Find the maximum number of handshakes

02 May, 2022
There are N persons in a room. Find the maximum number of Handshakes possible. Given the fact that any two persons shake hand exactly once.
Examples : 
Input : N = 2
Output : 1.
There are only 2 persons in the room. 1 handshake take place.

Input : N = 10
Output : 45.
To maximize the number of handshakes, each person should shake hand with every other person in the room. For the first person, there would be N-1 handshakes. For second person there would N-1 person available but he had already shaken hand with the first person. So there would be N-2 handshakes and so on. So, Total number of handshake = N-1 + N-2 +....+ 1 + 0, which is equivalent to ((N-1)*N)/2 (using the formula of sum of first N natural number).
Below is the implementation of this problem.  
C++
C
Java
Python3
C#
PHP
Javascript
// C++ program to find maximum number of handshakes.#include <bits/stdc++.h>using namespace std; // Calculating the maximum number of handshake using derived formula.int maxHandshake(int n) { return (n * (n - 1)) / 2; } // Driver Codeint main(){    int n = 10;    cout << maxHandshake(n) << endl;     return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
// C program to find maximum number of handshakes.#include <stdio.h>// Calculating the maximum number of handshake using derived formula.int maxHandshake(int n) { return (n * (n - 1)) / 2; } // Driver Codeint main(){    int n = 10;    printf("%d\n", maxHandshake(n));     return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
// Java program to find maximum number of handshakes. class GFG {    // Calculating the maximum number of handshake using    // derived formula.    static int maxHandshake(int n)    {        return (n * (n - 1)) / 2;    }     // Driver code    public static void main(String[] args)    {        int n = 10;        System.out.println(maxHandshake(n));    }} // This code is contributed by Aditya Kumar (adityakumar129)
# Python3 program to find maximum# number of handshakes. # Calculating the maximum number# of handshake using derived formula.def maxHandshake(n):     return int((n * (n - 1)) / 2) # Driver Coden = 10print(maxHandshake(n)) # This code is contributed by Smitha Dinesh Semwal.
// C# program to find maximum number of// handshakes.using System; class GFG{    // Calculating the maximum number of    // handshake using derived formula.    static int maxHandshake(int n)    {        return (n * (n - 1)) / 2;    }              // Driver code    public static void Main ()    {        int n = 10;        Console.Write( maxHandshake(n));    }} // This code is contributed by nitin mittal.
<?php// PHP program to// find maximum number// of handshakes. // Calculating the maximum// number of handshake// using derived formula.function maxHandshake($n){    return ($n * ($n - 1))/ 2;} // Driver Code$n = 10;echo maxHandshake($n); // This code is contributed by anuj_67.?>
<script> // JavaScript program to find maximum// number of handshakes. // Calculating the maximum number of// handshake using derived formula.function maxHandshake(n){    return (n * (n - 1)) / 2;}  // Driver Codelet n = 10; document.write( maxHandshake(n)); // This code is contributed by souravghosh0416  </script>
Output: 
45
Time Complexity : O(1)
This article is contributed by Anuj Chauhan(anuj0503). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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