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:**

```
class Solution {
public boolean check(int[] nums) {
int count =0;
for(int i=0;i<nums.length;i++)
{
if(nums[i]>nums[(i+1)%nums.length])
count++;
}
if(count>1) return false;
return true;
}
}
```

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