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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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)

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:

1. Store the count of elements having even set-bit and odd set-bit from the given array arr[].
2. 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.
3. 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 ` `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 `

Output:

```Even = 2, Odd = 4
```

Time Complexity: O(N)

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