<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 {
public:
    int findUnsortedSubarray(vector<int>& nums) {
      int n = nums.size();
      vector<int> l(n), r(n);
      int mn = 1e9, mx = -1e9;

      for(int i=0; i<n; i++) {
          mx = max(mx, nums[i]);
          l[i] = mx;

          mn = min(mn, nums[n - i - 1]);
          r[n - i - 1] = mn;
      }

      int s = -1, e = -1;
      for(int i=0; i<n; i++) {
          if(l[i] > r[i] && s == -1) s = i;
          if(l[n - i - 1] > r[n - i - 1] && e == -1) e = n - i - 1;
      }
      return s == -1 ? 0 : e - s + 1;
    }
};

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