Toggle all even bits of a number
Last Updated :
07 Mar, 2023
Given a number, the task is to Toggle all even bit of a number
Examples:
Input : 10
Output : 0
binary representation 1 0 1 0
after toggle 0 0 0 0
Input : 20
Output : 30
binary representation 1 0 1 0 0
after toggle 1 1 1 1 0
1. First generate a number that contains even position bits.
2. Take XOR with the original number. Note that 1 ^ 1 = 0 and 1 ^ 0 = 1.
Let's understand this approach with below code.
C++
// CPP code to Toggle all even
// bit of a number
#include <iostream>
using namespace std;
// Returns a number which has all even
// bits of n toggled.
int evenbittogglenumber(int n)
{
// Generate number form of 101010
// ..till of same order as n
int res = 0, count = 0;
for (int temp = n; temp > 0; temp >>= 1) {
// if bit is even then generate
// number and or with res
if (count % 2 == 1)
res |= (1 << count);
count++;
}
// return toggled number
return n ^ res;
}
// Driver code
int main()
{
int n = 11;
cout << evenbittogglenumber(n);
return 0;
}
// Java code to Toggle all
// even bit of a number
import java.io.*;
class GFG {
// Returns a number which has
// all even bits of n toggled.
static int evenbittogglenumber(int n)
{
// Generate number form of 101010
// ..till of same order as n
int res = 0, count = 0;
for (int temp = n; temp > 0;
temp >>= 1)
{
// if bit is even then generate
// number and or with res
if (count % 2 == 1)
res |= (1 << count);
count++;
}
// return toggled number
return n ^ res;
}
// Driver code
public static void main(String args[])
{
int n = 11;
System.out.println(evenbittogglenumber(n));
}
}
// This code is contributed by Nikita Tiwari.
# Python code to Toggle all
# even bit of a number
# Returns a number which has all even
# bits of n toggled.
def evenbittogglenumber(n) :
# Generate number form of 101010
# ..till of same order as n
res = 0
count = 0
temp = n
while (temp > 0) :
# if bit is even then generate
# number and or with res
if (count % 2 == 1) :
res = res | (1 << count)
count = count + 1
temp >>= 1
# return toggled number
return n ^ res
# Driver code
n = 11
print(evenbittogglenumber(n))
#This code is contributed by Nikita Tiwari.
// C# code to Toggle all
// even bit of a number
using System;
class GFG {
// Returns a number which has
// all even bits of n toggled.
static int evenbittogglenumber(int n)
{
// Generate number form of 101010
// ..till of same order as n
int res = 0, count = 0;
for (int temp = n; temp > 0;
temp >>= 1)
{
// if bit is even then generate
// number and or with res
if (count % 2 == 1)
res |= (1 << count);
count++;
}
// return toggled number
return n ^ res;
}
// Driver code
public static void Main()
{
int n = 11;
Console.WriteLine(evenbittogglenumber(n));
}
}
// This code is contributed by Anant Agarwal.
<?php
// php code to Toggle all
// even bit of a number
// Returns a number which has
// all even bits of n toggled.
function evenbittogglenumber($n)
{
// Generate number form of 101010
// ..till of same order as n
$res = 0;
$count = 0;
for ($temp = $n; $temp > 0; $temp >>= 1)
{
// if bit is even then generate
// number and or with res
if ($count % 2 == 1)
$res |= (1 << $count);
$count++;
}
// return toggled number
return $n ^ $res;
}
// Driver code
$n = 11;
echo evenbittogglenumber($n);
// This code is contributed by mits
?>
<script>
// JavaScript program to Toggle all
// even bit of a number
// Returns a number which has
// all even bits of n toggled.
function evenbittogglenumber(n)
{
// Generate number form of 101010
// ..till of same order as n
let res = 0, count = 0;
for (let temp = n; temp > 0;
temp >>= 1)
{
// if bit is even then generate
// number and or with res
if (count % 2 == 1)
res |= (1 << count);
count++;
}
// return toggled number
return n ^ res;
}
// Driver code
let n = 11;
document.write(evenbittogglenumber(n));
</script>
Time Complexity : O(log n)
Auxiliary Space: O(1)
Another Approach:
To toggle a bit, we can take XOR of 1 and that bit (as 1 ^ 1 = 0, and 1 ^ 0 = 1). Therefore, to set all odd bits of an n bit number, we need to use a bit mask which is an n bit binary number with all odd bits set. This mask can be generated in 1 step using the formula of sum of a geometric progression, as an n bit number ...1010 is equal to 21 + 23 + 25 + .... 2 (n - !(n % 1)) .
C++
//C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
//function to toggle all the even bits
long long int evenbittogglenumber(long long int n) {
//calculating number of bits using log
int numOfBits = 1 + (int)log2(n);
//if there is only one bit,
if (numOfBits == 1)
return n;
//calculating the max power of GP series
int m = (numOfBits / 2);
//calculating mask using GP sum
//which is a(r ^ n - 1) / (r - 1)
//where a = 2, r = 4, n = m
int mask = 2 * (pow(4, m) - 1) / 3;
//toggling all even bits using mask ^ n
return mask ^ n;
}
// Driver code
int main()
{
int n = 11;
//function call
cout << evenbittogglenumber(n);
return 0;
}
//this code is contributed by phasing17
import java.lang.Math;
public class Main {
// function to toggle all the even bits
public static long evenBitToggleNumber(long n) {
// calculating number of bits using log
int numOfBits = 1 + (int) (Math.log(n) / Math.log(2));
// if there is only one bit,
if (numOfBits == 1)
return n;
// calculating the max power of GP series
int m = (numOfBits / 2);
// calculating mask using GP sum
// which is a(r ^ n - 1) / (r - 1)
// where a = 2, r = 4, n = m
long mask = 2 * ((long) Math.pow(4, m) - 1) / 3;
// toggling all even bits using mask ^ n
return mask ^ n;
}
// Main function
public static void main(String[] args) {
long n = 11;
// function call
System.out.println(evenBitToggleNumber(n));
}
}
# Python implementation of the approach
import math
# function to toggle all the even bits
def even_bit_toggle_number(n):
# calculating number of bits using log
num_of_bits = 1 + int(math.log2(n))
# if there is only one bit, return n
if num_of_bits == 1:
return n
# calculating the max power of GP series
m = num_of_bits // 2
# calculating mask using GP sum
# which is a(r ^ n - 1) / (r - 1)
# where a = 2, r = 4, n = m
mask = int(2 * ((4 ** m) - 1) / 3)
# toggling all even bits using mask ^ n
return mask ^ n
# Driver code
n = 11
# function call
print(even_bit_toggle_number(n))
// C# implementation of the approach
using System;
class GFG {
// function to toggle all the even bits
static int evenbittogglenumber(int n)
{
// calculating number of bits using log
int numOfBits
= 1 + (int)(Math.Log(n) / Math.Log(2));
// if there is only one bit,
if (numOfBits == 1)
return n;
// calculating the max power of GP series
int m = (numOfBits / 2);
// calculating mask using GP sum
// which is a(r ^ n - 1) / (r - 1)
// where a = 2, r = 4, n = m
int mask = 2 * (int)(Math.Pow(4, m) - 1) / 3;
// toggling all even bits using mask ^ n
return mask ^ n;
}
// Driver code
public static void Main(string[] args)
{
int n = 11;
// Function call
Console.WriteLine(evenbittogglenumber(n));
}
}
// This code is contributed by phasing17
//JavaScript implementation of the approach
//function to toggle all the even bits
function evenbittogglenumber(n) {
//calculating number of bits using log
let numOfBits = 1 + Math.floor(Math.log2(n));
//if there is only one bit,
if (numOfBits == 1)
return n;
//calculating the max power of GP series
let m = Math.floor(numOfBits / 2);
//calculating mask using GP sum
//which is a(r ^ n - 1) / (r - 1)
//where a = 2, r = 4, n = m
let mask = Math.floor(2 * (Math.pow(4, m) - 1) / 3);
//toggling all even bits using mask ^ n
return mask ^ n;
}
// Driver code
let n = 11;
//function call
console.log(evenbittogglenumber(n));
//this code is contributed by phasing17
Time Complexity : O(1)
Auxiliary Space: O(1)
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