check.js

/**
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 B[i] == A[(i+x) % A.length] for every valid index i.

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 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

* @param {number[]} nums
* @return {boolean}
*/
var check = function (nums) {
  let n = nums.length;
  let irregularities = 0;

  for (let i = 0; i < n; i++) {
    if (nums[i] > nums[(i + 1) % n]) {
      irregularities++;
    }
    if (irregularities > 1) {
      return false;
    }
  }
  return true;
};