Count pairs with Odd XOR

Given an array of n integers. Find out number of pairs in array whose XOR is odd.

Examples :

Input : arr[] = { 1, 2, 3 }
Output : 2
All pairs of array
1 ^ 2 = 3
1 ^ 3 = 2
2 ^ 3 = 1

Input : arr[] = { 1, 2, 3, 4 }
Output : 4

Naive Approach: We can find pairs whose XOR is odd by running two loops. If XOR of two number is odd increase count of pairs.

Implementation:

Try It Yourself

C++

// C++ program to count pairs in array
// whose XOR is odd
#include <iostream>
using namespace std;

// A function will return number of pair
// whose XOR is odd
int countXorPair(int arr[], int n)
{
    // To store count of XOR pair
    int count = 0;

    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++)

            // If XOR is odd increase count
            if ((arr[i] ^ arr[j]) % 2 == 1)
                count++;
    }

    // Return count
    return count;
}

// Driver program to test countXorPair()
int main()
{
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countXorPair(arr, n);

    return 0;
}

Java

// Java program to count pairs in array whose
// XOR is odd
public class CountXor {
    // A function will return number of pair
    // whose XOR is odd
    static int countXorPair(int arr[], int n)
    {
        // To store count of XOR pair
        int count = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++)

                // If XOR is odd increase count
                if ((arr[i] ^ arr[j]) % 2 == 1)
                    count++;
        }

        // Return count
        return count;
    }

    // Driver program to test countXorPair()
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3 };
        System.out.println(countXorPair(arr, arr.length));
    }
}

Python 3

# Python 3 program to count 
# pairs in array whose XOR is odd

# A function will 
# return number of pair
# whose XOR is odd
def countXorPair(arr, n):

    # To store count of XOR pair
    count = 0

    for i in range(n):
        for j in range(i + 1, n):

            # If XOR is odd increase count
            if ((arr[i] ^ arr[j]) % 2 == 1):
                count += 1

    # Return count
    return count

# Driver Code
if __name__ == "__main__":
    arr= [ 1, 2, 3 ]
    n = len(arr)
    print(countXorPair(arr, n))

# This code is contributed
# by ChitraNayal

// C# program to count pairs in 
// array whose XOR is odd
using System;

public class CountXor {
    
    // A function will return number of pair
    // whose XOR is odd
    static int countXorPair(int[] arr, int n)
    {
        // To store count of XOR pair
        int count = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++)

                // If XOR is odd increase count
                if ((arr[i] ^ arr[j]) % 2 == 1)
                    count++;
        }

        // Return count
        return count;
    }

    // Driver program to test countXorPair()
    public static void Main()
    {
        int[] arr = {1, 2, 3};
        Console.WriteLine(countXorPair(arr, arr.Length));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to count
// pairs in array
// whose XOR is odd

// A function will
// return number of pair
// whose XOR is odd
function countXorPair($arr,$n)
{

    // To store count 
    // of XOR pair
    $count = 0;

    for ($i = 0; $i < $n; $i++)
    {
        for ($j = $i + 1; $j < $n; $j++)

            // If XOR is odd 
            // increase count
            if (($arr[$i] ^ $arr[$j]) % 2 == 1)
                $count++;
    }

    // Return count
    return $count;
}

    // Driver Code
    $arr = array(1, 2, 3);
    $n = count($arr);
    echo countXorPair($arr, $n);

// This code is contributed by anuj_67.
?>

JavaScript

<script>
    // Javascript program to count pairs in 
    // array whose XOR is odd
    
    // A function will return number of pair
    // whose XOR is odd
    function countXorPair(arr, n)
    {
        // To store count of XOR pair
        let count = 0;
  
        for (let i = 0; i < n; i++) {
            for (let j = i + 1; j < n; j++)
  
                // If XOR is odd increase count
                if ((arr[i] ^ arr[j]) % 2 == 1)
                    count++;
        }
  
        // Return count
        return count;
    }
    
    let arr = [1, 2, 3];
      document.write(countXorPair(arr, arr.length));
    
</script>

Output

Time Complexity : O(n*n)
Space Complexity - O(1)

Efficient Approach: We can observe that:

odd ^ odd = even
odd ^ even = odd
even ^ odd = odd
even ^ even = even

Therefore total pairs in array whose XOR is odd will be equal to count of odd numbers multiplied by count of even numbers.