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 approachclass 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 */ |
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.
Approach 2:
- Here’s another approach to solve the same problem:
- Iterate through all possible k values from 1 to n/2.
- For each k value, check if the string can be split into k cycles of length n/k. To do this, compare the substring of the original string from 0 to n/k with the substring of the original string from i*(n/k) to (i+1)*(n/k), for all i from 1 to k-1.
- If the string can be split into k cycles of length n/k, then it satisfies the given condition. Return “Yes”.
- If the string cannot be split into any cycles, then return “No”.
- Here’s the Java implementation of this approach:
Java
public class Main { public static String hasKCycles(int n, String s) { for (int k = 1; k <= n/2; k++) { if (n % k != 0) { continue; } int len = n/k; boolean flag = true; for (int i = 1; i < k; i++) { if (!s.substring((i-1)*len, i*len).equals(s.substring(i*len, (i+1)*len))) { flag = false; break; } } if (flag) { return "Yes"; } } return "No"; } public static void main(String[] args) { int n = 3; String s = "123"; System.out.println(hasKCycles(n, s)); }} |
Output:
Yes
Time Complexity: O(N^2), where N represents the length of the given string.
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!
