Toggle first and last bits of a number - Coding

Given a number n, the task is to toggle only first and last bits of a number
Examples :

Input : 10
Output : 3
Binary representation of 10 is
1010. After toggling first and
last bits, we get 0011.

Input : 15
Output : 6

Prerequisite : Find MSB of given number.
1) Generate a number which contains first and last bit as set. We need to change all middle bits to 0 and keep corner bits as 1.
2) Answer is XOR of generated number and original number.

C++

<?php
// PHP program to toggle first and last 
// bits of a number
 
// Returns a number which has same bit
// count as n and has only first and last 
// bits as set.
function takeLandFsetbits($n)
{
    // set all the bit of the number
    $n |= $n >> 1;
    $n |= $n >> 2;
    $n |= $n >> 4;
    $n |= $n >> 8;
    $n |= $n >> 16;
 
    // Adding one to n now unsets
    // all bits and moves MSB to 
    // one place. Now we shift 
    // the number by 1 and add 1.
    return (($n + 1) >> 1) + 1;
}
 
function toggleFandLbits(int $n)
{
    // if number is 1
    if ($n == 1)
        return 0;
 
    // take XOR with first and
    // last set bit number
    return $n ^ takeLandFsetbits($n);
}
 
// Driver code
$n = 10;
echo toggleFandLbits($n);
 
// This code is contributed by mits 
?>

Javascript

<script>
 
// JavaScript program to toggle first and last 
// bits of a number
 
    // Returns a number which has same bit
    // count as n and has only first and last 
    // bits as set.
    function takeLandFsetbits(n)
    {
        // set all the bit of the number
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
        
        // Adding one to n now unsets
        // all bits and moves MSB to 
        // one place. Now we shift 
        // the number by 1 and add 1.
        return ((n + 1) >> 1) + 1;
    }
        
    function toggleFandLbits(n)
    {
        // if number is 1
        if (n == 1)
            return 0;
        
        // take XOR with first and
        // last set bit number
        return n ^ takeLandFsetbits(n);
    }
 
// Driver code
 
        let n = 10;
        document.write(toggleFandLbits(n));
 
</script>

Output

Time Complexity: O(1).
Auxiliary Space: O(1).

Another Approach:

We can do this in two steps by:
To generate an bit mask for n, where only the last and first bits are set, we need to calculate 2^log_²ⁿ + 2⁰
Then, just take the XOR of the mask and n.

Below is the implementation of the above approach:

C++

// CPP program to toggle first and last
// bits of a number
#include <bits/stdc++.h>
using namespace std;
 
//function to toggle first and last bits
//of a number
int toggleFandLbits(int n)
{
    //calculating mask
    int mask = pow(2, (int)log2(n)) + 1;
    //taking xor of mask and n
    return mask ^ n;
}
 
// Driver code
int main()
{
    int n = 10;
    //function call
    cout << toggleFandLbits(n);
    return 0;
}
 
 
//this code is contributed by phasing17

Java

// Java program to toggle first and last
// bits of a number
class GFG
{
   
    // function to toggle first and last bits
    // of a number
    static int toggleFandLbits(int n)
    {
        // calculating mask
        int mask = (int)Math.pow(
                       2, (int)(Math.log(n) / Math.log(2)))
                   + 1;
 
        // taking xor of mask and n
        return mask ^ n;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 10;
 
        // Function call
        System.out.println(toggleFandLbits(n));
    }
}
 
// this code is contributed by phasing17

Python3

# Python3 program to toggle first and last
# bits of a number
from math import log
 
# Function to toggle first and last bits
# of a number
def toggleFandLbits(n):
 
    # calculating mask
    mask = 2 ** (int(log(n, 2))) + 1
     
    # taking xor of mask and n
    return mask ^ n
 
# Driver code
n = 10
 
# Function call
print(toggleFandLbits(n))
 
# This code is contributed by phasing17

C#

// C# program to toggle first and last
// bits of a number
using System;
 
class GFG {
 
  // function to toggle first and last bits
  // of a number
  static int toggleFandLbits(int n)
  {
    // calculating mask
    int mask = (int)Math.Pow(
      2, (int)(Math.Log(n) / Math.Log(2)))
      + 1;
 
    // taking xor of mask and n
    return mask ^ n;
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int n = 10;
 
    // Function call
    Console.WriteLine(toggleFandLbits(n));
  }
}
 
// This code is contributed by phasing17

Javascript

// JavaScript program to toggle first and last
// bits of a number
 
 
// function to toggle first and last bits
// of a number
function toggleFandLbits(n)
{
    // calculating mask
    let mask = Math.pow(2, Math.floor(Math.log2(n))) + 1;
     
    // taking xor of mask and n
    return mask ^ n;
}
 
// Driver code
let  n = 10;
     
// function call
console.log(toggleFandLbits(n));
 
// This code is contributed by phasing17

Output

Time Complexity: O(log₂(log₂n)), as pow(log(n)) will be calculated to complexity will be log(log(n)).
Auxiliary Space: O(1).

Reffered: https://www.geeksforgeeks.org

Bit Magic

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Reverse actual bits of the given number
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Category:	Coding
Sub Category:	Tutorial
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