Given an integer x, the task is to find if every k-cycle shift on the element produces a number greater than or equal to the same element.
A k-cyclic shift of an integer x is a function that removes the last k digits of x and inserts them in its beginning.
For example, the k-cyclic shifts of 123 are 312 for k=1 and 231 for k=2. Print Yes if the given condition is satisfied else print No.
Input: x = 123
Output : Yes
The k-cyclic shifts of 123 are 312 for k=1 and 231 for k=2.
Both 312 and 231 are greater than 123.
The k-cyclic shift of 2214 when k=2 is 1422 which is smaller than 2214
Approach: Simply find all the possible k cyclic shifts of the number and check if all are greater than the given number or not.
Below is the implementation of the above approach:
Time Complexity: O(N2), where N represents the length of the given string.
The time complexity of the program is O(N2) because first it runs a loop for traversing the string and inside that substring function is used.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Check whether all the rotations of a given number is greater than or equal to the given number or not for more details!