# 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 = 2Output:2

10 and 11 are the only possible binary numbers.Input:N = 4Output:8

**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 **2 ^{N – 1}**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `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` |

## Javascript

`<script>` `// JavaScript implementation of the approach` `// Function to return the count` `// of possible numbers` `function` `count(n)` `{` ` ` `return` `Math.pow(2, n - 1);` `}` `// Driver code` `var` `n = 4;` `document.write(count(n));` `</script>` |

**Output:**

8

Time Complexity: O(log n)

Auxiliary Space: O(1)