1752. Check if Array Is Sorted and Rotated

#Problem

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.

https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/

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

#Test Cases

class Solution:
    def check(self, nums: List[int]) -> bool:

import pytest

from solution import Solution


@pytest.mark.parametrize('nums, expected', [
    ([3,4,5,1,2], True),
    ([2,1,3,4], False),
    ([1,2,3], True),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums, expected):
    assert sol.check(nums) == expected

#Thoughts

遍历数组，统计数字变小的次数，如果超过一次就不是循环有序数组。要注意数组的第一个数字和最后一个数字也要比较。

时间复杂度 O(n)，空间复杂度 O(1)。

#Code

class Solution:
    def check(self, nums: list[int]) -> bool:
        prev = nums[-1]
        down = 0
        for num in nums:
            if num < prev:
                down += 1
                if down > 1:
                    return False
            prev = num

        return True