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# Count of elements in an Array whose set bits are in a multiple of K

• Last Updated : 12 May, 2021

Given an array arr[] of N elements and an integer K, the task is to count all the elements whose number of set bits is a multiple of K.
Examples:

Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output:
Explanation:
Two numbers whose setbits count is multiple of 2 are {3, 5}.
Input: arr[] = {10, 20, 30, 40}, K = 4
Output:
Explanation:
There number whose setbits count is multiple of 4 is {30}.

Approach:

1. Traverse the numbers in the array one by one.
2. Count the set bits of every number in the array.
3. Check if the setbits count is a multiple of K or not.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach` `#include ``using` `namespace` `std;` `// Function to find the count of numbers``int` `find_count(vector<``int``> arr, ``int` `k)``{` `    ``int` `ans = 0;``    ``for` `(``int` `i : arr) {` `        ``// Get the set-bits count of each element``        ``int` `x = __builtin_popcount(i);` `        ``// Check if the setbits count``        ``// is divisible by K``        ``if` `(x % k == 0)` `            ``// Increment the count``            ``// of required numbers by 1``            ``ans += 1;``    ``}` `    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``vector<``int``> arr = { 12, 345, 2, 68, 7896 };``    ``int` `K = 2;` `    ``cout << find_count(arr, K);` `    ``return` `0;``}`

## Java

 `// Java implementation of above approach` `class` `GFG{` `// Function to find the count of numbers``static` `int` `find_count(``int` `[]arr, ``int` `k)``{` `    ``int` `ans = ``0``;``    ``for` `(``int` `i : arr) {` `        ``// Get the set-bits count of each element``        ``int` `x = Integer.bitCount(i);` `        ``// Check if the setbits count``        ``// is divisible by K``        ``if` `(x % k == ``0``)` `            ``// Increment the count``            ``// of required numbers by 1``            ``ans += ``1``;``    ``}` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `[]arr = { ``12``, ``345``, ``2``, ``68``, ``7896` `};``    ``int` `K = ``2``;` `    ``System.out.print(find_count(arr, K));` `}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of above approach` `# Function to count total set bits``def` `bitsoncount(x):``    ``return` `bin``(x).count(``'1'``)` `# Function to find the count of numbers``def` `find_count(arr, k) :` `    ``ans ``=` `0``    ``for` `i ``in` `arr:` `        ``# Get the set-bits count of each element``        ``x ``=` `bitsoncount(i)` `        ``# Check if the setbits count``        ``# is divisible by K``        ``if` `(x ``%` `k ``=``=` `0``) :``            ``# Increment the count``            ``# of required numbers by 1``            ``ans ``+``=` `1``    ``return` `ans` `# Driver code``arr ``=` `[ ``12``, ``345``, ``2``, ``68``, ``7896` `]``K ``=` `2``print``(find_count(arr, K))` `# This code is contributed by Sanjit_Prasad`

## C#

 `// C# implementation of above approach``using` `System;` `class` `GFG{` `// Function to find the count of numbers``static` `int` `find_count(``int` `[]arr, ``int` `k)``{``    ``int` `ans = 0;``    ``foreach``(``int` `i ``in` `arr)``    ``{` `        ``// Get the set-bits count of each element``        ``int` `x = countSetBits(i);` `        ``// Check if the setbits count``        ``// is divisible by K``        ``if` `(x % k == 0)``            ` `            ``// Increment the count``            ``// of required numbers by 1``            ``ans += 1;``    ``}` `    ``return` `ans;``}` `static` `int` `countSetBits(``long` `x)``{``    ``int` `setBits = 0;``    ``while` `(x != 0)``    ``{``        ``x = x & (x - 1);``        ``setBits++;``    ``}` `    ``return` `setBits;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 12, 345, 2, 68, 7896 };``    ``int` `K = 2;` `    ``Console.Write(find_count(arr, K));``}``}``// This code is contributed by sapnasingh4991`

## Javascript

 ``
Output:
`3`

Time complexity: O(N * M), where N is the size of the array, and M is the bits count of the largest number in the array.
Auxiliary Space complexity: O(1)

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