0290EasyArrayGreedy

Locker Kiosk Change

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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 left-luggage kiosk rents one locker per traveller at a flat 5 credits. Travellers reach the counter in the order given by tokens, and tokens[i] is the single token the i-th traveller hands over: worth 5, 10 or 20 credits.

The kiosk opens with an empty till. Whenever a traveller pays more than 5 credits, the kiosk must hand back the difference using whole tokens it already holds, and the only tokens it holds are the ones earlier travellers paid with. A traveller paying 10 needs 5 credits back, and a traveller paying 20 needs 15 credits back.

Return true if every traveller can be given exact change in this order, and false if some traveller cannot.

Examples

Example 1

Input: tokens = [5,5,5,5,10,20,10,10]
Output: true

The fifth traveller is handed one 5 back. The sixth is handed a 10 and a 5, which is the 15 credits owed. The last two are each handed one 5 back, so nobody is short.

Example 2

Input: tokens = [5,5,20,5,10]
Output: false

When the third traveller pays 20, the till holds only two 5 tokens, which is 10 credits, so the 15 credits owed cannot be handed back.

Constraints

1 <= tokens.length <= 10^5
tokens[i] is either 5, 10, or 20.

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 can_serve_all(tokens: list[int]) -> bool:

Java

public boolean canServeAll(int[] tokens)

The intended approach

Time O(n)Space O(1)

Track how many 5 and 10 tokens sit in the till, since a 20 token can never be part of any refund. A 10 token is refunded from a 5. A 15-credit refund is paid as a 10 plus a 5 when a 10 is available and as three 5s otherwise, and the traveller is refused when neither combination is on hand.

Pythonentry point: can_serve_all

def can_serve_all(tokens):
    # Only 5s and 10s are ever usable as change, so the kiosk state is just the two
    # counts.  A 15-credit refund is paid with a 10 plus a 5 whenever a 10 is on
    # hand, keeping 5s for the refunds that have no other source.
    # O(n) time, O(1) space.
    fives = 0
    tens = 0
    for t in tokens:
        if t == 5:
            fives += 1
        elif t == 10:
            if fives == 0:
                return False
            fives -= 1
            tens += 1
        else:
            if tens > 0 and fives > 0:
                tens -= 1
                fives -= 1
            elif fives >= 3:
                fives -= 3
            else:
                return False
    return True

The obvious version

Carry the full set of reachable (five count, ten count) tills forward one traveller at a time, branching on both ways of paying 15 credits, and fail when the set becomes empty.

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_can_serve_all(tokens):
    states = {(0, 0)}
    for t in tokens:
        nxt = set()
        for fives, tens in states:
            if t == 5:
                nxt.add((fives + 1, tens))
            elif t == 10:
                if fives >= 1:
                    nxt.add((fives - 1, tens + 1))
            else:
                if tens >= 1 and fives >= 1:
                    nxt.add((fives - 1, tens - 1))
                if fives >= 3:
                    nxt.add((fives - 3, tens))
        if not nxt:
            return False
        states = nxt
    return True

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
2 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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