Given a number, toggle all bits of it after most significant bit including most significant bit.
Examples :
Input : 10
Output : 5
Binary representation of 10 is 1010
After toggling we get 0101
Input : 5
Output : 2
We can toggle a bit by doing XOR of it with 1 (Note that 1 ^ 0 = 1 and 1 ^ 1 = 0). The idea is to take a number temp with only one bit set. One by one move the only set bit of temp to left and do XOR of it with n until it crosses MSB (Most Significant Bit) of n.
C++
#include<bits/stdc++.h>
using namespace std;
void toggle(int &n)
{
int temp = 1;
while (temp <= n)
{
n = n ^ temp;
temp = temp << 1;
}
}
int main()
{
int n = 10;
toggle(n);
cout << n;
return 0;
}
|
Java
class GFG {
static int toggle(int n) {
int temp = 1;
while (temp <= n) {
n = n ^ temp;
temp = temp << 1;
}
return n;
}
public static void main(String arg[])
{
int n = 10;
n = toggle(n);
System.out.print(n);
}
}
|
Python3
def toggle(n):
temp = 1
while (temp <= n):
n = n ^ temp
temp = temp << 1
return n
n = 10
n=toggle(n)
print(n)
|
C#
using System;
class GFG {
static int toggle(int n) {
int temp = 1;
while (temp <= n) {
n = n ^ temp;
temp = temp << 1;
}
return n;
}
public static void Main()
{
int n = 10;
n = toggle(n);
Console.Write(n);
}
}
|
PHP
<?php
function toggle( &$n)
{
$temp = 1;
while ($temp <= $n)
{
$n = $n ^ $temp;
$temp = $temp << 1;
}
}
$n = 10;
toggle($n);
echo $n;
?>
|
Javascript
<script>
function toggle(n)
{
let temp = 1;
while (temp <= n)
{
n = n ^ temp;
temp = temp << 1;
}
return n;
}
let n = 10;
n = toggle(n);
document.write(n);
</script>
|
Time Complexity: O(log n)
Auxiliary Space: O(1)
The above solution can be optimized to work in O(1) time under the assumption that numbers are stored in 32 bits.
C++
#include<bits/stdc++.h>
using namespace std;
int setAllBitsAfterMSB(int n)
{
n |= n>>1;
n |= n>>2;
n |= n>>4;
n |= n>>8;
n |= n>>16;
return n;
}
void toggle(int &n)
{
n = n ^ setAllBitsAfterMSB(n);
}
int main()
{
int n = 10;
toggle(n);
cout << n;
return 0;
}
|
Java
class GFG {
static int setAllBitsAfterMSB(int n) {
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
return n;
}
static int toggle(int n)
{
n = n ^ setAllBitsAfterMSB(n);
return n;
}
public static void main(String arg[])
{
int n = 10;
n = toggle(n);
System.out.print(n);
}
}
|
Python3
def setAllBitsAfterMSB(n):
n |= n>>1
n |= n>>2
n |= n>>4
n |= n>>8
n |= n>>16
return n
def toggle(n):
n = n ^ setAllBitsAfterMSB(n)
return n
n = 10
n=toggle(n)
print(n)
|
C#
using System;
class GFG {
static int setAllBitsAfterMSB(int n)
{
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
return n;
}
static int toggle(int n)
{
n = n ^ setAllBitsAfterMSB(n);
return n;
}
public static void Main()
{
int n = 10;
n = toggle(n);
Console.WriteLine(n);
}
}
|
PHP
<?php
function setAllBitsAfterMSB($n)
{
$n |= $n >> 1;
$n |= $n >> 2;
$n |= $n >> 4;
$n |= $n >> 8;
$n |= $n >> 16;
return $n;
}
function toggle(&$n)
{
$n = $n ^ setAllBitsAfterMSB($n);
}
$n = 10;
toggle($n);
echo $n;
?>
|
Javascript
<script>
function setAllBitsAfterMSB(n)
{
n |= n>>1;
n |= n>>2;
n |= n>>4;
n |= n>>8;
n |= n>>16;
return n;
}
function toggle(n)
{
n = n ^ setAllBitsAfterMSB(n);
return n;
}
let n = 10;
document.write(toggle(n));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Thanks to Devanshu Agarwal for suggesting this approach.
Another Approach:
To toggle a specific bit, we can take XOR of that bit with 1.
Therefore, for an n – bit number, we can construct a binary mask of the form 1111….11111, ie, all n bits are set. This is nothing but 2n – 1.
The implementation is shown below:
C++
#include<bits/stdc++.h>
using namespace std;
int toggle(int num)
{
int n = (int)log2(num) + 1;
int mask = pow(2, n) - 1;
return num ^ mask;
}
int main()
{
int num = 10;
cout << toggle(num);
}
|
Java
import java.util.*;
class GFG {
static int toggle(int num)
{
int n = (int)(Math.log(num) / Math.log(2)) + 1;
int mask = (int)Math.pow(2, n) - 1;
return num ^ mask;
}
public static void main(String[] args)
{
int num = 10;
System.out.println(toggle(num));
}
}
|
Python3
from math import log
def toggle(num):
n = int(log(num, 2)) + 1
mask = pow(2, n) - 1
return num ^ mask
num = 10
print(toggle(num))
|
C#
using System;
class GFG {
static int toggle(int num)
{
int n = (int)(Math.Log(num) / Math.Log(2)) + 1;
int mask = (int)Math.Pow(2, n) - 1;
return num ^ mask;
}
public static void Main(string[] args)
{
int num = 10;
Console.WriteLine(toggle(num));
}
}
|
Javascript
function toggle(num)
{
let n = Math.floor((Math.log(num) / Math.log(2)) + 1);
let mask = Math.pow(2, n) - 1;
return num ^ mask;
}
let num = 10;
console.log(toggle(num));
|
Time Complexity: O(logn)
Auxiliary Space: O(1)
This article is contributed by Devanshu Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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