Number of ways to choose a pair containing an even and an odd number from 1 to N
Last Updated :
27 Jul, 2022
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
- if N is an even number of even numbers = number of odd numbers = N/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++
#include <iostream>
using namespace std;
int main()
{
int N = 6;
int Even = N / 2;
int Odd = N - Even;
cout << Even * Odd;
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
public static void main(String args[])
{
int N = 6 ;
int Even = N / 2 ;
int Odd = N - Even ;
System.out.println( Even * Odd );
}
}
|
Python3
N = 6
Even = N / / 2
Odd = N - Even
print (Even * Odd)
|
C#
using System;
class GFG
{
public static void Main()
{
int N = 6;
int Even = N / 2 ;
int Odd = N - Even ;
Console.WriteLine(Even * Odd);
}
}
|
PHP
<?php
$N = 6;
$Even = $N / 2 ;
$Odd = $N - $Even ;
echo $Even * $Odd ;
?>
|
Javascript
<script>
let N = 6;
let Even = Math.floor(N / 2) ;
let Odd = N - Even ;
document.write( Even * Odd );
</script>
|
Time Complexity: O(1)
Space Complexity: O(1)
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