# LeetCode 1752. Check if Array Is Sorted and Rotated

## 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: 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`

## 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).