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).