Related Articles

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

• Difficulty Level : Expert
• Last Updated : 11 Jun, 2021

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 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);` `    ``// 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

 ``
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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up