<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:
        """
        Return the length of the longest well-performing interval.

        well-performing interval:
            an interval of days that
                count(tiring days) > count(non-tiring days)

        tiring day := val > 8.
            Ex. [9,9, 0~8]  tiring days  = 2 > 1

        prefix sum => help getting interval count(tiring days)
            count(non-tiring days) = right - left - count(tiring days)
            => iter all left < right, find leftmost 
                (to be longest right - left)

        Hint: Make a new array A of +1/-1
            goal = to find the longest subarray with positive sum.
        """

        max_len = 0
        diff = 0
        diff_map_leftmost = {}
        for i, h in enumerate(hours):
            diff += 1 if h > 8 else -1
            # update length
            if diff > 0:
                max_len = i + 1
            elif diff - 1 in diff_map_leftmost:
                max_len = max(max_len, 
                    i - diff_map_leftmost[diff - 1])

            if diff not in diff_map_leftmost:
                diff_map_leftmost[diff] = i

        return max_len

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