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 represented as 10101 in binary representationInput : 16
Output : 1
16 represented as 10000 in binary representation
Approach:
- First, check the LSB of the number.
- If the LSB is 1, then we add 1 to our answer and divide the number by 2.
- If the LSB is 0, we add 0 to our answer and divide the number by 2.
- Then we recursively follow step (1) until the number is greater than 0.
Below is the implementation of the above approach :
C++
// CPP program 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 numberint CountSetBits(int n){ // Base condition if (n == 0) return 0; // If Least significant bit is set if((n & 1) == 1) return 1 + CountSetBits(n >> 1); // If Least significant bit is not set else return CountSetBits(n >> 1);}// Driver codeint main(){ int n = 21; // Function call cout << CountSetBits(n) << endl; return 0;} |
Java
// Java program to find number // of set bist in a numberclass 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 significant bit is set if((n & 1) == 1) return 1 + CountSetBits(n >> 1); // If Least significant 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 |
Python3
# Python3 program to find number # of set bist in a number# Recursive function to find # number of set bist in a numberdef CountSetBits(n): # Base condition if (n == 0): return 0; # If Least significant bit is set if((n & 1) == 1): return 1 + CountSetBits(n >> 1); # If Least significant bit is not set else: return CountSetBits(n >> 1);# Driver codeif __name__ == '__main__': n = 21; # Function call print(CountSetBits(n));# This code is contributed by 29AjayKumar |
C#
// C# program to find number // of set bist in a numberusing 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 significant bit is set if((n & 1) == 1) return 1 + CountSetBits(n >> 1); // If Least significant 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 |
Javascript
<script>// Javascript program to find number // of set bist in a number// Recursive function to find // number of set bist in a numberfunction CountSetBits(n){ // Base condition if (n == 0) return 0; // If Least significant bit is set if ((n & 1) == 1) return 1 + CountSetBits(n >> 1); // If Least significant bit is not set else return CountSetBits(n >> 1);}// Driver codevar n = 21;// Function calldocument.write(CountSetBits(n));// This code is contributed by Amit Katiyar </script> |
Output:
3
Time Complexity: O(log n)
Auxiliary Space: O(log n)
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