<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) {
        vector<int> prefix(hours.size()+1);
        prefix[0] = 0;
        vector<int> leastNums;
        leastNums.push_back(0);

        int maxInvLength = 0;
        for (int i=0; i<hours.size(); i++) {
            prefix[i+1] = (hours[i] > 8) - (hours[i] <= 8) + prefix[i];
            if (prefix[i+1] < leastNums.back())
                leastNums.push_back(i+1);
            
            if (i == hours.size()-1 || hours[i] <= 8 || hours[i+1] <= 8) 
                for (int j=0; j<leastNums.size(); j++) {
                    if (prefix[leastNums[j]] < prefix[i+1]) {
                        if (i+1 - leastNums[j] + 1 > maxInvLength)
                            maxInvLength = i+1 - leastNums[j];
                        break;
                    }   
                }
        }

        return maxInvLength;
    }
};
26700