<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:
    static constexpr int longestWPI(const std::vector<int>& hours) {
        const int N = hours.size();
        std::vector<int> PSA(N + 1, 0);
        for (int i = 0; i < N; i++) {
            PSA[i + 1] = PSA[i] + 1 - 2 * (hours[i] <= 8);
        }

        // ok, so by definition, we are now just trying to find the largest pair
        // of l, r in PSA, where l < r and PSA[l] < PSA[r]. Now how do we find
        // that interval.
        //
        // We do this in the following, by exploiting that if l1 < l2 and
        // PSA[l1] < PSA[l2], then if l2 < r and PSA[l2] < PSA[r], then we will
        // also have l1 < r and PSA[l1] < PSA[r]. This means that we first just
        // build up a montonic decreasing stack, of all the lowest points in the
        // series, going from left to right. We can then go from right to left
        // and just pop elements as long as They are smaller than the current
        // element. This will by definition be the biggest interval where PSA[l]
        // < PSA[r].

        std::vector<int> stk{0};
        for (int l = 1; l <= N; l++) {
            if (PSA[l] < PSA[stk.back()]) {
                stk.push_back(l);
            }
        }

        int ans = 0;
        for (int r = N; 0 <= r; r--) {
            while (!stk.empty() && PSA[stk.back()] < PSA[r]) {
                ans = std::max(ans, r - stk.back());
                stk.pop_back();
            }
        }

        return ans;
    }
};
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