<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 {
public:
    int longestWPI(vector<int>& hours) {
        int n = hours.size();
        vector<int> preSum(n + 1);
        unordered_map<int, int> valToIndex;

        preSum[0] = 0;
        int res = 0;

        for(int i = 1; i < n + 1; i++){
            preSum[i] = preSum[i - 1] + (hours[i - 1] > 8 ? 1 : -1);
            if(!valToIndex.count(preSum[i])){
                valToIndex[preSum[i]] = i;
            }

            if(preSum[i] > 0){
                res = max(res, i);
            }
            else{
                if(valToIndex.count(preSum[i] - 1)){
                    res = max(res, i - valToIndex[preSum[i] - 1]);
                }
            }
        }
        return res;
    }
};
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