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.
PHP
<?php // PHP 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. $x64 = $x | ( $x << 32); while ( $x64 >= $y ) { // comapring 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" ," "; else echo "no" ," "; // This code is contributed by aj_36 ?> |
yes
Time Complexity: O(log2n), where n is 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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