<p>Given an integer array <code>nums</code>, you need to find one <b>continuous subarray</b> such that if you only sort this subarray in non-decreasing order, then the whole array will be sorted in non-decreasing order.</p>

<p>Return <em>the shortest such subarray and output its length</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,6,4,8,10,9,15]
<strong>Output:</strong> 5
<strong>Explanation:</strong> You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you solve it in <code>O(n)</code> time complexity?
class Solution:
    def findUnsortedSubarray(self, nums: List[int]) -> int:
        i = 0
        curr_min = float("inf")
        while i < len(nums)-1:
            if nums[i] > nums[i+1]:
                curr_min = min(curr_min, nums[i+1])
            i+=1
        for j, num in enumerate(nums):
            if num > curr_min:
                start = j
                break
        else:
            return 0
                
        i = len(nums)-1
        curr_max = -float("inf")
        while i > 0:
            if nums[i-1] > nums[i]:
                curr_max = max(curr_max, nums[i-1])
            i-=1
        for j, num in enumerate(reversed(nums)):
            if num < curr_max:
                end = len(nums)-1-j
                break
                
        return end-start+1
18000