<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 result=0;
        if(hours[0]>8){
        hours[0]=1;
        result=1;}
        else
        hours[0]=-1;
        map<int,int>count;
        count.insert({hours[0],0});
    for(int i=1;i<hours.size();i++){
        if(hours[i]>8)
        hours[i]=hours[i-1]+1;
        else
        hours[i]=hours[i-1]-1;
        if(hours[i]>0)
        {
            if(result<i+1)
            result=i+1;
        }
        else{
            if(count.find(hours[i]-1)!=count.end()){
            int j=count.find(hours[i]-1)->second;
            if(result<i-(j+1)+1)
            result=i-(j+1)+1;
            }
        }
        if(count.find(hours[i])==count.end())
        count.insert({hours[i],i});
    }
    return result;
    }
};
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