"""
1752. Check if Array Is Sorted and Rotated
Easy

Given an array nums, return true if the array was originally sorted in non-decreasing order, then rotated some number of positions (including zero). Otherwise, return false.

There may be duplicates in the original array.

Note: An array A rotated by x positions results in an array B of the same length such that A[i] == B[(i+x) % A.length], where % is the modulo operation.

 

Example 1:

Input: nums = [3,4,5,1,2]
Output: true
Explanation: [1,2,3,4,5] is the original sorted array.
You can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2].
Example 2:

Input: nums = [2,1,3,4]
Output: false
Explanation: There is no sorted array once rotated that can make nums.
Example 3:

Input: nums = [1,2,3]
Output: true
Explanation: [1,2,3] is the original sorted array.
You can rotate the array by x = 0 positions (i.e. no rotation) to make nums.
Example 4:

Input: nums = [1,1,1]
Output: true
Explanation: [1,1,1] is the original sorted array.
You can rotate any number of positions to make nums.
Example 5:

Input: nums = [2,1]
Output: true
Explanation: [1,2] is the original sorted array.
You can rotate the array by x = 5 positions to begin on the element of value 2: [2,1].
 

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
"""
class Solution:
    def check(self, nums: List[int]) -> bool:
        return sum(nums[i] < nums[i-1] for i in range(len(nums))) <= 1