Python3 Program to 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 = 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: 

Python

# Python3 implementation of the approach
def CheckKCycles(n, s):
    ff = True
    for i in range(1, n):

        # Splitting the number at index i
        # and adding to the front
        x = int(s[i:] + s[0:i])

        # Checking if the value is greater than
        # or equal to the given value
        if (x >= int(s)):
            continue
        ff = False
        break
    if (ff):
        print("Yes")
    else:
        print("No")

n = 3
s = "123"
CheckKCycles(n, s)

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.

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 Python implementation of this approach:

Python3

def has_k_cycles(n, s):
    for k in range(1, n//2 + 1):
        if n % k != 0:
            continue

        length = n // k
        flag = True

        for i in range(1, k):
            if s[(i-1)*length:i*length] != s[i*length:(i+1)*length]:
                flag = False
                break

        if flag:
            return "Yes"

    return "No"

n = 3
s = "123"
print(has_k_cycles(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.

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