# Count of N-bit binary numbers without leading zeros

Given an integer N, the task is to find the count of N-bit binary numbers without leading zeros.

Examples:

Input: N = 2
Output: 2
10 and 11 are the only possible binary numbers.

Input: N = 4
Output: 8

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

Approach: Since the numbers cannot have leading zeros so the left-most bit has to be set to 1. Now for the rest of the N – 1 bits, there are two choices they can either be set to 0 or 1. So, the count of possible numbers will be 2N – 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count ` `// of possible numbers ` `int` `count(``int` `n) ` `{ ` `    ``return` `pow``(2, n - 1); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 4; ` ` `  `    ``cout << count(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `     `  `    ``// Function to return the count  ` `    ``// of possible numbers  ` `    ``static` `int` `count(``int` `n)  ` `    ``{  ` `        ``return` `(``int``)Math.pow(``2``, n - ``1``);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``int` `n = ``4``;  ` `     `  `        ``System.out.println(count(n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count ` `# of possible numbers ` `def` `count(n): ` `    ``return` `pow``(``2``, n ``-` `1``) ` ` `  `# Driver code ` `n ``=` `4` ` `  `print``(count(n)) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` `    ``// Function to return the count  ` `    ``// of possible numbers  ` `    ``static` `int` `count(``int` `n)  ` `    ``{  ` `        ``return` `(``int``)Math.Pow(2, n - 1);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (String[] args)  ` `    ``{  ` `        ``int` `n = 4;  ` `     `  `        ``Console.WriteLine(count(n));  ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```8
```

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.