Count Set-bits of number using Recursion

Given a number N. The task is to find the number of set bits in its binary representation using recursion.

Examples:

Input : 21
Output : 3
21 represesnted as 10101 in binray representation

Input : 16
Output : 1
16 represesnted as 10000 in binray representation

Approach:

  1. First, check the LSB of the number.
  2. If the LSB is 1, then we add 1 to our answer and divide the number by 2.
  3. If the LSB is 0, we add 0 to our answer and divide the number by 2.
  4. Then we recursively follow step (1) until the number is greater than 0.

Below is the implementation of the above approach :

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP progarm to find number 
// of set bist in a number
#include <bits/stdc++.h>
using namespace std;
  
// Recursive function to find 
// number of set bist in a number
int CountSetBits(int n)
{
    // Base condition
    if (n == 0)
        return 0;
          
    // If Least signifiant bit is set
    if((n & 1) == 1)
        return 1 + CountSetBits(n >> 1);
      
    // If Least signifiant bit is not set
    else
        return CountSetBits(n >> 1);
}
  
// Driver code
int main()
{
    int n = 21;
      
    // Function call
    cout << CountSetBits(n) << endl;
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java progarm to find number 
// of set bist in a number
class GFG
{
    // Recursive function to find 
    // number of set bist in a number
    static int CountSetBits(int n)
    {
        // Base condition
        if (n == 0)
            return 0;
              
        // If Least signifiant bit is set
        if((n & 1) == 1)
            return 1 + CountSetBits(n >> 1);
          
        // If Least signifiant bit is not set
        else
            return CountSetBits(n >> 1);
    }
      
    // Driver code
    public static void main (String [] args)
    {
        int n = 21;
          
        // Function call
        System.out.println(CountSetBits(n));
    }
}
  
// This code is contributed by ihritik

chevron_right


Python3

# Python3 progarm to find number
# of set bist in a number

# Recursive function to find
# number of set bist in a number
def CountSetBits(n):

# Base condition
if (n == 0):
return 0;

# If Least signifiant bit is set
if((n & 1) == 1):
return 1 + CountSetBits(n >> 1);

# If Least signifiant bit is not set
else:
return CountSetBits(n >> 1);

# Driver code
if __name__ == ‘__main__’:
n = 21;

# Function call
print(CountSetBits(n));

# This code is contributed by 29AjayKumar

C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# progarm to find number 
// of set bist in a number
using System;
  
class GFG
{
    // Recursive function to find 
    // number of set bist in a number
    static int CountSetBits(int n)
    {
        // Base condition
        if (n == 0)
            return 0;
              
        // If Least signifiant bit is set
        if((n & 1) == 1)
            return 1 + CountSetBits(n >> 1);
          
        // If Least signifiant bit is not set
        else
            return CountSetBits(n >> 1);
    }
      
    // Driver code
    public static void Main ()
    {
        int n = 21;
          
        // Function call
        Console.WriteLine(CountSetBits(n));
    }
}
  
// This code is contributed by ihritik

chevron_right


Output:

3


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : ihritik



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.