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

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 = 123Output :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:2214Output: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

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