# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// CPP program to find number  ` `// of set bist in a number ` `#include ` `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 significant bit is not set ` `    ``else` `        ``return` `CountSetBits(n >> 1); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 21; ` `     `  `    ``// Function call ` `    ``cout << CountSetBits(n) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program 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 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 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#

 `// C# program 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 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 `

Output:

```3
```

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