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Count of numbers whose 0th and Nth bits are set

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  • Last Updated : 31 May, 2022
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Given a positive integer N, the task is to count the numbers that can be represented with N bits and whose 0th and Nth bits are set.

Examples:

Input: N = 2 
Output:
All possible 2-bit integers are 00, 01, 10 and 11. 
Out of which only 11 has 0th and Nth bit set.
Input: N = 4 
Output: 4  

Approach: Out of the given N bits, only two bits need to be set i.e. the 0th and the Nth bit. So, setting these 2 bits as 1 we are left with the rest N – 2 bits every single of which can either be 0 or 1 and there are 2N – 2 ways of doing that.
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 n-bit
// numbers whose 0th and nth bits are set
int countNum(int n)
{
    if (n == 1)
        return 1;
    int count = pow(2, n - 2);
    return count;
}
 
// Driver code
int main()
{
    int n = 3;
    cout << countNum(n);
 
    return 0;
}


Java




// Java implementation of the approach
import java.io.*;
 
class GFG
{
    // Function to return the count of n-bit
    // numbers whose 0th and nth bits are set
    static int countNum(int n)
    {
        if (n == 1)
            return 1;
             
        int count = (int) Math.pow(2, n - 2);
        return count;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 3;
        System.out.println(countNum(n));
    }
}
 
// This code is contributed by ajit


Python




# Python3 implementation of the approach
 
# Function to return the count of n-bit
# numbers whose 0th and nth bits are set
def countNum(n):
    if (n == 1):
        return 1
    count = pow(2, n - 2)
    return count
 
# Driver code
 
n = 3
print(countNum(n))
 
# This code is contributed by mohit kumar 29


C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the count of n-bit
    // numbers whose 0th and nth bits are set
    static int countNum(int n)
    {
        if (n == 1)
            return 1;
             
        int count = (int) Math.Pow(2, n - 2);
        return count;
    }
     
    // Driver code
    static public void Main ()
    {
        int n = 3;
        Console.WriteLine(countNum(n));
    }
}
 
// This code is contributed by AnkitRai01


Javascript




<script>
// Javascript implementation of the approach
 
// Function to return the count of n-bit
// numbers whose 0th and nth bits are set
function countNum(n)
{
    if (n == 1)
        return 1;
    let count = Math.pow(2, n - 2);
    return count;
}
 
// Driver code
    let n = 3;
    document.write(countNum(n));
 
</script>


Output: 

2

 

Time Complexity: O(logn)

Auxiliary Space: O(1)


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