# 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

**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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

8

## Recommended Posts:

- Count numbers having N 0's and and M 1's with no leading zeros
- Comparing leading zeros in binary representations of two numbers
- All possible numbers of N digits and base B without leading zeros
- Number of leading zeros in binary representation of a given number
- Count unique numbers that can be generated from N by adding one and removing trailing zeros
- Count number of trailing zeros in Binary representation of a number using Bitset
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- Count number of trailing zeros in product of array
- Count numbers < = N whose difference with the count of primes upto them is > = K
- Count numbers which are divisible by all the numbers from 2 to 10
- Count numbers which can be constructed using two numbers
- Count numbers that don't contain 3
- Count number of binary strings of length N having only 0's and 1's

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