2483. minimum penalty for shop

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 ith character is 'Y', it means that customers come at the ith hour
• whereas 'N' indicates that no customers come at the ith hour.

If the shop closes at the jth 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 jth hour, it means the shop is closed at the hour j.

solution

def bestClosingTime(self, customers: str) -> int:
	counts = Counter(customers)
	n_before, y_after = counts["N"], 0
	best_penalty = n_before
	time_to_close = len(customers)
 
	for i in reversed(range(len(customers))):
		if customers[i] == "Y":
			y_after += 1
		else:
			n_before -= 1
 
		penalty_if_close = n_before + y_after
		if penalty_if_close <= best_penalty:
			best_penalty = penalty_if_close
			time_to_close = i
	return time_to_close