<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) {
        map<int,int> mp;
        int n=hours.size(),nc=0,pc=0,ans=0;
        mp[0]=-1;
        for(int i=0;i<n;i++){
            if(hours[i]>8) pc++;
            else nc++;
            int diff=pc-nc;
            if(!mp.count(diff)) mp[diff]=i;
        }
        int pr=n+1;
        for(auto& p:mp){
            p.second=min(p.second,pr);
            pr=p.second;
        }
        pc=nc=0;
        for(int i=0;i<n;i++){
            if(hours[i]>8) pc++;
            else nc++;
            int diff=pc-nc;
            auto it=mp.lower_bound(diff);
            if(it==mp.begin()) continue;
            it=prev(it,1);
            ans=max(ans,(i)-(*it).second);
        }
        return ans;
    }
};
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