<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>

/**
If s[i] > 0: j = 0 is left bounder, and i is right bounder
If s[i] < 0: j = s[i]-1 is left bounder, and i is right bounder
*/
class Solution {
public:
    int longestWPI(vector<int>& hours) {
        for (auto& h : hours) {
            if (h > 8) {
                h = 1;
            } else {
                h = -1;
            }
        }
        int n = hours.size();
        vector<int> prefix(n+1,0);
        stack<int> stk;
        prefix[0] = 0;
        stk.push(prefix[0]);
        int res = 0;
        for (int i = 1; i <= n; i++) {
            prefix[i] = prefix[i-1] + hours[i-1];
            if (prefix[i] < prefix[stk.top()]) stk.push(i);
        }
        for (int i = n; i > 0; i--) {
            while (!stk.empty() && prefix[i] > prefix[stk.top()]) {
                res = max(res,i-stk.top());
                stk.pop();
            }
        }
        return res;
    }
};
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