# 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

**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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

Even = 2, Odd = 4

**Time Complexity:** O(N)

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