Count pairs with Bitwise XOR as ODD number

Given an array of N integers, the task is to find the number of pairs (i, j) such that A[i] ^ A[j] is odd.
Examples:

Input : N = 5
        A[] =  { 5, 4, 7, 2, 1}
Output :6
Since pair of A[] =
( 5, 4 ) = 1( 5, 7 ) = 2( 5, 2 ) = 7( 5, 1 ) = 4
( 4, 7 ) = 3( 4, 2 ) = 6( 4, 1 ) = 5
( 7, 2 ) = 5( 7, 1 ) = 6
( 2, 1 ) = 3
Total XOR ODD pair  = 6

Input : N = 7
        A[] = { 7, 2, 8, 1, 0, 5, 11 }
Output :12
Since pair of A[] =
( 7, 2 ) = 5( 7, 8 ) = 15( 7, 1 ) = 6( 7, 0 ) = 7( 7, 5 ) = 2( 7, 11 ) = 12
( 2, 8 ) = 10( 2, 1 ) = 3( 2, 0 ) = 2( 2, 5 ) = 7( 2, 11 ) = 9
( 8, 1 ) = 9( 8, 0 ) = 8( 8, 5 ) = 13( 8, 11 ) = 3
( 1, 0 ) = 1( 1, 5 ) = 4( 1, 11 ) = 10
( 0, 5 ) = 5( 0, 11 ) = 11
( 5, 11 ) = 14

A naive approach is to check for every pair and print the count of pairs.
Below is the implementation of the above approach:

C++

// C++ program to count pairs
// with XOR giving a odd number
#include <iostream>

using namespace std;
 
// Function to count number of odd pairs

int findOddPair(int A[], int N)
{

    int i, j;
 
    // variable for counting odd pairs

    int oddPair = 0;
 
    // find all pairs

    for (i = 0; i < N; i++) {

        for (j = i + 1; j < N; j++) {
 
            // find XOR operation

            // check odd or even

            if ((A[i] ^ A[j]) % 2 != 0)

                oddPair++;

        }

        cout << endl;

    }
 
    // return number of odd pair

    return oddPair;
}
 
// Driver Code

int main()
{
 
    int A[] = { 5, 4, 7, 2, 1 };

    int N = sizeof(A) / sizeof(A[0]);
 
    // calling function findOddPair

    // and print number of odd pair

    cout << findOddPair(A, N) << endl;
 
    return 0;
}

Java

// Java program to count pairs
// with XOR giving a odd number
 
class GFG
{

// Function to count 
// number of odd pairs

static int findOddPair(int A[],

                       int N)
{

    int i, j;
 
    // variable for counting 

    // odd pairs

    int oddPair = 0;
 
    // find all pairs

    for (i = 0; i < N; i++)

    {

        for (j = i + 1; j < N; j++)

        {
 
            // find XOR operation

            // check odd or even

            if ((A[i] ^ A[j]) % 2 != 0)

                oddPair++;

        }

    }
 
    // return number 

    // of odd pair

    return oddPair;
}
 
// Driver Code

public static void main(String args[])
{

    int A[] = { 5, 4, 7, 2, 1 };

    int N = A.length;
 
    // calling function findOddPair

    // and print number of odd pair

    System.out.println(findOddPair(A, N)); 
}
}
 
// This code is contributed 
// by Kirti_Mangal

Python3

# Python3 program to count pairs 
# with XOR giving a odd number 
 
# Function to count number of odd pairs 

def findOddPair(A, N) :
 
    # variable for counting odd pairs 

    oddPair = 0
 
    # find all pairs 

    for i in range(0, N) : 

        for j in range(i+1, N) :
 
            # find XOR operation 

            # check odd or even 

            if ((A[i] ^ A[j]) % 2 != 0): 

                oddPair+=1
 
    # return number of odd pair 

    return oddPair 
 
# Driver Code

if __name__=='__main__':

    A = [5, 4, 7, 2, 1 ] 

    N = len(A)
 
# calling function findOddPair 
# and print number of odd pair 

    print(findOddPair(A, N)) 
 
# This code is contributed by Smitha Dinesh Semwal

// C# program to count pairs
// with XOR giving a odd number

using System;
 
class GFG
{

// Function to count 
// number of odd pairs

static int findOddPair(int[] A,

                    int N)
{

    int i, j;
 
    // variable for counting 

    // odd pairs

    int oddPair = 0;
 
    // find all pairs

    for (i = 0; i < N; i++)

    {

        for (j = i + 1; j < N; j++)

        {
 
            // find XOR operation

            // check odd or even

            if ((A[i] ^ A[j]) % 2 != 0)

                oddPair++;

        }

    }
 
    // return number 

    // of odd pair

    return oddPair;
}
 
// Driver Code

public static void Main()
{

    int[] A = { 5, 4, 7, 2, 1 };

    int N = A.Length;
 
    // calling function findOddPair

    // and print number of odd pair

    Console.WriteLine(findOddPair(A, N)); 
}
}
 
// This code is contributed 
// by Akanksha Rai(Abby_akku)
 
# calling function findOddPair 
# and print number of odd pair 
print(findOddPair(a, n))

PHP

<?php
//PHP  program to count pairs 
// with XOR giving a odd number 
 
// Function to count number of odd pairs 
 
function  findOddPair($A, $N) 
{ 

     $i; $j; 
 
    // variable for counting odd pairs 

     $oddPair = 0; 
 
    // find all pairs 

    for ($i = 0; $i < $N; $i++) { 

        for ($j = $i + 1; $j < $N; $j++) { 
 
            // find XOR operation 

            // check odd or even 

            if (($A[$i] ^ $A[$j]) % 2 != 0) 

                $oddPair++; 

        } 

    } 
 
    // return number of odd pair 

    return $oddPair; 
} 
 
// Driver Code 
 
    $A = array( 5, 4, 7, 2, 1 ); 

    $N = sizeof($A) / sizeof($A[0]); 
 
    // calling function findOddPair 

    // and print number of odd pair 

    echo  findOddPair($A, $N),"\n"; 
 
// This Code is Contributed by ajit    
?>

Javascript

<script>
// JavaScript program to count pairs 
// with XOR giving a odd number  
 
// Function to count number of odd pairs 

function findOddPair(A, N) 
{ 

    let i, j; 
 
    // variable for counting odd pairs 

    let oddPair = 0; 
 
    // find all pairs 

    for (i = 0; i < N; i++) { 

        for (j = i + 1; j < N; j++) { 
 
            // find XOR operation 

            // check odd or even 

            if ((A[i] ^ A[j]) % 2 != 0) 

                oddPair++; 

        } 

    } 
 
    // return number of odd pair 

    return oddPair; 
} 
 
// Driver Code 
 
    let A = [ 5, 4, 7, 2, 1 ]; 

    let N = A.length; 
 
    // calling function findOddPair 

    // and print number of odd pair 

    document.write(findOddPair(A, N) + "<br>"); 
 
// This code is contributed by Surbhi Tyagi.
 
</script>

OUTPUT:

Time Complexity: O(N^2) as two nested loops are being used.
Auxiliary Space: O(1), as constant space is being used by the algorithm.

An efficient solution is to count the even numbers. Then return count * (N – count).

C++

// C++ program to count pairs
// with XOR giving a odd number
#include <iostream>

using namespace std;
 
// Function to count number of odd pairs

int findOddPair(int A[], int N)
{

    int i, count = 0;
 
    // find all pairs

    for (i = 0; i < N; i++) {

        if (A[i] % 2 == 0)

            count++;

    }
 
    // return number of odd pair

    return count * (N - count);
}
 
// Driver Code

int main()
{

    int a[] = { 5, 4, 7, 2, 1 };

    int n = sizeof(a) / sizeof(a[0]);
 
    // calling function findOddPair

    // and print number of odd pair

    cout << findOddPair(a, n) << endl;
 
    return 0;
}

Java

// Java program to count pairs
// with XOR giving a odd number

class GFG
{
// Function to count
// number of odd pairs

static int findOddPair(int A[], 

                       int N)
{

    int i, count = 0;
 
    // find all pairs

    for (i = 0; i < N; i++) 

    {

        if (A[i] % 2 == 0)

            count++;

    }
 
    // return number of odd pair

    return count * (N - count);
}
 
// Driver Code

public static void main(String[] arg)
{

    int a[] = { 5, 4, 7, 2, 1 };

    int n = a.length ;
 
    // calling function findOddPair

    // and print number of odd pair

    System.out.println(findOddPair(a, n));
}
}
 
// This code is contributed
// by Smitha

Python3

# Python3 program to count pairs 
# with XOR giving a odd number 
 
# Function to count number of odd pairs 

def findOddPair(A, N) :
 
    count = 0
 
    # find all pairs 

    for i in range(0 , N) : 

        if (A[i] % 2 == 0) :

            count+=1

    # return number of odd pair 

    return count * (N - count) 
 
# Driver Code

if __name__=='__main__':

    a = [5, 4, 7, 2, 1] 

    n = len(a)

    print(findOddPair(a,n))
 
# this code is contributed by Smitha Dinesh Semwal

// C# program to count pairs
// with XOR giving a odd number

using System;
 
class GFG
{
// Function to count
// number of odd pairs

static int findOddPair(int []A,

                       int N)
{

    int i, count = 0;
 
    // find all pairs

    for (i = 0; i < N; i++) 

    {

        if (A[i] % 2 == 0)

            count++;

    }
 
    // return number of odd pair

    return count * (N - count);
}
 
// Driver Code

public static void Main()
{

    int []a = { 5, 4, 7, 2, 1 };

    int n = a.Length ;
 
    // calling function findOddPair

    // and print number of odd pair

    Console.Write(findOddPair(a, n));
}
}
 
// This code is contributed
// by Smitha

PHP

<?php
// PHP program to count pairs 
// with XOR giving a odd number 
 
// Function to count number
// of odd pairs 

function findOddPair($A, $N) 
{ 

    $count = 0; 
 
    // find all pairs 

    for ($i = 0; $i < $N; $i++) 

    { 

        if ($A[$i] % 2 == 0) 

            $count++; 

    } 
 
    // return number of odd pair 

    return $count * ($N - $count); 
} 
 
// Driver Code 

$a = array( 5, 4, 7, 2, 1 ); 

$n = count($a);
 
// calling function findOddPair 
// and print number of odd pair 

echo( findOddPair($a, $n)); 
 
// This code is contributed 
// by Smitha
?>

Javascript

<script>
 
// Javascript program to count pairs
// with XOR giving a odd number
 
// Function to count number of odd pairs

function findOddPair(A, N)
{

    let i, count = 0;
 
    // find all pairs

    for (i = 0; i < N; i++) {

        if (A[i] % 2 == 0)

            count++;

    }
 
    // return number of odd pair

    return count * (N - count);
}
 
// Driver Code

    let a = [ 5, 4, 7, 2, 1 ];

    let n = a.length;
 
    // calling function findOddPair

    // and print number of odd pair

    document.write(findOddPair(a, n));
 
</script>

Output:

Time Complexity: O(N), since one traversal of the array is required to complete all operations.
Auxiliary Space: O(1), as constant space is being used.

Article Tags :

Bit Magic

DSA

Bitwise-XOR