Count number of ordered pairs with Even and Odd Sums

Given an array of n positive numbers, the task is to count number of ordered pairs with even and odd sum.

Examples:

Input: arr[] = {1, 2, 4} 
Output: Even sum Pairs = 2, Odd sum Pairs = 4 
The ordered pairs are (1, 2), (1, 4), (2, 1), (4, 1), (2, 4), (4, 2) 
Pairs with Even sum: (2, 4), (4, 2) 
Pairs with Odd sum: (1, 2), (1, 4), (2, 1), (4, 1)

Input: arr[] = {2, 4, 5, 9, 1, 8} 
Output: Even sum Pairs = 12, Odd sum Pairs = 18 

Approach: 

The sum of two numbers is odd if one number is odd and another one is even. So now we have to count the even and odd numbers. As in order pair (a, b) and (b, a) both treated as different pair, therefore 

Number of odd sum pairs = (count of even numbers) * (count of odd numbers) * 2

This is because every even number can pair with every odd number, and every odd number can pair with every even number. Thus multiply 2 is done in the answer.
And the number of even sum pairs will be an inversion of the number of odd sum pairs. Therefore:

Number of even sum pairs = Total Number of pairs – Number of odd sum 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 sum 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 * even * 2;

    return ans;
}

// function to count even sum pair
int count_even_pair(int odd_sum_pairs, int n)
{
    int total_pairs = (n * (n - 1));
    int ans = total_pairs - odd_sum_pairs;

    return ans;
}

// Driver code
int main()
{

    int n = 6;
    int a[] = { 2, 4, 5, 9, 1, 8 };

    int odd_sum_pairs = count_odd_pair(n, a);

    int even_sum_pairs = count_even_pair(
        odd_sum_pairs, n);

    cout << "Even Sum Pairs = "
         << even_sum_pairs
         << endl;
    cout << "Odd Sum Pairs= "
         << odd_sum_pairs
         << endl;

    return 0;
}

Java

// Java implementation of the above approach

class GFG
{
    // function to count odd sum 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 * even * 2;

        return ans;
    }

    // function to count even sum pair
    static int count_even_pair(int odd_sum_pairs, int n)
    {
        int total_pairs = (n * (n - 1));
        int ans = total_pairs - odd_sum_pairs;

        return ans;
    }

    // Driver code
    public static void main(String []args)
    {

        int n = 6;
        int []a = { 2, 4, 5, 9, 1, 8 };

        int odd_sum_pairs = count_odd_pair(n, a);

        int even_sum_pairs = count_even_pair( odd_sum_pairs, n);

        System.out.println("Even Sum Pairs = " + even_sum_pairs);

        System.out.println("Odd Sum Pairs= " + odd_sum_pairs);


    }

}

// This code is contributed by ihritik

Python3

# Python3 implementation of the above approach

# function to count odd sum 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 *  even * 2

    return  ans


# function to count even sum pair
def count_even_pair( odd_sum_pairs,  n):

    total_pairs = ( n * ( n - 1))
    ans =  total_pairs -  odd_sum_pairs
    return ans


# Driver code

n = 6
a = [2, 4, 5, 9, 1, 8]

odd_sum_pairs = count_odd_pair( n,  a)

even_sum_pairs = count_even_pair( odd_sum_pairs,  n)

print("Even Sum Pairs =", even_sum_pairs)
print("Odd Sum Pairs=", odd_sum_pairs)

# This code is contributed by ihritik

C#

// C# implementation of the above approach

using System;
class GFG
{
    // function to count odd sum 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 * even * 2;

        return ans;
    }

    // function to count even sum pair
    static int count_even_pair(int odd_sum_pairs, int n)
    {
        int total_pairs = (n * (n - 1));
        int ans = total_pairs - odd_sum_pairs;

        return ans;
    }

    // Driver code
    public static void Main()
    {

        int n = 6;
        int []a = { 2, 4, 5, 9, 1, 8 };

        int odd_sum_pairs = count_odd_pair(n, a);

        int even_sum_pairs = count_even_pair( odd_sum_pairs, n);

        Console.WriteLine("Even Sum Pairs = " + even_sum_pairs);

        Console.WriteLine("Odd Sum Pairs= " + odd_sum_pairs);

    }

}

// This code is contributed by ihritik

PHP

<?php
// PHP implementation of the above approach

// function to count odd sum 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 * $even * 2;

    return $ans;
}

// function to count even sum pair
function count_even_pair($odd_sum_pairs, $n)
{
    $total_pairs = ($n * ($n - 1));
    $ans = $total_pairs - $odd_sum_pairs;
    return $ans;
}

// Driver code

$n = 6;
$a = array( 2, 4, 5, 9, 1, 8 );

$odd_sum_pairs = count_odd_pair($n, $a);

$even_sum_pairs = count_even_pair($odd_sum_pairs, $n);

echo "Even Sum Pairs = $even_sum_pairs \n";
echo "Odd Sum Pairs=  $odd_sum_pairs \n";

// This code is contributed by ihritik
?>

Javascript

<script>

// JavaScript implementation of the above approach

// function to count odd sum pair
function count_odd_pair(n, a)
{
    var odd = 0, even = 0;

    for (var 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
    var ans = odd * even * 2;

    return ans;
}

// function to count even sum pair
function count_even_pair(odd_sum_pairs, n)
{
    var total_pairs = (n * (n - 1));
    var ans = total_pairs - odd_sum_pairs;

    return ans;
}

// Driver code
var n = 6;
var a = [2, 4, 5, 9, 1, 8];
var odd_sum_pairs = count_odd_pair(n, a);
var even_sum_pairs = count_even_pair(
    odd_sum_pairs, n);

document.write( "Even Sum Pairs = "
     + even_sum_pairs + "<br>");

document.write( "Odd Sum Pairs = "
     + odd_sum_pairs + "<br>");


</script>

Output

Even Sum Pairs = 12
Odd Sum Pairs= 18

Complexity Analysis:

• Time Complexity: O(n), to find the number of odd and even numbers in the given array
• Auxiliary Space: O(1), as no extra space is used