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 } }