<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:
    def longestWPI(self, hours: List[int]) -> int:
        [1,2,1,0,-1,-2,-1]
        pre = [0]*len(hours)
        for i in range(len(hours)):
            pre[i]= 1 if hours[i]>8 else -1
            pre[i]+=pre[i-1]
        h = {}
        res = 0
        for i in range(len(hours)):
            if pre[i]>0:
                res = i+1
            elif pre[i]-1 in h:
                res = max(res,i - h[pre[i]-1])
            
            if pre[i] not in h:
                h[pre[i]] = i
            
        return res




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