2483 - Minimum Penalty for a Shop.

Medium

You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters 'N' and 'Y':

• if the <code>i^th</code> character is 'Y', it means that customers come at the <code>i^th</code> hour

• whereas 'N' indicates that no customers come at the <code>i^th</code> hour.

If the shop closes at the <code>j^th</code> hour (0 <= j <= n), the penalty is calculated as follows:

• For every hour when the shop is open and no customers come, the penalty increases by 1.

• For every hour when the shop is closed and customers come, the penalty increases by 1.

Return the earliest hour at which the shop must be closed to incur a minimum penalty.

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

Example 1:

Input: customers = "YYNY"

Output: 2

Explanation:

• Closing the shop at the 0^th hour incurs in 1+1+0+1 = 3 penalty.

• Closing the shop at the 1^st hour incurs in 0+1+0+1 = 2 penalty.

• Closing the shop at the 2^nd hour incurs in 0+0+0+1 = 1 penalty.

• Closing the shop at the 3^rd hour incurs in 0+0+1+1 = 2 penalty.

• Closing the shop at the 4^th hour incurs in 0+0+1+0 = 1 penalty.

Closing the shop at 2^nd or 4^th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.

Example 2:

Input: customers = "NNNNN"

Output: 0

Explanation: It is best to close the shop at the 0^th hour as no customers arrive.

Example 3:

Input: customers = "YYYY"

Output: 4

Explanation: It is best to close the shop at the 4^th hour as customers arrive at each hour.

Constraints:

• <code>1 <= customers.length <= 10⁵</code>

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