<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) {

        for(int i = hours.size()-1; i >=0; i--){
            if(hours[i] > 8) hours[i]=1;
            else hours[i]=-1;
        }
        int sum = 0;
        unordered_map<int,int> hash;
        hash[0]=-1;
        int ans = INT_MIN;
        for(int i = 0; i < hours.size(); i++){
            sum += hours[i];
            if(hash[sum]==0) hash[sum]=i+1;
            if(sum > 0){
                ans = max(ans,i-hash[0]);
            }
            if(sum <= 0){
                if(hash[sum-1]!=0){
                    ans = max(ans,i+1-hash[sum-1]);
                }
            }        
        }
        if(ans == INT_MIN) return 0;
        return ans;

    }
};
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