<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();
        int left = 0;
        int right = n - 1;
        
        while (left < n - 1 && nums[left] <= nums[left + 1]) {
            left++;
        }
        
        if (left == n - 1) {
            return 0;
        }
        
        while (right > 0 && nums[right] >= nums[right - 1]) {
            right--;
        }
        
        int subarray_min = INT_MAX, subarray_max = INT_MIN;
        for (int i = left; i <= right; i++) {
            subarray_min = min(subarray_min, nums[i]);
            subarray_max = max(subarray_max, nums[i]);
        }
        
        while (left > 0 && nums[left - 1] > subarray_min) {
            left--;
        }
        
        while (right < n - 1 && nums[right + 1] < subarray_max) {
            right++;
        }
        
        return right - left + 1;
    }
};

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