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`

Solution: