value of
        // [n * (n - 1) *---* (n - k + 1)] /
        // [k * (k - 1) * ... * 1]
        for (int i = 0; i < k; ++i)
        {
            res *= (n - i);
            res /= (i + 1);
        }
        return res;
    }
 
// Driver Code
public static void main(String[] args)
{
    int N = 5, M = 1;
    int P = (N * (N - 1)) / 2;
 
    System.out.println(binomialCoeff(P, M));
}
}
 
// This code is contributed by Shivi_Aggarwal

Python 3




# Python 3 implementation of the approach
 
# Function to return the value of
# Binomial Coefficient C(n, k)
def binomialCoeff(n, k):
 
    if (k > n):
        return 0
 
    res = 1
 
    # Since C(n, k) = C(n, n-k)
    if (k > n - k):
        k = n - k
 
    # Calculate the value of
    # [n * (n - 1) *---* (n - k + 1)] /
    # [k * (k - 1) * ... * 1]
    for i in range( k):
        res *= (n - i)
        res //= (i + 1)
 
    return res
 
# Driver Code
if __name__=="__main__":
     
    N = 5
    M = 1
 
    P = (N * (N - 1)) // 2
 
    print(binomialCoeff(P, M))
 
# This code is contributed by ita_c

C#




// C# implementation of the approach
using System;
 
class GFG
{
// Function to return the value of
// Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
 
    if (k > n)
        return 0;
 
    int res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate the value of
    // [n * (n - 1) *---* (n - k + 1)] /
    // [k * (k - 1) * ... * 1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// Driver Code
public static void Main()
{
    int N = 5, M = 1;
 
    int P = (N * (N - 1)) / 2;
 
    Console.Write(binomialCoeff(P, M));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP




<?php
// PHP implementation of the approach
 
// Function to return the value of
// Binomial Coefficient C(n, k)
function binomialCoeff($n, $k)
{
    if ($k > $n)
        return 0;
 
    $res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if ($k > $n - $k)
        $k = $n - $k;
 
    // Calculate the value of
    // [n * (n - 1) *---* (n - k + 1)] /
    // [k * (k - 1) * ... * 1]
    for ($i = 0; $i < $k; ++$i)
    {
        $res *= ($n - $i);
        $res /= ($i + 1);
    }
 
    return $res;
}
 
// Driver Code
$N = 5;
$M = 1;
 
$P = floor(($N * ($N - 1)) / 2);
 
echo binomialCoeff($P, $M);
 
// This code is contributed by Ryuga
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the value of
// Binomial Coefficient C(n, k)
function binomialCoeff(n, k)
{
 
    if (k > n)
        return 0;
 
    var res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate the value of
    // [n * (n - 1) *---* (n - k + 1)] / [k * (k - 1) * ... * 1]
    for (var i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// Driver Code
var N = 5, M = 1;
var P = (N * (N - 1)) / 2;
document.write( binomialCoeff(P, M));
 
</script>

Output: 

10

 

Time Complexity: O(M), where M is the number of edges.

Space Complexity: O(1), since no extra space has been taken.


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