<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) {
        /*
            for each number if u can figure out the next greater, and prev less
        pop anythign smaller
        
        */

        vector<int> st;
        int lower = nums.size();
        int upper = 0;

        for (int i = 0; i < nums.size(); ++i) {
            while (st.size() && nums[i] < nums[st.back()]) {
                lower = min(st.back(), lower);
                st.pop_back();
            }
            st.push_back(i);
        }
        st.clear();
        for (int i = nums.size() - 1; i >= 0; --i) {
            while (st.size() && nums[i] > nums[st.back()]) {
                upper = max(st.back(), upper);
                st.pop_back();
            }
            st.push_back(i);
        }

        if (upper < lower) {
            return 0;
        }
        return upper - lower  + 1;
    }
};
30300