LeetCode 1752. Check if Array Is Sorted and Rotated

Description

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 <= nums.length <= 100
  • 1 <= nums[i] <= 100

Explanation

Find if there exists rotation position and whether from that rotation position, two parts of numbers can form the original sorted number list.

Python Solution

class Solution:
    def check(self, nums: List[int]) -> bool:
        
        sorted_nums = sorted(nums)
        
        prev = None        
        rotation_position = None
        
        for i, num in enumerate(nums):
            if prev != None and num < prev:                
                rotation_position = i
                break
                
            prev = num
        
        if not rotation_position:
            return True
        
        return (nums[rotation_position:] + nums[:rotation_position]) == sorted_nums
  • Time Complexity: O(Nlog(N)).
  • Space Complexity: O(1).

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