<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 n=hours.size();
        vector<int>prefix(n,0);
        for(int i=0;i<hours.size();i++){
            hours[i]=hours[i]>8?1:-1;
            prefix[i]+=hours[i];
            if(i>0){
                prefix[i]+=prefix[i-1];
            }     
        }
        //it similar subarray sum greater than k
        int ans=0; 
        map<int,int>mp;
        for(int i=0;i<n;i++){
            if(prefix[i]>0){
                ans=i+1;
            }
            else{
                if(mp.find(prefix[i]-1)!=mp.end()){
                    ans=max(ans,i-mp[prefix[i]-1]);
                }
                if(mp.find(prefix[i])==mp.end()){
                    mp[prefix[i]]=i;
                }

            }
        }

        return ans;

        
       
    }
};
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