Skip to content
Related Articles

Related Articles

Improve Article

Count number of ordered pairs with Even and Odd Product

  • Difficulty Level : Basic
  • Last Updated : 03 Mar, 2021

Given an array of n positive numbers, the task is to count number of ordered pairs with even and odd product. Ordered pairs means (a, b) and (b,a) will be considered as different.
Examples: 
 

Input: n = 3, arr[] = {1, 2, 7} 
Output: Even product Pairs = 4, Odd product Pairs = 2 
The ordered pairs are (1, 2), (1, 7), (2, 1), (7, 1), (2, 7), (7, 2) 
Pairs with Odd product: (1, 7), (7, 1) 
Pairs with Even product: (1, 2), (2, 7), (2, 1), (7, 2)
Input: n = 6, arr[] = {2, 4, 5, 9, 1, 8} 
Output: Even product Pairs = 24, Odd product Pairs = 6 
 

 

Approach:
The product of two numbers is odd only if both are numbers are odd. Therefore: 
 

Number of odd product pairs = (count of odd numbers) * (count of odd numbers – 1)



And the number of even product pairs will be an inversion of number of odd product pairs. Therefore:
 

Number of even product pairs = Total Number of pairs – Number of odd product pairs

Below is the implementation of above approach: 
 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// function to count odd product pair
int count_odd_pair(int n, int a[])
{
    int odd = 0, even = 0;
 
    for (int i = 0; i < n; i++) {
 
        // if number is even
        if (a[i] % 2 == 0)
            even++;
 
        // if number is odd
        else
            odd++;
    }
 
    // count of ordered pairs
    int ans = odd * (odd - 1);
 
    return ans;
}
 
// function to count even product pair
int count_even_pair(int odd_product_pairs, int n)
{
    int total_pairs = (n * (n - 1));
    int ans = total_pairs - odd_product_pairs;
    return ans ;
}
 
// Driver code
int main()
{
 
    int n = 6;
    int a[] = { 2, 4, 5, 9, 1, 8 };
 
    int odd_product_pairs = count_odd_pair(n, a);
 
    int even_product_pairs = count_even_pair(
        odd_product_pairs, n);
 
    cout << "Even Product Pairs = "
         << even_product_pairs
         << endl;
    cout << "Odd Product Pairs= "
         << odd_product_pairs
         << endl;
 
    return 0;
}

Java




// Java  implementation of the above approach
import java.io.*;
 
class GFG {
     
     
// function to count odd product pair
static int count_odd_pair(int n, int a[])
{
    int odd = 0, even = 0;
 
    for (int i = 0; i < n; i++) {
 
        // if number is even
        if (a[i] % 2 == 0)
            even++;
 
        // if number is odd
        else
            odd++;
    }
 
    // count of ordered pairs
    int ans = odd * (odd - 1);
 
    return ans;
}
 
// function to count even product pair
static int count_even_pair(int odd_product_pairs, int n)
{
    int total_pairs = (n * (n - 1));
    int ans = total_pairs - odd_product_pairs;
    return ans;
}
 
// Driver code
    public static void main (String[] args) {
 
        int n = 6;
        int []a = { 2, 4, 5, 9, 1, 8 };
 
        int odd_product_pairs = count_odd_pair(n, a);
 
        int even_product_pairs = count_even_pair(
            odd_product_pairs, n);
 
        System.out.println( "Even Product Pairs = "+
            even_product_pairs );
          
        System.out.println("Odd Product Pairs= "+
             odd_product_pairs );
     
    }
}
//This Code is Contributed by ajit

Python3




# Python3 implementation of
# above approach
 
# function to count odd product pair
def count_odd_pair(n, a):
    odd = 0
    even = 0
    for i in range(0,n):
         
        # if number is even
        if a[i] % 2==0:
            even=even+1
        # if number is odd
        else:
            odd=odd+1
     
    # count of ordered pairs
    ans = odd * (odd - 1)
    return ans
 
# function to count even product pair
def count_even_pair(odd_product_pairs, n):
    total_pairs = (n * (n - 1))
    ans = total_pairs - odd_product_pairs
    return ans
 
#Driver code
if __name__=='__main__':
    n = 6
    a = [2, 4, 5, 9, 1 ,8]
 
    odd_product_pairs = count_odd_pair(n, a)
    even_product_pairs = (count_even_pair
                       (odd_product_pairs, n))
 
    print("Even Product Pairs = "
          ,even_product_pairs)
    print("Odd Product Pairs= "
          ,odd_product_pairs)
 
# This code is contributed by
# Shashank_Sharma

C#




// C#  implementation of the above approach
using System;
 
public class GFG{
     
         
// function to count odd product pair
static int count_odd_pair(int n, int []a)
{
    int odd = 0, even = 0;
 
    for (int i = 0; i < n; i++) {
 
        // if number is even
        if (a[i] % 2 == 0)
            even++;
 
        // if number is odd
        else
            odd++;
    }
 
    // count of ordered pairs
    int ans = odd * (odd - 1);
 
    return ans;
}
 
// function to count even product pair
static int count_even_pair(int odd_product_pairs, int n)
{
    int total_pairs = (n * (n - 1));
    int ans = total_pairs - odd_product_pairs;
    return ans;
}
 
// Driver code
     
static public void Main (){
        int n = 6;
        int []a = { 2, 4, 5, 9, 1, 8 };
 
        int odd_product_pairs = count_odd_pair(n, a);
 
        int even_product_pairs = count_even_pair(
            odd_product_pairs, n);
 
        Console.WriteLine( "Even Product Pairs = "+
            even_product_pairs );
         
        Console.WriteLine("Odd Product Pairs= "+
            odd_product_pairs );
    }
}

PHP




<?php
// function to count odd product pair
function count_odd_pair($n, $a)
{
    $odd = 0 ;
    $even = 0 ;
 
    for ($i = 0; $i < $n; $i++)
    {
 
        // if number is even
        if ($a[$i] % 2 == 0)
            $even++;
 
        // if number is odd
        else
            $odd++;
    }
 
    // count of ordered pairs
    $ans = $odd * ($odd - 1);
 
    return $ans;
}
 
// function to count even product pair
function count_even_pair($odd_product_pairs, $n)
{
    $total_pairs = ($n * ($n - 1));
    $ans = $total_pairs - $odd_product_pairs;
     
    return $ans ;
}
 
// Driver code
$n = 6;
$a = array( 2, 4, 5, 9, 1, 8 );
 
$odd_product_pairs = count_odd_pair($n, $a);
 
$even_product_pairs =
      count_even_pair($odd_product_pairs, $n);
 
echo "Even Product Pairs = ",
      $even_product_pairs, "\n";
echo "Odd Product Pairs = ",
      $odd_product_pairs, "\n";
 
// This code is contributed
// by ANKITRAI1
?>

Javascript




<script>
 
// JavaScript implementation of the above approach
 
// function to count odd product pair
function count_odd_pair(n, a)
{
    let odd = 0, even = 0;
 
    for (let i = 0; i < n; i++) {
 
        // if number is even
        if (a[i] % 2 == 0)
            even++;
 
        // if number is odd
        else
            odd++;
    }
 
    // count of ordered pairs
    let ans = odd * (odd - 1);
 
    return ans;
}
 
// function to count even product pair
function count_even_pair(odd_product_pairs, n)
{
    let total_pairs = (n * (n - 1));
    let ans = total_pairs - odd_product_pairs;
    return ans ;
}
 
// Driver code
    let n = 6;
    let a = [ 2, 4, 5, 9, 1, 8 ];
 
    let odd_product_pairs = count_odd_pair(n, a);
 
    let even_product_pairs = count_even_pair(
        odd_product_pairs, n);
 
    document.write("Even Product Pairs = "
        + even_product_pairs
        + "<br>");
    document.write("Odd Product Pairs= "
        + odd_product_pairs
        + "<br>");
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output: 
Even Product Pairs = 24
Odd Product Pairs= 6

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :