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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