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Find the maximum number of handshakes

  • Difficulty Level : Basic
  • Last Updated : 25 Mar, 2021

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++ 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 Code
int main()
{
  int n = 10;
  cout << maxHandshake(n) <<endl;
 
  return 0;
}

Java




// 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 Anant Agarwal.

Python3




# 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 Code
n = 10
print(maxHandshake(n))
 
# This code is contributed by Smitha Dinesh Semwal.

C#




// 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
// 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.
?>

Javascript




<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 Code
let 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 contribute.geeksforgeeks.org or mail your article to contribute@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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