Count integers in an Array which are multiples their bits counts

Given an array arr[] of N elements, the task is to count all the elements which are a multiple of their set bits count.

Examples:

Input : arr[] = { 1, 2, 3, 4, 5, 6 }
Output : 4
Explanation :
There numbers which are multiple of their setbits count are { 1, 2, 4, 6 }.

Input : arr[] = {10, 20, 30, 40}
Output : 3
Explanation :
There numbers which are multiple of their setbits count are { 10, 20, 40 }

Approach: Loop through each array elements one by one. Count the set bits of every number in the array. Check if the current integer is a multiple of its set bits count or not. If ‘yes’ then increment the counter by 1, else skip that integer.

Below is the implementation of above approach:

C++

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#include <bits/stdc++.h>
using namespace std;
  
// Function to find the count of numbers
// which are multiple of its set bits count
int find_count(vector<int>& arr)
{
    // variable to store count
    int ans = 0;
  
    // iterate over elements of array
    for (int i : arr) {
  
        // Get the set-bits count of each element
        int x = __builtin_popcount(i);
  
        // Check if the setbits count
        // divides the integer i
        if (i % x == 0)
  
            // Increment the count
            // of required numbers by 1
            ans += 1;
    }
  
    return ans;
}
  
// Driver code
int main()
{
    vector<int> arr
        = { 1, 2, 3, 4, 5, 6 };
  
    cout << find_count(arr);
  
    return 0;
}

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Java

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class GFG{
   
// Function to find the count of numbers
// which are multiple of its set bits count
static int find_count(int []arr)
{
    // variable to store count
    int ans = 0;
   
    // iterate over elements of array
    for (int i : arr) {
   
        // Get the set-bits count of each element
        int x = Integer.bitCount(i);
   
        // Check if the setbits count
        // divides the integer i
        if (i % x == 0)
   
            // Increment the count
            // of required numbers by 1
            ans += 1;
    }
   
    return ans;
}
   
// Driver code
public static void main(String[] args)
{
    int []arr
        = { 1, 2, 3, 4, 5, 6 };
   
    System.out.print(find_count(arr));
   
}
}
  
// This code contributed by Princi Singh

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Python3

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# Python3 implementation of above approach
  
# function to return set bits count
def bitsoncount(x):
    return bin(x).count('1')
  
# Function to find the count of numbers 
# which are multiple of its set bits count 
def find_count(arr) :
    # variable to store count 
    ans = 0 
  
    # iterate over elements of array 
    for i in arr :
  
        # Get the set-bits count of each element 
        x = bitsoncount(i)
  
        # Check if the setbits count 
        # divides the integer i 
        if (i % x == 0):
  
            # Increment the count 
            # of required numbers by 1 
            ans += 1
  
    return ans
  
# Driver code 
arr = [ 1, 2, 3, 4, 5, 6 ]
  
print(find_count(arr))
  
# This code is contributed by Sanjit_Prasad

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C#

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using System;
  
public class GFG{
    
// Function to find the count of numbers
// which are multiple of its set bits count
static int find_count(int []arr)
{
    // Variable to store count
    int ans = 0;
    
    // Iterate over elements of array
    foreach (int i in arr) {
    
        // Get the set-bits count of each element
        int x = bitCount(i);
    
        // Check if the setbits count
        // divides the integer i
        if (i % x == 0)
    
            // Increment the count
            // of required numbers by 1
            ans += 1;
    }
    
    return ans;
}
static int bitCount(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
        = { 1, 2, 3, 4, 5, 6 };
    
    Console.Write(find_count(arr));
    
}
}
// This code contributed by Princi Singh

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Output :

4

Time complexity:- O(nlog(max(arr[])), where n is the size of the array,
Space complexity:- O(1)

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