Minimum Penalty for a Shop Python Solution

lcpy gen -n 2483   # by problem number
lcpy gen -s minimum_penalty_for_a_shop   # by problem name

Input: customers = "YYNY"
Output: 2
Explanation:
- Closing the shop at the 0th hour incurs in 1+1+0+1 = 3 penalty.
- Closing the shop at the 1st hour incurs in 0+1+0+1 = 2 penalty.
- Closing the shop at the 2nd hour incurs in 0+0+0+1 = 1 penalty.
- Closing the shop at the 3rd hour incurs in 0+0+1+1 = 2 penalty.
- Closing the shop at the 4th hour incurs in 0+0+1+0 = 1 penalty.
Closing the shop at 2nd or 4th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.

Input: customers = "NNNNN"
Output: 0
Explanation: It is best to close the shop at the 0th hour as no customers arrive.

Input: customers = "YYYY"
Output: 4
Explanation: It is best to close the shop at the 4th hour as customers arrive at each hour.

class Solution:
    # Time: O(n)
    # Space: O(1)
    def best_closing_time(self, customers: str) -> int:
        penalty = customers.count("Y")
        best_penalty = penalty
        best_hour = 0
        for hour, cust in enumerate(customers, start=1):
            if cust == "Y":
                penalty -= 1
            else:
                penalty += 1
            if penalty < best_penalty:
                best_penalty = penalty
                best_hour = hour
        return best_hour