<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>& h) {
        int n=h.size();
        int flag = 0;
        for(int i=0; i<n; i++){
            if(h[i]>8){
                h[i]=1;
                flag = 1;
            }
            else h[i]=-1; 
        }
        vector<int>p(n);
        p[0]=h[0];
        for(int i=1; i<n; i++){
            p[i]+=p[i-1];
        }
        unordered_map<int,int>mp;
        int ans = 0;
        int psum = 0;
        for(int i=0; i<n; i++){
            psum+=h[i];
            if(psum>0)ans = max(ans,i+1);
            else {
                  if( mp.find(psum)==mp.end()){
              mp[psum]=i;
                 }
             if( mp.find(psum-1)!=mp.end()){
                ans = max(ans,i-mp[psum-1]);
            }
            }
           
            
          
        }
        if(flag==1 && ans ==0 )ans = 1;
        return ans;
    }
};
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