<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):
        n = len(hours)
        running_sum = 0
        dp = 0  # We don't need an array for this problem; we'll just store the max length found.
        sum_to_index = {}  # Hash map to store the first occurrence of each running sum.
        
        for i in range(n):
            # Convert hours into 1 or -1
            if hours[i] > 8:
                running_sum += 1
            else:
                running_sum -= 1
            
            # If the running sum is positive, the interval from 0 to i is well-performing
            if running_sum > 0:
                dp = i + 1  # The longest interval ending at i is the whole array [0, i]
            else:
                # Check if we have seen `running_sum - 1` before, indicating a well-performing interval.
                if running_sum - 1 in sum_to_index:
                    dp = max(dp, i - sum_to_index[running_sum - 1])
                
                # Store the first occurrence of this running sum if not already stored
                if running_sum not in sum_to_index:
                    sum_to_index[running_sum] = i
        
        return dp

17100