<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();
        stack<int> stack;
        hours[0] = hours[0] > 8 ? 1 : -1;
        for (int i = 1; i < hours.size(); i++) {
            hours[i] = hours[i] > 8 ? hours[i - 1] + 1 : hours[i - 1] - 1;
        }

        // to find farthest j > i, such that hours[j] - hours[i] > 0

        for (int i = n - 1; i >= 0; i--) {
            if (stack.empty() || hours[i] > hours[stack.top()]) {
                stack.push(i);
            }
        }

        int ans = 0;
        while (!stack.empty() && 0 < hours[stack.top()]) {
            ans = max(ans, stack.top() + 1);
            stack.pop();
        }
        for (int i = 0; i < n; i++) {
            while (!stack.empty() && hours[i] < hours[stack.top()]) {
                ans = max(ans, stack.top() - i);
                stack.pop();
            }
        }

        return ans;
    }
};

            //   6 6 9
            //   -1 -2 -1
            //         -1
24700