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 the implementation of the above algorithm.
Javascript
<script>// javascript program to check if two numbers are bit rotations // of each other.// function to check if two numbers are equal // after bit rotation function isRotation(x, y){ // x64 has concatenation of x with itself. var x64 = x | (x << 32); while (x64 >= y) { // comapring only last 32 bits if (x64 == y) { return true; } // right shift by 1 unit x64 >>= 1; } return false;}// driver code to test above function var x = 122;var y = 2147483678;if (isRotation(x, y) == false) { document.write("Yes");} else { document.write("No");}// This code is contributed by 29AjayKumar </script> |
Output:
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!
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