0422EasyHash TableString

Refundable Drops

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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 bottle-return depot stamps every container it accepts with a single letter code. The string refundable lists the codes the depot pays a deposit on, one letter per code, and no code appears in that list twice.

A customer empties a sack into the chute. The string dropped gives the code stamped on each container that went in, one letter per container, in the order they fell.

Codes are letters and the depot's stamping machine uses both cases, so 'd' and 'D' are two different codes and a deposit on one says nothing about the other.

Return how many of the dropped containers earn a deposit. A code that the depot pays on may of course turn up on several containers, and each of them earns its own deposit.

Examples

Example 1

Input: refundable = "aA", dropped = "aAAbb"
Output: 3

Both cases of a are paid on. Three of the five containers carry one of them, and the two b containers carry a code the depot does not pay on.

Example 2

Input: refundable = "z", dropped = "ZZZ"
Output: 0

The depot pays on lowercase z, and all three containers carry the uppercase code, which is a different code.

Example 3

Input: refundable = "A", dropped = "bAcAb"
Output: 2

Two of the five containers are stamped A, and each of those earns its own deposit.

Example 4

Input: refundable = "ABCDEFGHIJKLMNOPQRSTUVWXYZ", dropped = "abcdefghijklmnopqrstuvwxyz"
Output: 0

Every capital letter is paid on and every container carries a small letter, so nothing in the sack matches.

Constraints

1 <= refundable.length, dropped.length <= 50
refundable and dropped consist of English letters only.
All the characters of refundable are unique.

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 num_jewels_in_stones(refundable: str, dropped: str) -> int:

Java

public int numJewelsInStones(String refundable, String dropped)

The intended approach

Time O(j + s) for j refundable codes and s dropped containers: one pass to build the set and one pass over the containers.Space O(j) for the set of refundable codes.

Load the refundable codes into a set so that asking whether a code pays out costs the same no matter how long the list is, then walk the dropped containers once and add one for each hit. Nothing is case-folded, because the two cases of a letter are separate codes.

Pythonentry point: num_jewels_in_stones

def num_jewels_in_stones(refundable, dropped):
    # Put the refundable codes in a set once, then charge through the bin
    # counting the items whose code is in it. Case is part of the code, so no
    # folding anywhere.
    payable = set(refundable)
    refunds = 0
    for code in dropped:
        if code in payable:
            refunds += 1
    return refunds

The obvious version

For every dropped container, walk the whole list of refundable codes looking for a match and count one when found. O(j * s).

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_num_jewels_in_stones(refundable, dropped):
    refunds = 0
    for pos in range(len(dropped)):
        for other in range(len(refundable)):
            if dropped[pos] == refundable[other]:
                refunds += 1
                break
    return refunds

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
4 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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