Given a number N which represents the number of players participating in a Badminton match. The task is to determine the number of matches required to determine the Winner. In each Match 1 player is knocked out.
Examples:
Input: N = 4
Output: Matches played = 3
(As after each match only N - 1 players left)
Input: N = 9
Output: Matches played = 8
Approach: Since, after each match, one player is knocked out. So to get the winner, n-1 players should be knocked out and for which n-1 matches to be played.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int noOfMatches( int N)
{
return N - 1;
}
int main()
{
int N = 8;
cout << "Matches played = " << noOfMatches(N);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int noOfMatches( int N)
{
return N - 1 ;
}
public static void main (String[] args)
{
int N = 8 ;
System.out.println ( "Matches played = " +
noOfMatches(N));
}
}
|
Python3
def noOfMatches(N) :
return N - 1
if __name__ = = "__main__" :
N = 8
print ( "Matches played =" ,
noOfMatches(N))
|
C#
using System;
public class GFG{
static int noOfMatches( int N)
{
return N - 1;
}
static public void Main (){
int N = 8;
Console.WriteLine( "Matches played = " +
noOfMatches(N));
}
}
|
PHP
<?php
function noOfMatches( $N )
{
return ( $N - 1);
}
$N = 8;
echo "Matches played = " ,
noOfMatches( $N );
?>
|
Javascript
<script>
function noOfMatches(N)
{
return N - 1;
}
var N = 8;
document.write( "Matches played = " +
noOfMatches(N));
</script>
|
Output: Matches played = 7
Time Complexity: O(1)
Auxiliary Space: O(1)