<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:

        if len(nums) < 2:
            return 0

        first = -1
        last = -1
        largest = -10**6
        smallest = 10**6


        for i in range(1, len(nums)):
            if nums[i] >= nums[i-1]:
                if nums[i] < largest:
                    last = i    
                continue
            
            if first == -1:
                first = i-1

            largest = max(largest,nums[i-1])
            smallest = min(smallest, nums[i])

            last = i

        print(first, last)

        f = first
        for i in range(f):
            if nums[i] > smallest:
                first = i
                break
 

        print(first, last)


        if first == -1:
            return 0

        return last-first + 1
            

            





        
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