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 2N – 1.

Below is the implementation of the above approach:

C++

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// 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














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// 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














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# 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#














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// 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






          
          
          
          

            

    

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