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Count of N-bit binary numbers without leading zeros

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  • Last Updated : 10 Mar, 2022
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Given an integer N, the task is to find the count of N-bit binary numbers without leading zeros.
Examples: 
 

Input: N = 2 
Output:
10 and 11 are the only possible binary numbers.
Input: N = 4 
Output:
 

 

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 <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)


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