<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:
        a = sorted(nums)
        n = len(nums)
        if a==nums or n==1:
            return 0
        l = 0
        movel=True
        r=n-1
        mover=True
        for i in range(n):
            if a[i]==nums[i] and movel:
                l+=1
            else:
                movel=False
            if a[n-i-1]==nums[n-i-1] and mover:
                r-=1
            else:
                mover = False
        return r-l+1
17700