Count of even and odd power pairs in an Array

Last Updated : 10 Jun, 2021

Given an array arr[] of length N, the task is to count the number of pairs (X, Y) such that X^Y is even and count the number of pairs such that X^Y is odd.
Examples:

Input: arr[] = {2, 3, 4, 5}
Output:
6
6
Explanation: (2, 3), (2, 4), (2, 5), (4, 2), (4, 3) and (4, 5) are the pairs with even values
and (3, 2), (3, 4), (3, 5), (5, 2), (5, 3) and (5, 4) are the pairs with odd values.
Input: arr[] = {10, 11, 20, 60, 70}
Output:
16
4
Explanation: (10, 11), (10, 20), (10, 60), (10, 70), (20, 10), (20, 11), (20, 60), (20, 70), (60, 10), (60, 11), (60, 20), (60, 70), (70, 10), (70, 11), (70, 20), (70, 60) are the pairs with even values and (11, 10), (11, 20), (11, 60), (11, 70) are the pairs with odd values.

Naive approach: Calculate the powers for every single pair possible and find whether the calculated value is even or odd.
Efficient approach: Count the even and odd elements in the array and then use the concept pow (even, any element except itself) is even and pow (odd, any element except itself) is odd.
So, the number of pairs (X, Y) are,

pow(X, Y) is even = (count of even number * (n – 1))
pow(X, Y) is odd = (count of odd number * (n – 1))

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to find and print the
// required count of pairs
void countPairs(int arr[], int n)
{
 
    // Find the count of even and
    // odd elements in the array
    int even = 0, odd = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0)
            even++;
        else
            odd++;
    }
 
    // Print the required count of pairs
    cout << (even) * (n - 1) << endl;
    cout << (odd) * (n - 1) << endl;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    countPairs(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
    // Function to find and print the
    // required count of pairs
    static void countPairs(int arr[], int n)
    {
 
        // Find the count of even and
        // odd elements in the array
        int even = 0, odd = 0;
        for (int i = 0; i < n; i++)
        {
            if (arr[i] % 2 == 0)
                even++;
            else
                odd++;
        }
 
        // Print the required count of pairs
        System.out.println((even) * (n - 1));
        System.out.println((odd) * (n - 1));
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 2, 3, 4, 5 };
        int n = arr.length;
 
        countPairs(arr, n);
    }
}
 
// This code is contributed by ANKITUMAR34

Python3

# Python3 implementation of the approach
 
# Function to find and print the
# required count of pairs
def countPairs(arr, n):
     
    # Find the count of even and
    # odd elements in the array
    odd = 0
    even = 0
    for i in range(n):
        if (arr[i] % 2 == 0):
            even += 1
        else:
            odd += 1
             
    # Count the number of odd pairs
    odd_pairs = odd*(n-1)
 
    # Count the number of even pairs
    even_pairs = even*(n-1)
 
    print(odd_pairs)
    print(even_pairs)
 
# Driver code
if __name__ == '__main__':
    arr = [2, 3, 4, 5]
    n = len(arr)
    countPairs(arr, n)

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to find and print the
    // required count of pairs
    static void countPairs(int []arr, int n)
    {
 
        // Find the count of even and
        // odd elements in the array
        int even = 0, odd = 0;
        for (int i = 0; i < n; i++)
        {
            if (arr[i] % 2 == 0)
                even++;
            else
                odd++;
        }
 
        // Print the required count of pairs
        Console.WriteLine((even) * (n - 1));
        Console.WriteLine((odd) * (n - 1));
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 3, 4, 5 };
        int n = arr.Length;
 
        countPairs(arr, n);
    }
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
 
    // Javascript implementation of the approach
     
    // Function to find and print the
    // required count of pairs
    function countPairs(arr, n)
    {
  
        // Find the count of even and
        // odd elements in the array
        let even = 0, odd = 0;
        for (let i = 0; i < n; i++)
        {
            if (arr[i] % 2 == 0)
                even++;
            else
                odd++;
        }
  
        // Print the required count of pairs
        document.write((even) * (n - 1) + "</br>");
        document.write((odd) * (n - 1));
    }
     
    let arr = [ 2, 3, 4, 5 ];
    let n = arr.length;
 
    countPairs(arr, n);
     
</script>

Output:

6
6

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