Minimum Penalty for a Shop
<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>
String; Prefix Sum
class Solution:
    def bestClosingTime(self, customers: str) -> int:
        n = len(customers)
        s = [0] * (n + 1)
        for i, c in enumerate(customers):
            s[i + 1] = s[i] + int(c == 'Y')
        ans, cost = 0, inf
        for j in range(n + 1):
            t = j - s[j] + s[-1] - s[j]
            if cost > t:
                ans, cost = j, t
        return ans
