<p>We are given <code>hours</code>, a list of the number of hours worked per day for a given employee.</p>

<p>A day is considered to be a <em>tiring day</em> if and only if the number of hours worked is (strictly) greater than <code>8</code>.</p>

<p>A <em>well-performing interval</em> is an interval of days for which the number of tiring days is strictly larger than the number of non-tiring days.</p>

<p>Return the length of the longest well-performing interval.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> hours = [9,9,6,0,6,6,9]
<strong>Output:</strong> 3
<strong>Explanation: </strong>The longest well-performing interval is [9,9,6].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> hours = [6,6,6]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= hours.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= hours[i] &lt;= 16</code></li>
</ul>

class Solution {
public:
    int longestWPI(vector<int>& hours) {
        int n = hours.size();
        vector<int>data(n);
        int sum = 0;
        for (int i = 0; i < n; i++){
            int cur = hours[i] > 8 ? 1 : -1;
            sum += cur;
            data[i] = sum;
        }
        data.insert(data.begin(), 0);
        n = data.size();
        stack<int>sk;
        sk.push(0);
        for (int i = 1; i < n; i++){
            if (data[i] < data[sk.top()]){sk.push(i);}
        }
        int res  = 0;
        for (int i = n - 1; i >= 0 && !sk.empty(); i--){
            while (!sk.empty() && data[sk.top()] < data[i]){
                res = max(res, i - sk.top());
                sk.pop();
            }
        }
        return res;
    }
};
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