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

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  • Last Updated : 25 May, 2022
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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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