# Count of numbers having only 1 set bit in the range [0, n]

Given an integer n, the task is to count the numbers having only 1 set bit in the range [0, n].

Examples:

Input: n = 7
Output: 3
000, 001, 010, 011, 100, 101, 110 and 111 are the binary representation of all the numbers upto 7.
And there are only 3 numbers ( 001, 010 and 100 ) having only 1 set bit.

Input: n = 3
Output: 2

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

Approach: If k bits are required to represent n then there are k numbers possible as 1 can be positioned at k different positions each time.

Below is the implementation of the above approach

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the required count ` `int` `count(``int` `n) ` `{ ` ` `  `    ``// To store the count of numbers ` `    ``int` `cnt = 0; ` `    ``int` `p = 1; ` `    ``while` `(p <= n) { ` `        ``cnt++; ` ` `  `        ``// Every power of 2 contains ` `        ``// only 1 set bit ` `        ``p *= 2; ` `    ``} ` `    ``return` `cnt; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 7; ` `    ``cout << count(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function to return the required count ` `    ``static` `int` `count(``int` `n) ` `    ``{ ` ` `  `        ``// To store the count of numbers ` `        ``int` `cnt = ``0``; ` `        ``int` `p = ``1``; ` `        ``while` `(p <= n) { ` `            ``cnt++; ` ` `  `            ``// Every power of 2 contains ` `            ``// only 1 set bit ` `            ``p *= ``2``; ` `        ``} ` `        ``return` `cnt; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``7``; ` `        ``System.out.print(count(n)); ` `    ``} ` `} `

## C#

 `// C# implementation of the approach ` `using` `System; ` `class` `GFG { ` ` `  `    ``// Function to return the required count ` `    ``static` `int` `count(``int` `n) ` `    ``{ ` ` `  `        ``// To store the count of numbers ` `        ``int` `cnt = 0; ` `        ``int` `p = 1; ` `        ``while` `(p <= n) { ` `            ``cnt++; ` ` `  `            ``// Every power of 2 contains ` `            ``// only 1 set bit ` `            ``p *= 2; ` `        ``} ` `        ``return` `cnt; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 7; ` `        ``Console.Write(count(n)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the required count ` `def` `count(n): ` `     `  `    ``# To store the count of numbers ` `    ``cnt ``=` `0` `    ``p ``=` `1` `    ``while` `(p <``=` `n): ` `        ``cnt ``=` `cnt ``+` `1` `         `  `        ``# Every power of 2 contains  ` `        ``# only 1 set bit ` `        ``p ``*``=` `2` `    ``return` `cnt ` ` `  `# Driver code ` `n ``=` `7` `print``(count(n)); `

## PHP

 ` `

Output:

```3
```

Time Complexity: O(log n)

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Improved By : jit_t