<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 len = hours.size();
        vector<int> prefixSum(len+1, 0);
        for (int i = 1; i <= len; ++i) {
            prefixSum[i] = prefixSum[i-1] + (hours[i-1] > 8 ? 1 : -1);
        }
        stack<int> smdStack;
        for (int i = 0; i <= len; ++i) {
            if (smdStack.empty() || prefixSum[smdStack.top()] > prefixSum[i]) {
				// Trick, store index than value.
                smdStack.push(i);
            }
        }
        int res = 0;
        for (int j = len; j >= 0; --j) {
		    // For generalized problem:
		    // while (!smdStack.empty() && prefixSum[smdStack.top()]+K <= prefixSum[j]) {
            
			// For this problem:
			while (!smdStack.empty() && prefixSum[smdStack.top()] < prefixSum[j]) {
                res = max(res, j - smdStack.top());
                smdStack.pop();
            }
        }
        return res;
    }
};
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