<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) {
       
        // deqeue<int>dq;
        int maxi=0;

        // for(int i=0;i<hours.size();i++){
        //     if(hours[i]>8){
        //         dq.push_back(hours[i]);
        //         x++;
        //     } 
        //     else{
        //         dq.push_back(hours[i]);
        //         y++;
        //     }

        //     //update maximum
        //     if(x>y){
        //         maxi=max(maxi,dq.size());
        //     }
        // }
        // //elliminate extra y
        // while(x<y){
        //     if(x>y) return maxi;

        //     else{

        //     }
        // }


        for(int i=0;i<hours.size();i++){
            int x=0;//greater than 8
            int y=0;//less than 9
            for(int j=i;j<hours.size();j++){
                if(hours[j]>8) x++;
                else y++;
                if(x>y){
                    maxi=max(maxi,x+y);
                }
            }
        }
        return maxi;
    }
};
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