<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 {
    int presum[10005];
public:
    int longestWPI(vector<int>& hours) {
        // 1. find the longest interval sum > 0 if we convert it to 1/-1 vector.
        // 2. interval[i:j] sum = presum[j] - presum[i-1] 
        //    => we want to maximize j-i+1, where presum[j] > presum[i-1]
        // 3. maintain decreasing stack to find the leftmost end point, then we
        //    iterate it from end to find the rightmost end point.
        //    we find an interval if right end point value > stack value.

        int n = hours.size(), ret = 0;

        stack<int> st;
        st.push(0);
        
        for (int i = 1; i <= n; ++i) {
            presum[i] = presum[i-1] + (hours[i-1] > 8 ? 1 : -1);
            if (presum[i] < presum[st.top()]) st.push(i);
        }
        
        for (int j = n; j > 0; --j) {
            while (!st.empty() and presum[j] > presum[st.top()]) {
                int i = st.top(); st.pop();
                ret = max(ret, j-i);
            }
        }

        return ret;
    }
};
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