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Count pairs with Bitwise XOR as ODD number

  • Difficulty Level : Medium
  • Last Updated : 23 Apr, 2021

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

 

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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#




// 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: 

6

Time Complexity: O(N^2)
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#




// 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: 

6

Time Complexity: O(N)
 




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