Check if Array Is Sorted and Rotated - LeetCode javascript solutions

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

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var check = function(nums) {
    let hasBreak = false;
    for (var i = 0; i < nums.length - 1; i++) {
        if (nums[i + 1] < nums[i]) {
            if (hasBreak) return false;
            if (nums[i + 1] <= nums[0] && nums[nums.length - 1] <= nums[0]) {
                hasBreak = true;
                continue;
            }
            return false;
        }
    }
    return true;
};

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(1).