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

Last Updated : 04 Oct, 2023

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:

Traverse the numbers in the array one by one.
Count the set bits of every number in the array.
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# 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

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)

Using Bit Manipulation and Brute Force in python:

Approach:

Define a function named countSetBits that takes an integer num as input.
- Initialize a variable count to 0.
- Loop through the binary representation of num using bit manipulation.
- If the last bit is 1, increment the count variable.
- Right shift the binary representation of num by 1 bit.
- Repeat steps 4-5 until the binary representation of num becomes 0.
- Return the count variable.
Define a function named countMultiples that takes a list arr and an integer K as input.
- Initialize a variable res to 0.
- Loop through each element num in arr.
- Call the countSetBits function with num as input and store the result in a variable count.
- If count is a multiple of K, increment the res variable.
- Repeat steps 10-12 until all elements in arr have been processed.
- Return the res variable.

C++

#include <iostream>
using namespace std;

// Function to count the number of set bits in a number
int countSetBits(int num)
{
    int count = 0;
    while (num) {
        count += num
                 & 1; // Count the number of set bits in num
        num >>= 1; // Right shift num to check the next bit
    }
    return count;
}

// Function to count the numbers in the array for which the
// count of set bits is divisible by K
int countMultiples(int arr[], int n, int K)
{
    int res = 0;
    for (int i = 0; i < n; i++) {
        if (countSetBits(arr[i]) % K
            == 0) { // Check if the number of set bits is
                    // divisible by K
            res += 1;
        }
    }
    return res;
}

int main()
{
    // Example inputs
    int arr1[] = { 1, 2, 3, 4, 5 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int K1 = 2;

    int arr2[] = { 10, 20, 30, 40 };
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    int K2 = 4;

    // Example outputs
    cout << countMultiples(arr1, n1, K1)
         << endl; // Output: 2
    cout << countMultiples(arr2, n2, K2)
         << endl; // Output: 1

    return 0;
}

Java

public class CountSetBits {

    // Function to count the number of set bits in a number
    public static int countSetBits(int num)
    {
        int count = 0;
        while (num > 0) {
            count += num & 1; // Count the number of set
                              // bits in num
            num >>= 1; // Right shift num to check the next
                       // bit
        }
        return count;
    }

    // Function to count the numbers in the array for which
    // the count of set bits is divisible by K
    public static int countMultiples(int[] arr, int K)
    {
        int res = 0;
        for (int num : arr) {
            if (countSetBits(num) % K
                == 0) { // Check if the number of set bits
                        // is divisible by K
                res += 1;
            }
        }
        return res;
    }

    public static void main(String[] args)
    {
        // Example inputs
        int[] arr1 = { 1, 2, 3, 4, 5 };
        int K1 = 2;

        int[] arr2 = { 10, 20, 30, 40 };
        int K2 = 4;

        // Example outputs
        System.out.println(
            countMultiples(arr1, K1)); // Output: 2
        System.out.println(
            countMultiples(arr2, K2)); // Output: 1
    }
}

Python3

def countSetBits(num):
    count = 0
    while num:
        count += num & 1
        num >>= 1
    return count

def countMultiples(arr, K):
    res = 0
    for num in arr:
        if countSetBits(num) % K == 0:
            res += 1
    return res

# Example inputs
arr1 = [1, 2, 3, 4, 5]
K1 = 2
arr2 = [10, 20, 30, 40]
K2 = 4

# Example outputs
print(countMultiples(arr1, K1)) # 2
print(countMultiples(arr2, K2)) # 1

using System;

class Program {
    // Function to count the number of set bits in a number
    static int CountSetBits(int num)
    {
        int count = 0;
        while (num != 0) {
            count += num & 1; // Count the number of set
                              // bits in num
            num >>= 1; // Right shift num to check the next
                       // bit
        }
        return count;
    }

    // Function to count the numbers in the array for which
    // the count of set bits is divisible by K
    static int CountMultiples(int[] arr, int n, int K)
    {
        int res = 0;
        for (int i = 0; i < n; i++) {
            if (CountSetBits(arr[i]) % K
                == 0) // Check if the number of set bits is
                      // divisible by K
            {
                res += 1;
            }
        }
        return res;
    }

    static void Main(string[] args)
    {
        // Example inputs
        int[] arr1 = { 1, 2, 3, 4, 5 };
        int n1 = arr1.Length;
        int K1 = 2;

        int[] arr2 = { 10, 20, 30, 40 };
        int n2 = arr2.Length;
        int K2 = 4;

        // Example outputs
        Console.WriteLine(
            CountMultiples(arr1, n1, K1)); // Output: 2
        Console.WriteLine(
            CountMultiples(arr2, n2, K2)); // Output: 1
    }
}

JavaScript

function countSetBits(num) {
    let count = 0;
    while (num) {
        count += num & 1;
        num >>= 1;
    }
    return count;
}

function countMultiples(arr, K) {
    let res = 0;
    for (let num of arr) {
        if (countSetBits(num) % K === 0) {
            res += 1;
        }
    }
    return res;
}

// Example inputs
const arr1 = [1, 2, 3, 4, 5];
const K1 = 2;
const arr2 = [10, 20, 30, 40];
const K2 = 4;

// Example outputs
console.log(countMultiples(arr1, K1)); // 2
console.log(countMultiples(arr2, K2)); // 1

Output

2
1

Time Complexity: O(n * log(max(arr)))
Space Complexity: O(1)

Count arrays of length K whose product of elements is same as that of given array

Sanjit_Prasad

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

Using Bit Manipulation and Brute Force in python:

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