Given n number, the task is to toggle odd bit of the number.
Examples:
Input : 10 Output : 15 binary representation 1 0 1 0 after toggle 1 1 1 1 Input : 20 Output : 1 binary representation 1 0 1 0 0 after toggle 0 0 0 0 1
1. First generate a number that contains odd 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++
// Toggle all odd bit of a number #include <iostream> using namespace std;
// Returns a number which has all odd // 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 odd, then generate
// number and or with res
if (count % 2 == 0)
res |= (1 << count);
count++;
}
// return toggled number
return n ^ res;
} // Driver code int main()
{ int n = 11;
cout << evenbittogglenumber(n);
return 0;
} |
Java
// Toggle all odd bit of a number import java.io.*;
class GFG {
// Returns a number which has all odd
// 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 odd, then generate
// number and or with res
if (count % 2 == 0 )
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.*/ |
Python3
# Python3 code for Toggle all odd bit of a number # Returns a number which has all odd # 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 odd, then generate
# number and or with res
if (count % 2 = = 0 ) :
res = res | ( 1 << count)
count = count + 1
temp >> = 1
# Return toggled number
return n ^ res
# Driver code if __name__ = = '__main__' :
n = 11
print (evenbittogglenumber(n))
# This code is contributed by Nikita Tiwari. |
C#
// C# code for Toggle all odd bit of a number using System;
class GFG {
// Returns a number which has all odd
// 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 odd, then generate
// number and or with res
if (count % 2 == 0)
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 // php implementation of Toggle // all odd bit of a number // Returns a number which has // all odd 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 odd, then generate
// number and or with res
if ( $count % 2 == 0)
$res |= (1 << $count );
$count ++;
}
// return toggled number
return $n ^ $res ;
} // Driver code
$n = 11;
echo evenbittogglenumber( $n );
// This code is contributed by mits ?> |
Javascript
<script> // JavaScript program Toggle all odd bit of a number // Returns a number which has all odd
// 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 odd, then generate
// number and or with res
if (count % 2 == 0)
res |= (1 << count);
count++;
}
// return toggled number
return n ^ res;
}
// Driver code let n = 11;
document.write(evenbittogglenumber(n));
</script> |
Output :
14
Time Complexity : O(log n)
Space Complexity : O(1)
Recommended Articles