<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:
        smallest_dip = -1
        end_dip = -1
        prev_peak=0
        for index, ele in enumerate(nums[1:]):
            index+=1
            if nums[prev_peak]<=ele:
                prev_peak = index
                print(prev_peak)
            if prev_peak!=-1:
                if ele<nums[prev_peak]:
                    end_dip = index
                    
                    if smallest_dip==-1:
                        smallest_dip = index
                    elif (ele<nums[smallest_dip]):
                        smallest_dip = index
        if smallest_dip==-1 and end_dip ==-1:
            return 0
        start=0
        while(nums[start]<=nums[smallest_dip]):
            start+=1
        return end_dip-start+1

18100