Find the maximum number of handshakes

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++

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// 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;

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Java

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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.
?>

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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.

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Improved By : nitin mittal, vt_m



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