<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)
        presum = [0] * (n + 1)  # 前缀和数组，长度为n+1
        hash_map = {}  # 记录前缀和首次出现的位置
        max_len = 0  # 最大表现良好区间的长度
        
        # 计算前缀和数组
        for i in range(n):
            if hours[i] > 8:
                presum[i + 1] = presum[i] + 1  # 疲劳天 +1
            else:
                presum[i + 1] = presum[i] - 1  # 非疲劳天 -1
        
        # 遍历前缀和数组，寻找最长的表现良好区间
        for i in range(n + 1):
            if presum[i] > 0:
                # 如果前缀和为正，说明从起点到当前是一个表现良好区间
                max_len = i
            else:
                # 如果前缀和小于或等于0，检查是否存在presum[i]-1
                if (presum[i] - 1) in hash_map:
                    max_len = max(max_len, i - hash_map[presum[i] - 1])
                
                # 如果当前前缀和第一次出现，记录它的位置
                if presum[i] not in hash_map:
                    hash_map[presum[i]] = i
        
        return max_len

17400