<p>You are given the customer visit log of a shop represented by a <strong>0-indexed</strong> string <code>customers</code> consisting only of characters <code>&#39;N&#39;</code> and <code>&#39;Y&#39;</code>:</p>

<ul>
	<li>if the <code>i<sup>th</sup></code> character is <code>&#39;Y&#39;</code>, it means that customers come at the <code>i<sup>th</sup></code> hour</li>
	<li>whereas <code>&#39;N&#39;</code> indicates that no customers come at the <code>i<sup>th</sup></code> hour.</li>
</ul>

<p>If the shop closes at the <code>j<sup>th</sup></code> hour (<code>0 &lt;= j &lt;= n</code>), the <strong>penalty</strong> is calculated as follows:</p>

<ul>
	<li>For every hour when the shop is open and no customers come, the penalty increases by <code>1</code>.</li>
	<li>For every hour when the shop is closed and customers come, the penalty increases by <code>1</code>.</li>
</ul>

<p>Return<em> the <strong>earliest</strong> hour at which the shop must be closed to incur a <strong>minimum</strong> penalty.</em></p>

<p><strong>Note</strong> that if a shop closes at the <code>j<sup>th</sup></code> hour, it means the shop is closed at the hour <code>j</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> customers = &quot;YYNY&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
- Closing the shop at the 0<sup>th</sup> hour incurs in 1+1+0+1 = 3 penalty.
- Closing the shop at the 1<sup>st</sup> hour incurs in 0+1+0+1 = 2 penalty.
- Closing the shop at the 2<sup>nd</sup> hour incurs in 0+0+0+1 = 1 penalty.
- Closing the shop at the 3<sup>rd</sup> hour incurs in 0+0+1+1 = 2 penalty.
- Closing the shop at the 4<sup>th</sup> hour incurs in 0+0+1+0 = 1 penalty.
Closing the shop at 2<sup>nd</sup> or 4<sup>th</sup> hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> customers = &quot;NNNNN&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> It is best to close the shop at the 0<sup>th</sup> hour as no customers arrive.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> customers = &quot;YYYY&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> It is best to close the shop at the 4<sup>th</sup> hour as customers arrive at each hour.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= customers.length &lt;= 10<sup>5</sup></code></li>
	<li><code>customers</code> consists only of characters <code>&#39;Y&#39;</code> and <code>&#39;N&#39;</code>.</li>
</ul>

class Solution {
public:
    
//     int close_shop(string customers , int i)
//     {
//      int n = customers.size();
//      if(i ==n) return 0;   int count = 0;  
//      while(i<n)
//      {
//      if(customers[i] == 'Y')
//      count +=1;
     
//      i++;
//      }


// return count;


//     }
    
//     int open_shop(string customers , int i)
//     {
//     if(i<0)  return 0;
//     int count = 0;
//     while(i>=0)
//     {
//     if( customers[i] == 'N')
//     count +=1;

//     i -= 1;

//     }
// return count;

//     }
    
    
    
    
    int bestClosingTime(string customers) {
    
 

// ************************************************* Memory limit exceeded ****************************************************



   int check = INT_MAX , indx =-1; 
int n = customers.size();
// for(int i =0; i<=n; i++)
// {
// int close = close_shop(customers , i);

// int open = open_shop(customers , i-1);


// if((close +open)  < check)
// {
//     check = close +open;
//     indx = i;
// }

// }
 
    
// return indx;




vector <int> close_shop;
vector <int>  open_shop;

for(int i=0; i<n; i++)
{
if(customers[i] == 'Y')
close_shop.push_back(i);

else
open_shop.push_back(i);

}

int count_openshop =0;

int not_come = open_shop.size();

for(int i =0; i<n; i++)
{
int count_closeshop = 0;
if(close_shop.size() >0)
{
if(i> close_shop[0])
close_shop.erase(close_shop.begin());

count_closeshop = close_shop.size();


}

if(i>0 and open_shop.size()>0)
{
if(i-1 == open_shop[0]) {
count_openshop +=1;


open_shop.erase(open_shop.begin());
}

}

if((count_openshop + count_closeshop)  < check)
{
    check = count_openshop + count_closeshop;
    indx = i;
}


}


  // when shop close at n hour 
   
 if(not_come < check)
 return n;




 return indx;  
   
   
   
   
   
   
   
   
   
   
   
   
    }
};
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