1182MediumStringPrefix Sum

Choosing When to Shut the Stall

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A stall keeps a log log of its hours, one character per hour: Y when a customer came in that hour and N when nobody did.

The stall may shut at the start of any hour from 0 up to the number of logged hours. Shutting at hour j means the stall was open for hours 0 through j - 1 and closed for the rest.

The cost of shutting at hour j is the number of open hours with no customer plus the number of closed hours with a customer.

Return the earliest hour whose cost is the smallest.

Examples

Example 1

Input: log = "YYNN"
Output: 2

Shutting at hour 2 leaves the two customer hours open and the two empty hours closed, so nothing is charged at all.

Example 2

Input: log = "NNYY"
Output: 0

Never opening costs two, one for each later customer hour, and nothing beats that, so the earliest such hour is reported.

Example 3

Input: log = "Y"
Output: 1

Staying open for the single customer hour costs nothing.

Constraints

1 <= log.length <= 10^5
every character of log is Y or N

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def best_closing_time(log: str) -> int:

Java

public int bestClosingTime(String log)

The intended approach

Time O(n) over the log.Space O(1).

Working the cost out from scratch for every shutting hour costs the log squared, but the cost barely changes when the hour slides along by one. Moving the shutting hour past an hour turns it from closed to open, so an hour with a customer stops costing and an hour without one starts costing. So begin with the stall never opening, whose cost is the number of customer hours in the whole log, and then slide the shutting hour forward, taking one off the cost for each `Y` passed and adding one for each `N`. Watch for a new smallest as it goes. The answer wants the earliest hour with the smallest cost, so the running best is only replaced on a strictly lower cost, never on a tie.

Pythonentry point: best_closing_time

def best_closing_time(log):
    cost = 0
    for ch in log:
        if ch == "Y":
            cost += 1
    best = cost
    at = 0
    for i in range(len(log)):
        if log[i] == "Y":
            cost -= 1
        else:
            cost += 1
        if cost < best:
            best = cost
            at = i + 1
    return at

The obvious version

Try every shutting hour and count its cost from scratch, walking the open hours for the empty ones and the closed hours for the customer ones, keeping the earliest hour with the smallest cost. Costs the log squared.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_best_closing_time(log):
    n = len(log)
    best = None
    at = 0
    for j in range(n + 1):
        cost = 0
        for i in range(j):
            if log[i] == "N":
                cost += 1
        for i in range(j, n):
            if log[i] == "Y":
                cost += 1
        if best is None or cost < best:
            best = cost
            at = j
    return at

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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