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

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: 2 
Explanation:
Two numbers whose setbits count is multiple of 2 are {3, 5}.
Input: arr[] = {10, 20, 30, 40}, K = 4 
Output: 1 
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 <bits/stdc++.h>
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

<script>
// Javascript implementation of above approach

// Function to find the count of numbers
function find_count(arr, k)
{
    var ans = 0;
    for(var i = 0; i <= arr.length; i++)
    {

        // Get the set-bits count of each element
        var 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;
}

function countSetBits(x)
{
    var setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }

    return setBits;
}


    var arr = [ 12, 345, 2, 68, 7896 ];
    var K = 2;

    document.write(find_count(arr, K));

// This code is contributed by SoumikMondal
</script>

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)