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[] n) {
int l= n.length; // Calculating length of given array.
int nRotation=0;// Declaring a variable to keep a count of number of rotations.
//Running a for loop to identify any rotation in given array.
for(int i=0;i<l;i++){
if(n[i]>n[(i+1)%l]){
nRotation++;
}
}
//If number of rotations are greater than 1 return false.
if(nRotation>1)
return false;
//Else return True.
return true;
}
}
```

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