feat(581-shortest-unsorted-continuous-subarray): add py3 solution
This commit is contained in:
parent
1abc610390
commit
ee2167f554
29
0581_shortest-unsorted-continuous-subarray/README.md
Normal file
29
0581_shortest-unsorted-continuous-subarray/README.md
Normal file
@ -0,0 +1,29 @@
|
||||
Given an integer array `nums`, you need to find one **continuous subarray** that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order.
|
||||
|
||||
Return _the shortest such subarray and output its length_.
|
||||
|
||||
**Example 1:**
|
||||
|
||||
Input: nums = [2,6,4,8,10,9,15]
|
||||
Output: 5
|
||||
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.
|
||||
|
||||
|
||||
**Example 2:**
|
||||
|
||||
Input: nums = [1,2,3,4]
|
||||
Output: 0
|
||||
|
||||
|
||||
**Example 3:**
|
||||
|
||||
Input: nums = [1]
|
||||
Output: 0
|
||||
|
||||
|
||||
**Constraints:**
|
||||
|
||||
* `1 <= nums.length <= 104`
|
||||
* `-105 <= nums[i] <= 105`
|
||||
|
||||
**Follow up:** Can you solve it in `O(n)` time complexity?
|
@ -0,0 +1,39 @@
|
||||
class Solution:
|
||||
def findUnsortedSubarray(self, nums: List[int]) -> int:
|
||||
min_unsorted = float("inf")
|
||||
max_unsorted = float("-inf")
|
||||
|
||||
# Find min and max element in the unsorted portion
|
||||
for i, num in enumerate(nums):
|
||||
if not self.is_sorted(i, nums):
|
||||
min_unsorted = min(min_unsorted, num)
|
||||
max_unsorted = max(max_unsorted, num)
|
||||
|
||||
if min_unsorted == float("inf"):
|
||||
return 0
|
||||
|
||||
# Find right places for min and max elements from previous
|
||||
# step in the "sorted" section of the array
|
||||
left = 0
|
||||
while nums[left] <= min_unsorted:
|
||||
left += 1
|
||||
|
||||
right = len(nums) - 1
|
||||
while nums[right] >= max_unsorted:
|
||||
right -= 1
|
||||
|
||||
return right - left + 1
|
||||
|
||||
def is_sorted(self, i: int, nums: List[int]) -> bool:
|
||||
n = len(nums)
|
||||
|
||||
if n <= 1:
|
||||
return True
|
||||
|
||||
if i == 0:
|
||||
return nums[i] <= nums[i + 1]
|
||||
|
||||
if i == n - 1:
|
||||
return nums[i - 1] <= nums[i]
|
||||
|
||||
return nums[i - 1] <= nums[i] <= nums[i + 1]
|
Loading…
Reference in New Issue
Block a user