Number of ways to choose a pair containing an even and an odd number from 1 to N

Given a number N the task is to find the number of pairs containing an even and an odd number from numbers between 1 and N inclusive. 

Note: The order of numbers in the pair does not matter. That is (1, 2) and (2, 1) are the same.

Examples: 

Input: N = 3
Output: 2
The pairs are (1, 2) and (2, 3).

Input: N = 6
Output: 9
The pairs are (1, 2), (1, 4), (1, 6), (2, 3),
(2, 5), (3, 4), (3, 6), (4, 5), (5, 6). 

Approach: The number of ways to form the pairs is (Total number of Even numbers*Total number of Odd numbers).
Thus  

1. if N is an even number of even numbers = number of odd numbers = N/2
2. if N is an odd number of even numbers = N/2 and the number of odd numbers = N/2+1

Below is the implementation of the above approach: 

// C++ implementation of the above approach
#include <iostream>
using namespace std;

// Driver code
int main()
{
    int N = 6;

    int Even = N / 2;

    int Odd = N - Even;

    cout << Even * Odd;

    return 0;
    // This code is contributed
    // by ANKITRAI1
}

// Java implementation of the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{

// Driver code
public static void main(String args[])
{
  int N = 6;
 
  int Even = N / 2 ;
 
  int Odd = N - Even ;
   
  System.out.println( Even * Odd );
   
}
}

# Python implementation of the above approach
N = 6

 # number of even numbers
Even = N//2

# number of odd numbers
Odd = N-Even 
print(Even * Odd)

// C# implementation of the 
// above approach
using System;

class GFG
{

// Driver code
public static void Main()
{
    int N = 6;
    
    int Even = N / 2 ;
    
    int Odd = N - Even ;
        
    Console.WriteLine(Even * Odd);
}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

<?php 
// PHP implementation of the 
// above approach 

// Driver code
$N = 6;

$Even = $N / 2 ;

$Odd = $N - $Even ;
    
echo $Even * $Odd ;
    
// This code is contributed
// by ChitraNayal
?>

<script>
// Javascript implementation of the above approach    
    
    // Driver code
    let N = 6;
   
      let Even = Math.floor(N / 2) ;
   
      let Odd = N - Even ;
     
     document.write( Even * Odd );
    

// This code is contributed by avanitrachhadiya2155
</script>

9

Time Complexity: O(1)

Space Complexity: O(1)