Count of numbers whose 0th and Nth bits are set

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: 1
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++

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// 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;
}

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Java

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// 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

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Python

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# 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

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C#

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// 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 

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Output:

2


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