Check if Array Is Sorted and Rotated

#leetcode #dsa #theabbie

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.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

SOLUTION:

class Solution:
    def check(self, nums: List[int]) -> bool:
        n = len(nums)
        pivot = 0
        for i in range(1, n):
            if nums[i] < nums[i - 1]:
                pivot = i
                break
        for i in range(pivot, pivot + n - 1):
            if nums[(i + 1) % n] < nums[i % n]:
                return False
        return True