Problem

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.

https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/

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.

Constraints:

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

Test Cases

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class Solution:
def check(self, nums: List[int]) -> bool:
solution_test.py
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import pytest

from solution import Solution


@pytest.mark.parametrize('nums, expected', [
([3,4,5,1,2], True),
([2,1,3,4], False),
([1,2,3], True),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums, expected):
assert sol.check(nums) == expected

Thoughts

遍历数组,统计数字变小的次数,如果超过一次就不是循环有序数组。要注意数组的第一个数字和最后一个数字也要比较。

时间复杂度 O(n),空间复杂度 O(1)

Code

solution.py
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class Solution:
def check(self, nums: list[int]) -> bool:
prev = nums[-1]
down = 0
for num in nums:
if num < prev:
down += 1
if down > 1:
return False
prev = num

return True