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Tested Python solution for LeetCode 1752 with 24 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 1752 # by problem number
lcpy gen -s check_if_array_is_sorted_and_rotated # by problem name
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 B[i] == A[(i+x) % A.length] for every valid index i.
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 = 2 positions to begin on the element of value 3: [3,4,5,1,2].
Input: nums = [2,1,3,4]
Output: false
Explanation: There is no sorted array once rotated that can make nums.
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.
class Solution:
# Time: O(n)
# Space: O(1)
def check(self, nums: list[int]) -> bool:
breaks = 0
for i in range(len(nums)):
if nums[i] > nums[(i + 1) % len(nums)]:
breaks += 1
if breaks > 1:
return False
return True
| Time | Space |
|---|---|
| O(n) | O(1) |