Java Program to Check if two numbers are bit rotations of each other or not

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.
More information on bit rotation can be found here
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.
 

// 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

Yes

Time Complexity: O(log₂n), where n is the given number.
Auxiliary Space: O(1)

Please refer complete article on Check if two numbers are bit rotations of each other or not for more details!