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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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)) 

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PHP

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<?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    
?>

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OUTPUT:

6

Time Complexity: O(N^2)

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

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

6

Time Complexity: O(N)



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