<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?
using namespace std;
class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
        stack<pair<int,int>> st;
        int l = nums.size()-1, r = 0;
        for(int i = 0;i<nums.size();i++){
            while(!st.empty() && st.top().first > nums[i]){
                l = min(l,st.top().second);
                st.pop();
            }
            st.push(make_pair(nums[i],i));
        }
        stack<pair<int,int>> st2;
        for(int i = nums.size()-1;i>=0;i--){
            while(!st2.empty() && st2.top().first < nums[i]){
                r = max(r,st2.top().second);
                st2.pop();
            }
            st2.push(make_pair(nums[i],i));
        }
        return r - l  <= 0 ? 0 : r - l + 1;
    }
};
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