Number of handshakes such that a person shakes hands only once
There is N number of people at a party. Find the total number of handshakes such that a person can handshake only once.
Examples:
Input : 5 Output : 10 Input : 9 Output : 36
We can see a recursive nature in the problem.
// n-th person has (n-1) choices and after // n-th person chooses a person, problem // recurs for n-1. handshake(n) = (n-1) + handshake(n-1) // Base case handshake(0) = 0
Below is the implementation of the above recursive formula.
C++
// Recursive C++ program to count total number of handshakes // when a person can shake hand with only one. #include <bits/stdc++.h> using namespace std; // Function to find all possible handshakes int handshake( int n) { // When n becomes 0 that means all the persons have done // handshake with other if (n == 0) return 0; else return (n - 1) + handshake(n - 1); } // Driver code int main() { int n = 9; cout << " " << handshake(n); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
C
// Recursive C program to count total number of handshakes // when a person can shake hand with only one. #include <stdio.h> // function to find all possible handshakes int handshake( int n) { // when n becomes 0 that means all the persons have done // handshake with other if (n == 0) return 0; else return (n - 1) + handshake(n - 1); } int main() { int n = 9; printf ( "%d" , handshake(n)); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Recursive Java program to count total number of // handshakes when a person can shake hand with only one. import java.io.*; class GFG { // function to find all possible handshakes static int handshake( int n) { // when n becomes 0 that means all the persons have // done handshake with other if (n == 0 ) return 0 ; else return (n - 1 ) + handshake(n - 1 ); } // Driver Code public static void main(String[] args) { int n = 9 ; System.out.print(handshake(n)); } } // This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Recursive Python program # to count total number of # handshakes when a person # can shake hand with only one. # function to find all # possible handshakes def handshake(n): # when n becomes 0 that means # all the persons have done # handshake with other if (n = = 0 ): return 0 else : return (n - 1 ) + handshake(n - 1 ) # Driver Code n = 9 print (handshake(n)) # This code is contributed # by Shivi_Aggarwal |
C#
// Recursive C# program to // count total number of // handshakes when a person // can shake hand with only one. using System; class GFG { // function to find all // possible handshakes static int handshake( int n) { // when n becomes 0 that // means all the persons // have done handshake // with other if (n == 0) return 0; else return (n - 1) + handshake(n - 1); } // Driver Code public static void Main (String []args) { int n = 9; Console.WriteLine(handshake(n)); } } // This code is contributed // by Arnab Kundu |
PHP
<?php // Recursive PHP program to // count total number of // handshakes when a person // can shake hand with only one. // function to find all // possible handshakes function handshake( $n ) { // when n becomes 0 that means // all the persons have done // handshake with other if ( $n == 0) return 0; else return ( $n - 1) + handshake( $n - 1); } // Driver Code $n = 9; echo (handshake( $n )); // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script> // Recursive JavaScript program to // count total number of // handshakes when a person // can shake hand with only one. // function to find all // possible handshakes function handshake(n) { // when n becomes 0 that // means all the persons // have done handshake // with other if (n === 0) return 0; else return n - 1 + handshake(n - 1); } // Driver Code var n = 9; document.write(handshake(n)); </script> |
Output:
36
Time Complexity: O(n)
Auxiliary Space: O(1), As the function is tail recursive the extra stack space will not be used.
We can come up with a direct formula by expanding the recursion.
handshake(n) = (n-1) + handshake(n-1) = (n-1) + (n-2) + handshake(n-2) = (n-1) + (n-2) + .... 1 + 0 = n * (n - 1)/2
C++
// Recursive CPP program to count total number of handshakes // when a person can shake hand with only one. #include <bits/stdc++.h> using namespace std; // function to find all possible handshakes int handshake( int n) { return n * (n - 1) / 2; } int main() { int n = 9; cout << handshake(n) << endl; return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
C
// Recursive CPP program to count total number of handshakes // when a person can shake hand with only one. #include <stdio.h> // function to find all possible handshakes int handshake( int n) { return n * (n - 1) / 2; } int main() { int n = 9; printf ( "%d" , handshake(n)); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Recursive Java program to count total number of // handshakes when a person can shake hand with only one. class GFG { // function to find all possible handshakes static int handshake( int n) { return n * (n - 1 ) / 2 ; } // Driver code public static void main(String args[]) { int n = 9 ; System.out.println(handshake(n)); } } // This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Recursive Python program # to count total number of # handshakes when a person # can shake hand with only one. # function to find all # possible handshakes def handshake(n): return int (n * (n - 1 ) / 2 ) # Driver Code n = 9 print (handshake(n)) # This code is contributed # by Shivi_Aggarwal |
C#
// Recursive C# program to // count total number of // handshakes when a person // can shake hand with only one. using System; class GFG { // function to find all // possible handshakes static int handshake( int n) { return n * (n - 1) / 2; } // Driver code static public void Main () { int n = 9; Console.WriteLine(handshake(n)); } } // This code is contributed by Sachin |
PHP
<?php // Recursive PHP program to // count total number of // handshakes when a person // can shake hand with only one. // function to find all // possible handshakes function handshake( $n ) { return $n * ( $n - 1) / 2; } // Driver Code $n = 9; echo (handshake( $n )); // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script> // Recursive Javascript program to // count total number of // handshakes when a person // can shake hand with only one. // Function to find all // possible handshakes function handshake(n) { return n * parseInt((n - 1) / 2, 10); } // Driver code let n = 9; document.write(handshake(n)); // This code is contributed by rameshtravel07 </script> |
Output:
36
Time Complexity: O(1)
Auxiliary Space: O(1)
Please Login to comment...