You are given a customer visit log represented by a string customers of length $n$. For each hour $i$ ($0 \leq i < n$), the character customers[i] indicates whether a customer arrived:

• 'Y' indicates that a customer arrived at the $i^{ ext{th}}$ hour.

• 'N' indicates that no customer arrived at the $i^{ ext{th}}$ hour.

The shop can be closed at any hour $j$ between $0$ and $n$, inclusive. Closing the shop incurs a penalty calculated as follows:

• For each hour $i$ where $0 \leq i < j$, if the shop is open and no customer arrives (customers[i] == 'N'), the penalty increases by $1$.

• For each hour $i$ where $j \leq i < n$, if the shop is closed and a customer arrives (customers[i] == 'Y'), the penalty increases by $1$.

Return the earliest closing hour $j$ that achieves the minimum possible penalty.

Constraints:

• $1 \leq$ customers.length $\leq 10^5$

• customers consists only of characters 'Y' and 'N'.

Input: customers = "YNNY"-str
Output: 1-int
Explanation: Closing at hour 1 gives the minimum penalty: the shop avoids the no-customer hour at 1 while only incurring a penalty for the customer arriving at hour 3 after closing.

Input: customers = "NYNYN"-str
Output: 0-int
Explanation: Closing at hour 0 gives the minimum penalty, with penalties only for the customers arriving at hours 1 and 3 after closing.

Input: customers = "N"-str
Output: 0-int
Explanation: Closing at hour 0 incurs no penalty because there are no customers after closing.

Input: customers = "Y"-str
Output: 1-int
Explanation: Closing at hour 1 keeps the shop open for the only customer, resulting in a penalty of zero.

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