<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>

// hours[i]: # of hours worked per day
// hours[i] > 8 => tiring day
// well-performing interval => tiring days > non-tiring days

// Return: the max length of well-perform interval


// Constraints:
// 1 <= hours.length <= 10^4
// 0 <= hours[i]     <= 16

class Solution
{
public:
    int longestWPI(vector<int>& hours)
    {
        // O(n). Use prefix sum to check the longest interval
        int n         = hours.size();
        int maxLength = 0;
        vector<int> vPrefix(n + 1, 0);
        map<int, int> mValueToLeftMostIdx;
        mValueToLeftMostIdx[0] = 0;

        for (int i = 0; i < hours.size(); i++)
        {
            vPrefix[i + 1] = vPrefix[i] + (hours[i] > 8 ? 1 : -1);

//            cout << vPrefix[i + 1] << endl;

            if (!mValueToLeftMostIdx.count(vPrefix[i + 1]))
            {
                mValueToLeftMostIdx[vPrefix[i + 1]] = i + 1;
            }
        }

        int leftValue;
        int leftIdx;
#if 0
        for (auto& n : mValueToLeftMostIdx)
        {
            cout << n.first << " " << n.second << endl;
        }
#endif
        for (int i = n; i > 0 && mValueToLeftMostIdx.size();)
        {
            leftValue = mValueToLeftMostIdx.begin()->first;
            leftIdx   = mValueToLeftMostIdx.begin()->second;
            mValueToLeftMostIdx.erase(leftValue);

//            cout << leftIdx << " " << leftValue << endl; 

            for ( ; vPrefix[i] <= leftValue && leftIdx < i; --i)
            {
                // find nums[i] > left
            }

            maxLength = max(maxLength, i - leftIdx);
        }

        return maxLength;
    }
};
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