# 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.

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 {
public boolean check(int[] nums) {
var breaks = 0;
if (nums[0] < nums[nums.length - 1]) {
breaks += 1;
}
for (int i = 1; i < nums.length; i++) {
if (nums[i] < nums[i - 1]) {
breaks += 1;
}
if (breaks > 1) {
return false;
}
}
return true;
}
}
```