# Check if two numbers are bit rotations of each other or not

• Difficulty Level : Medium
• Last Updated : 14 Jun, 2022

Given two positive integers x and y, check if one integer is obtained by rotating bits of other.

`Input constraint: 0 < x, y < 2^32 `

Bit Rotation: A rotation (or circular shift) is an operation similar to shift except that the bits that fall off at one end are put back to the other end.
Example 1 :

```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```

Example 2 :

```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```

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 it 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 concatenation of x to x ie..
x64 has first 32 bits same as bits of x and last 32 bits are also same as bits of x64.
Then we keep on shifting x64 by 1 on 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 combination due to rotation.
Here is implementation of above algorithm.

## C++

 `// C++ program to check if two numbers are bit rotations``// of each other.``#include ``using` `namespace` `std;` `// function to check if  two numbers are equal``// after bit rotation``bool` `isRotation(unsigned ``int` `x, unsigned ``int` `y)``{``    ``// x64 has concatenation of x with itself.``    ``unsigned ``long` `long` `int` `x64 = x | ((unsigned ``long` `long` `int``)x << 32);` `    ``while` `(x64 >= y)``    ``{``        ``// comparing only last 32 bits``        ``if` `(unsigned(x64) == y)``            ``return` `true``;` `        ``// right shift by 1 unit``        ``x64 >>= 1;``    ``}``    ``return` `false``;``}` `// driver code to test above function``int` `main()``{``    ``unsigned ``int` `x = 122;``    ``unsigned ``int` `y = 2147483678;` `    ``if` `(isRotation(x, y))``        ``cout << ``"yes"` `<< endl;``    ``else``        ``cout << ``"no"` `<< endl;` `    ``return` `0;``}`

## Java

 `// Java program to check if two numbers are bit rotations``// of each other.``class` `GFG {` `// function to check if two numbers are equal``// after bit rotation``    ``static` `boolean` `isRotation(``long` `x, ``long` `y) {``        ``// x64 has concatenation of x with itself.``        ``long` `x64 = x | (x << ``32``);` `        ``while` `(x64 >= y) {``            ``// comparing only last 32 bits``            ``if` `(x64 == y) {``                ``return` `true``;``            ``}` `            ``// right shift by 1 unit``            ``x64 >>= ``1``;``        ``}``        ``return` `false``;``    ``}` `// driver code to test above function``    ``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"``);``        ``}``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to check if two``# numbers are bit rotations of each other.` `# function to check if two numbers``# are equal after bit rotation``def` `isRotation(x, y) :``    ` `    ``# x64 has concatenation of x``    ``# with itself.``    ``x64 ``=` `x | (x << ``32``)``    ` `    ``while` `(x64 >``=` `y) :``        ` `        ``# comparing only last 32 bits``        ``if` `((x64) ``=``=` `y) :``            ``return` `True` `        ``# right shift by 1 unit``        ``x64 >>``=` `1` `    ``return` `False` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``x ``=` `122``    ``y ``=` `2147483678``    ` `    ``if` `(isRotation(x, y) ``=``=` `False``) :``        ``print``(``"yes"``)``    ``else` `:``        ``print``(``"no"``)` `# This code is contributed by Ryuga`

## C#

 `// C# program to check if two numbers``// are bit rotations of each other.``using` `System;` `class` `GFG``{` `// function to check if two numbers``// are equal after bit rotation``static` `bool` `isRotation(``long` `x, ``long` `y)``{``    ``// x64 has concatenation of``    ``// x with itself.``    ``long` `x64 = x | (x << 32);` `    ``while` `(x64 >= y)``    ``{``        ``// comparing only last 32 bits``        ``if` `(x64 == y)``        ``{``            ``return` `true``;``        ``}` `        ``// right shift by 1 unit``        ``x64 >>= 1;``    ``}``    ``return` `false``;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``long` `x = 122;``    ``long` `y = 2147483678L;` `    ``if` `(isRotation(x, y) == ``false``)``    ``{``        ``Console.Write(``"Yes"``);``    ``}``    ``else``    ``{``        ``Console.Write(``"No"``);``    ``}``}``}` `// This code is contributed``// by 29AjayKumar`

## PHP

 `= ``\$y``)``    ``{``        ``// comparing only last 32 bits``        ``if` `((``\$x64``) == ``\$y``)``            ``return` `1;` `        ``// right shift by 1 unit``        ``\$x64` `>>= 1;``    ``}``    ``return` `-1;``}` `// Driver Code``\$x` `= 122;``\$y` `= 2147483678;` `if` `(isRotation(``\$x``, ``\$y``))``    ``echo` `"yes"` `,``"\n"``;``else``    ``echo` `"no"` `,``"\n"``;` `// This code is contributed by aj_36``?>`

## Javascript

 ``

Output :

`yes`

Time Complexity: O(logn)

Auxiliary Space: O(1)
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