## Description

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

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:trueExplanation:[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:falseExplanation:There is no sorted array once rotated that can make nums.

**Example 3:**

Input:nums = [1,2,3]Output:trueExplanation:[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:trueExplanation:[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:trueExplanation:[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`

## Explanation

Find if there exists rotation position and whether from that rotation position, two parts of numbers can form the original sorted number list.

## Python Solution

```
class Solution:
def check(self, nums: List[int]) -> bool:
sorted_nums = sorted(nums)
prev = None
rotation_position = None
for i, num in enumerate(nums):
if prev != None and num < prev:
rotation_position = i
break
prev = num
if not rotation_position:
return True
return (nums[rotation_position:] + nums[:rotation_position]) == sorted_nums
```

- Time Complexity: O(Nlog(N)).
- Space Complexity: O(1).