Count of even and odd set bit with array element after XOR with K

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Given an array arr[] and a number K. The task is to find the count of the element having odd and even number of the set-bit after taking XOR of K with every element of the given arr[].
Examples:

Input: arr[] = {4, 2, 15, 9, 8, 8}, K = 3
Output: Even = 2, Odd = 4
Explanation:
The binary representation of the element after taking XOR with K are:
4 ^ 3 = 7 (111)
2 ^ 3 = 1 (1)
15 ^ 3 = 12 (1100)
9 ^ 3 = 10 (1010)
8 ^ 3 = 11 (1011)
8 ^ 3 = 11 (1011)
No of elements with even no of 1’s in binary representation : 2 (12, 10)
No of elements with odd no of 1’s in binary representation : 4 (7, 1, 11, 11)
Input: arr[] = {4, 2, 15, 9, 8, 8}, K = 6
Output: Even = 4, Odd = 2

Naive Approach: The naive approach is to take bitwise XOR of K with each element of the given array arr[] and then, count the set-bit for each element in the array after taking XOR with K.
Time Complexity: O(N*K)

Auxiliary space: O(1)
Efficient Approach:
With the help of the following observation, we have:

For Example:
If A = 4 and K = 3
Binary Representation:
A = 100
K = 011
A^K = 111
Therefore, the XOR of number A(which has an odd number of set-bit) with the number K(which has an even number of set-bit) results in an odd number of set-bit.
And If A = 4 and K = 7
Binary Representation:
A = 100
K = 111
A^K = 011
Therefore, the XOR of number A(which has an odd number of set-bit) with the number K(which has an odd number of set-bit) results in an even number of set-bit.

From the above observations:

If K has an odd number of set-bit, then the count of elements of array arr[] with even set-bit and odd set-bit after taking XOR with K, will be same as the count of even set-bit and odd set-bit int the array before XOR.
If K has an even number of set-bit, then the count of elements of array arr[] with even set-bit and odd set-bit after taking XOR with K, will reverse as the count of even set-bit and odd set-bit in the array before XOR.

Steps:

Store the count of elements having even set-bit and odd set-bit from the given array arr[].
If K has odd set-bit, then the count of even and odd set-bit after XOR with K will be the same as the count of even and odd set-bit calculated above.
If K has even set-bit, then the count of even and odd set-bit after XOR with K will be the count of odd and even set-bit calculated above.

Below is the implementation of the above approach:

C++

// C++ program to count the set
// bits after taking XOR with a
// number K
#include <bits/stdc++.h>
using namespace std;
 
// Function to store EVEN and odd variable
void countEvenOdd(int arr[], int n, int K)
{
    int even = 0, odd = 0;
 
    // Store the count of even and
    // odd set bit
    for (int i = 0; i < n; i++) {
 
        // Count the set bit using
        // in built function
        int x = __builtin_popcount(arr[i]);
        if (x % 2 == 0)
            even++;
        else
            odd++;
    }
 
    int y;
 
    // Count of set-bit of K
    y = __builtin_popcount(K);
 
    // If y is odd then, count of
    // even and odd set bit will
    // be interchanged
    if (y & 1) {
        cout << "Even = " << odd
             << ", Odd = " << even;
    }
 
    // Else it will remain same as
    // the original array
    else {
        cout << "Even = " << even
             << ", Odd = " << odd;
    }
}
 
// Driver's Code
int main(void)
{
    int arr[] = { 4, 2, 15, 9, 8, 8 };
    int K = 3;
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call to count even
    // and odd
    countEvenOdd(arr, n, K);
    return 0;
}

Java

// Java program to count the set
// bits after taking XOR with a
// number K
class GFG {
 
     
    /* Function to get no of set  
    bits in binary representation  
    of positive integer n */
    static int __builtin_popcount(int n) 
    { 
        int count = 0; 
        while (n > 0) { 
            count += n & 1; 
            n >>= 1; 
        } 
        return count; 
    }
     
    // Function to store EVEN and odd variable
    static void countEvenOdd(int arr[], int n, int K)
    {
        int even = 0, odd = 0;
     
        // Store the count of even and
        // odd set bit
        for (int i = 0; i < n; i++) {
     
            // Count the set bit using
            // in built function
            int x = __builtin_popcount(arr[i]);
            if (x % 2 == 0)
                even++;
            else
                odd++;
        }
     
        int y;
     
        // Count of set-bit of K
        y = __builtin_popcount(K);
     
        // If y is odd then, count of
        // even and odd set bit will
        // be interchanged
        if ((y & 1) != 0) {
            System.out.println("Even = "+ odd + ", Odd = " + even);
        }
     
        // Else it will remain same as
        // the original array
        else {
            System.out.println("Even = " + even + ", Odd = " + odd);
        }
    }
     
    // Driver's Code
    public static void main (String[] args)
    {
        int arr[] = { 4, 2, 15, 9, 8, 8 };
        int K = 3;
        int n = arr.length;
     
        // Function call to count even
        // and odd
        countEvenOdd(arr, n, K);
    }
  
}
// This code is contributed by Yash_R

Python3

# Python3 program to count the set 
# bits after taking XOR with a 
# number K 
 
# Function to store EVEN and odd variable 
def countEvenOdd(arr, n, K) : 
 
    even = 0; odd = 0; 
 
    # Store the count of even and 
    # odd set bit 
    for i in range(n) :
 
        # Count the set bit using 
        # in built function 
        x = bin(arr[i]).count('1'); 
        if (x % 2 == 0) :
            even += 1; 
        else :
            odd += 1; 
 
    # Count of set-bit of K 
    y = bin(K).count('1'); 
 
    # If y is odd then, count of 
    # even and odd set bit will 
    # be interchanged 
    if (y & 1) :
        print("Even =",odd ,", Odd =", even); 
 
    # Else it will remain same as 
    # the original array 
    else :
        print("Even =" , even ,", Odd =", odd); 
 
 
# Driver's Code 
if __name__ == "__main__" :
     
    arr = [ 4, 2, 15, 9, 8, 8 ]; 
    K = 3; 
    n = len(arr); 
 
    # Function call to count even 
    # and odd 
    countEvenOdd(arr, n, K); 
     
# This code is contributed by Yash_R

C#

// C# program to count the set
// bits after taking XOR with a
// number K
using System;
 
public class GFG {
 
     
    /* Function to get no of set  
    bits in binary representation  
    of positive integer n */
    static int __builtin_popcount(int n) 
    { 
        int count = 0; 
        while (n > 0) { 
            count += n & 1; 
            n >>= 1; 
        } 
        return count; 
    }
     
    // Function to store EVEN and odd variable
    static void countEvenOdd(int []arr, int n, int K)
    {
        int even = 0, odd = 0;
     
        // Store the count of even and
        // odd set bit
        for (int i = 0; i < n; i++) {
     
            // Count the set bit using
            // in built function
            int x = __builtin_popcount(arr[i]);
            if (x % 2 == 0)
                even++;
            else
                odd++;
        }
     
        int y;
     
        // Count of set-bit of K
        y = __builtin_popcount(K);
     
        // If y is odd then, count of
        // even and odd set bit will
        // be interchanged
        if ((y & 1) != 0) {
            Console.WriteLine("Even = "+ odd + ", Odd = " + even);
        }
     
        // Else it will remain same as
        // the original array
        else {
            Console.WriteLine("Even = " + even + ", Odd = " + odd);
        }
    }
     
    // Driver's Code
    public static void Main (string[] args)
    {
        int []arr = { 4, 2, 15, 9, 8, 8 };
        int K = 3;
        int n = arr.Length;
     
        // Function call to count even
        // and odd
        countEvenOdd(arr, n, K);
    }
  
}
// This code is contributed by Yash_R

Javascript

<script>
// Javascript program to count the set
// bits after taking XOR with a
// number K
 
/* Function to get no of set
bits in binary representation
of positive integer n */
function __builtin_popcount(n) {
    let count = 0;
    while (n > 0) {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
// Function to store EVEN and odd variable
function countEvenOdd(arr, n, K) {
    let even = 0, odd = 0;
 
    // Store the count of even and
    // odd set bit
    for (let i = 0; i < n; i++) {
 
        // Count the set bit using
        // in built function
        let x = __builtin_popcount(arr[i]);
        if (x % 2 == 0)
            even++;
        else
            odd++;
    }
 
    let y;
 
    // Count of set-bit of K
    y = __builtin_popcount(K);
 
    // If y is odd then, count of
    // even and odd set bit will
    // be interchanged
    if ((y & 1) != 0) {
        document.write("Even = " + odd + ", Odd = " + even);
    }
 
    // Else it will remain same as
    // the original array
    else {
        document.write("Even = " + even + ", Odd = " + odd);
    }
}
 
// Driver's Code
 
let arr = [4, 2, 15, 9, 8, 8];
let K = 3;
let n = arr.length;
 
// Function call to count even
// and odd
countEvenOdd(arr, n, K);
 
// This code is contributed by _saurabh_jaiswal
</script>

Output:

Even = 2, Odd = 4

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

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