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)

**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 numberA(which has an odd number of set-bit) with the numberK(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 numberA(which has an odd number of set-bit) with the numberK(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; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

**Output:**

Even = 2, Odd = 4

**Time Complexity:** O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Count elements in first Array with absolute difference greater than K with an element in second Array
- Count distinct elements after adding each element of First Array with Second Array
- Count of multiples in an Array before every element
- Count of greater elements for each element in the Array
- Count pairs in an array such that at least one element is prime
- Count of triplets from the given Array such that sum of any two elements is the third element
- Count unequal element pairs from the given Array
- Count pairs in array such that one element is reverse of another
- Count pairs in array such that one element is power of another
- Count of right shifts for each array element to be in its sorted position
- Count the number of pop operations on stack to get each element of the array
- Count the number of elements which are greater than any of element on right side of an array
- Count of distinct index pair (i, j) such that element sum of First Array is greater
- Minimize swaps required to maximize the count of elements replacing a greater element in an Array
- Largest element smaller than current element on left for every element in Array
- Count of pairs from arrays A and B such that element in A is greater than element in B at that index
- Array formed from difference of each element from the largest element in the given array
- Count of distinct possible pairs such that the element from A is greater than the element from B
- Replace each element by the difference of the total size of the array and frequency of that element
- Check if minimum element in array is less than or equals half of every other element

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.