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
}
}
