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.

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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.


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Constraints:


	1 &lt;= nums.length &lt;= 100
	1 &lt;= nums[i] &lt;= 100


class Solution:
    def check(self, num: List[int]) -> bool:
        ct=0
        for i in range(1,len(num)):
            if num[i-1]>num[i]:
                ct+=1
        if num[len(num)-1]>num[0]:
            ct+=1
        return ct<=1
class Solution {
    public boolean check(int[] nums) {
        // here we compare all the neighbouring elemnts and check whether they are in somewhat sorted
        // there will be a small change due to rotation in the array at only one place.
        // so if there are irregularities more than once, return false
        // else return true;
        int irregularities = 0;
        int length = nums.length;
        for (int i=0; i<length; i++) {
            if (nums[i] > nums[(i + 1) % length])
                irregularities += 1;
        }
        return irregularities > 1 ? false : true;
    }
}
class Solution {
public:
    bool check(vector<int>& nums) {
        int count=0;
        for(int i=0;i<nums.size();i++){
            if(nums[i]>nums[(i+1)%nums.size()])
                count++;
        }
        return (count<=1);
    }
};
var check = function(nums) {
  let decreased = false
  for (let i = 1; i < nums.length; i += 1) {
    if (nums[i] < nums[i - 1]) {
      if (decreased) {
        return false
      }
      decreased = true
    }
  }
  return decreased ? nums[0] >= nums[nums.length - 1] : true
};
Check if Array Is Sorted and Rotated