Check if two numbers are bit rotations of each other or not
Last Updated :
13 Jul, 2022
Given two positive integers x and y (0 < x, y < 2^32), check if one integer is obtained by rotating bits of the other.
Bit Rotation: A rotation (or circular shift) is an operation similar to a shift except that the bits that fall off at one end are put back to the other end.
Examples:
Input : a = 8, b = 1
Output : yes
Explanation : Representation of a = 8 : 0000 0000 0000 0000 0000 0000 0000 1000 ,Representation of b = 1 : 0000 0000 0000, 0000 0000 0000 0000 0001. If we rotate a by 3 units right we get b, hence answer is yes.
Input : a = 122, b = 2147483678
Output : yes
Explanation :Representation of a = 122 : 0000 0000 0000 0000 0000 0000 0111 1010,Representation of b = 2147483678 : 1000 0000 0000 0000 0000 0000 0001 1110, If we rotate a by 2 units right we get b, hence answer is yes.
Approach:
- Since total bits in which x or y can be represented is 32 since x, y > 0 and x, y < 2^32.
- So we need to find all 32 possible rotations of x and compare them with y till x and y are not equal.
- To do this we use a temporary variable x64 with 64 bits, which is result of the concatenation of x to x ie. x64 has the first 32 bits the same as bits of x and the last 32 bits are also the same as bits of x64.
- Then we keep on shifting x64 by 1 on the right side and compare the rightmost 32 bits of x64 with y.
- In this way, we’ll be able to get all the possible bits combinations due to rotation.
Here is implementation of above algorithm.
C++
#include <iostream>
using namespace std;
bool isRotation(unsigned int x, unsigned int y)
{
unsigned long long int x64 = x | ((unsigned long long int )x << 32);
while (x64 >= y)
{
if (unsigned(x64) == y)
return true ;
x64 >>= 1;
}
return false ;
}
int main()
{
unsigned int x = 122;
unsigned int y = 2147483678;
if (isRotation(x, y))
cout << "yes" << endl;
else
cout << "no" << endl;
return 0;
}
|
Java
class GFG {
static boolean isRotation( long x, long y) {
long x64 = x | (x << 32 );
while (x64 >= y) {
if (x64 == y) {
return true ;
}
x64 >>= 1 ;
}
return false ;
}
public static void main(String[] args) {
long x = 122 ;
long y = 2147483678L;
if (isRotation(x, y) == false ) {
System.out.println( "Yes" );
} else {
System.out.println( "No" );
}
}
}
|
Python3
def isRotation(x, y) :
x64 = x | (x << 32 )
while (x64 > = y) :
if ((x64) = = y) :
return True
x64 >> = 1
return False
if __name__ = = "__main__" :
x = 122
y = 2147483678
if (isRotation(x, y) = = False ) :
print ( "yes" )
else :
print ( "no" )
|
C#
using System;
class GFG
{
static bool isRotation( long x, long y)
{
long x64 = x | (x << 32);
while (x64 >= y)
{
if (x64 == y)
{
return true ;
}
x64 >>= 1;
}
return false ;
}
public static void Main()
{
long x = 122;
long y = 2147483678L;
if (isRotation(x, y) == false )
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
}
}
|
PHP
<?php
function isRotation( $x , $y )
{
$x64 = $x | ( $x << 32);
while ( $x64 >= $y )
{
if (( $x64 ) == $y )
return 1;
$x64 >>= 1;
}
return -1;
}
$x = 122;
$y = 2147483678;
if (isRotation( $x , $y ))
echo "yes" , "\n" ;
else
echo "no" , "\n" ;
?>
|
Javascript
<script>
function isRotation(x, y)
{
var x64 = x | (x << 32);
while (x64 >= y)
{
if (x64 == y) {
return true ;
}
x64 >>= 1;
}
return false ;
}
var x = 122;
var y = 2147483678;
if (isRotation(x, y) == false ) {
document.write( "Yes" );
} else {
document.write( "No" );
}
</script>
|
Time Complexity: O(logn)
Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...