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.

// 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;
}

Output:

2

Time Complexity : O(n*n)

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.

// 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 odd and even
    // numbers
    int odd = 0, even = 0;
    
    for (int i = 0; i < n; i++)
    {
        // Increase even if number is
        // even otherwise increase odd
        if (arr[i] % 2 == 0)
            even++;
        else
            odd++;
    }
    
    // Return number of pairs
    return odd * even;
}
// 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;

Output:

2

Time Complexity : O(n)

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/count-pairs-odd-xor/
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rakesh

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