# Java Program for Check whether all the rotations of a given number is greater than or equal to the given number or not

• Last Updated : 25 May, 2022

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.
Examples:

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.
Input: 2214
Output: No
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:

## Java

 `// Java implementation of the approach` `class` `GFG ` `{`   `    ``static` `void` `CheckKCycles(``int` `n, String s) ` `    ``{` `        ``boolean` `ff = ``true``;` `        ``int` `x = ``0``;` `        ``for` `(``int` `i = ``1``; i < n; i++) ` `        ``{`   `            ``// Splitting the number at index i ` `            ``// and adding to the front ` `            ``x = (s.substring(i) + s.substring(``0``, i)).length();`   `            ``// Checking if the value is greater than ` `            ``// or equal to the given value ` `            ``if` `(x >= s.length()) ` `            ``{` `                ``continue``;` `            ``}` `            ``ff = ``false``;` `            ``break``;` `        ``}` `        ``if` `(ff) ` `        ``{` `            ``System.out.println(``"Yes"``);` `        ``}` `        ``else` `        ``{` `            ``System.out.println(``"No"``);` `        ``}`   `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``int` `n = ``3``;` `        ``String s = ``"123"``;` `        ``CheckKCycles(n, s);` `    ``}` `}`   `/* This code contributed by PrinciRaj1992 */`

Output:

`Yes`

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!

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