Count Number of Pairs where Bitwise AND and Bitwise XOR is Equal

Given an integer array arr of size N, the task is to count the number of pairs whose BITWISE AND and BITWISE XOR are equal.

Example:

Input: N = 3, arr[] = {0,0,1}
Output: 1
Explanation: There is only one pair arr[0] and arr[1] as 0&0=0 and 0^0=0

Input: N = 4, arr[] = {1, 2, 4, 8}
Output: 0
Explanation: There are no pairs satisfying the condition.

Approach: This can be solved with the following idea:

To make Bitwise XOR and Bitwise AND equal, it is only possible when both bits of first and second element are 0 at each bit place. Therefore, it boils down to calculate number of pairs possible where both elements are 0.

Below are the steps involved:

• Initialize a count variable, count = 0.
• Iterate over array arr:
  • if arr[i] == 0, increment in count by 1.
• To count number of pairs possible:
  • (count * (count -1 ) ) / 2, will be the final ans.

Below is the implementation of the above code:

// C++ Implementation
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function to count number of pairs
// whose bitwise XOR and AND are equal
int BitByBit(int arr[], int n)
{

    int i = 0;
    int count = 0;

    // Iterate over array
    while (i < n) {

        // If arr[i] = 0
        if (arr[i] == 0) {

            count++;
        }
        i++;
    }

    // Number of pairs
    return (count * (count - 1)) / 2;
}

// Driver code
int main()
{

    int n = 5;
    int arr[] = { 1, 0, 0, 2, 2 };

    // Function call
    cout << BitByBit(arr, n);
    return 0;
}

public class BitwisePairs {

    // Function to count number of pairs
    // whose bitwise XOR and AND are equal
    static int bitByBit(int[] arr, int n) {
        int i = 0;
        int count = 0;

        // Iterate over array
        while (i < n) {
            // If arr[i] = 0
            if (arr[i] == 0) {
                count++;
            }
            i++;
        }

        // Number of pairs
        return (count * (count - 1)) / 2;
    }

    // Driver code
    public static void main(String[] args) {
        int n = 5;
        int[] arr = {1, 0, 0, 2, 2};

        // Function call
        System.out.println(bitByBit(arr, n));
    }
}

# Python Implementation

# Function to count number of pairs
# whose bitwise XOR and AND are equal
def BitByBit(arr, n):
    i = 0
    count = 0

    # Iterate over array
    while i < n:
        # If arr[i] = 0
        if arr[i] == 0:
            count += 1
        i += 1

    # Number of pairs
    return (count * (count - 1)) // 2


# Driver code
if __name__ == "__main__":
    n = 5
    arr = [1, 0, 0, 2, 2]

    # Function call
    print(BitByBit(arr, n))

    
# This code is contributed by Sakshi

using System;

public class Solution
{
    // Function to count number of pairs
    // whose bitwise XOR and AND are equal
    static int BitByBit(int[] arr, int n)
    {
        int i = 0;
        int count = 0;

        // Iterate over array
        while (i < n)
        {
            // If arr[i] = 0
            if (arr[i] == 0)
            {
                count++;
            }
            i++;
        }

        // Number of pairs
        return (count * (count - 1)) / 2;
    }

    // Driver code
    public static void Main()
    {
        int n = 5;
        int[] arr = { 1, 0, 0, 2, 2 };

        // Function call
        Console.WriteLine(BitByBit(arr, n));
    }
}


// This code is contributed by akshitaguprzj3

// JS Implementation

// Function to count number of pairs
// whose bitwise XOR and AND are equal
function bitByBit(arr) {
    let count = 0;

    // Iterate over array
    for (let i = 0; i < arr.length; i++) {
        // If arr[i] = 0
        if (arr[i] === 0) {
            count++;
        }
    }

    // Number of pairs
    return (count * (count - 1)) / 2;
}

// Driver code

const arr = [1, 0, 0, 2, 2];

// Function call
console.log(bitByBit(arr));



// This code is contributed by Sakshi

1

Time Complexity: O(N)
Auxiliary Space: O(1)

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Article Tags :

• Bit Magic
• DSA
• Bitwise-XOR
• Data Structures
• Bitwise-AND
• Geeks Premier League 2023

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Practice Tags :

• Bit Magic
• Data Structures