<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) {
        if (hours.empty()) {
            return 0;
        }
        vector<int> prefix_sum(hours.size(), 0);
        if (hours[0] > 8) {
            prefix_sum[0] = 1;
        }
        for (int i = 1; i < hours.size(); ++i) {
            if (hours[i] > 8) {
                prefix_sum[i] = 1;
            }
            prefix_sum[i] += prefix_sum[i - 1];
        }

        int len = 0;
        for (int i = hours.size() - 1; i > - 1; --i) {
            for (int j = - 1; j < i; ++j) {
                int ones = prefix_sum[i];
                if (j >= 0) {
                    ones -= prefix_sum[j];
                }
                int zeroes = i - j - ones;
                if (ones > zeroes && len < i - j) {
                    len = i - j;
                }
            }
        }
        //  [1, 1, 0]
        // |       |

        return len;
    }
};
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