<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) {
    const int n = nums.size();
    int mn = INT_MAX;
    int mx = INT_MIN;
    bool meetDecrease = false;
    bool meetIncrease = false;

    for (int i = 1; i < n; ++i) {
      if (nums[i] < nums[i - 1])
        meetDecrease = true;
      if (meetDecrease)
        mn = min(mn, nums[i]);
    }

    for (int i = n - 2; i >= 0; --i) {
      if (nums[i] > nums[i + 1])
        meetIncrease = true;
      if (meetIncrease)
        mx = max(mx, nums[i]);
    }

    int l;
    for (l = 0; l < n; ++l)
      if (nums[l] > mn)
        break;

    int r;
    for (r = n - 1; r >= 0; --r)
      if (nums[r] < mx)
        break;

    return l < r ? r - l + 1 : 0;
  }
};
29500