<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:
        n = len(hours)
        arr = [-1]*n
        table = {}

        for i in range(n):
            if hours[i]>8:
                arr[i]=1
        
        prev = 0
        total = 0
        for i in range(n):
            prev+=arr[i]
            if prev>0:
                total = max(total,(i+1))
            
            elif prev-1 in table:
                total = max(total,(i - table[prev-1]))
            
            if prev not in table:
                table[prev] = i
        return total
            



        
        






        # # brute force each element
        # max_interval = 0

        # for i in range(n):
        #     curr = new_arr[i]
        #     next_ele = curr
        #     for j in range(i+1,n):
        #         next_ele = next_ele + new_arr[j]
        #         if next_ele>0:
        #             max_interval = max((j-i)+1,max_interval)
        #     if curr>0:
        #         max_interval = max(1,max_interval)
        # return max_interval


17000