1752. Check if Array Is Sorted and Rotated

package LeetCode_1752

/**
 * 1752. Check if Array Is Sorted and Rotated
 * https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/
 * 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. 1 <= nums.length <= 100
2. 1 <= nums[i] <= 100
 * */
class Solution {
    /*
    * solution 1: follow by rule: A[i] == B[(i+x) % A.length], Time:O(nlogn), Space:O(n)
    * */
    fun check(nums: IntArray): Boolean {
        var x = 0
        val n = nums.size
        for (i in 0 until n - 1) {
            if (nums[i] > nums[i + 1]) {
                x = i + 1
            }
        }
        val B = nums.clone()
        nums.sort()
        for (i in 0 until n) {
            if (nums[i] != B[(i + x) % n]) {
                return false
            }
        }
        return true
    }
}