0691MediumArrayStringDynamic ProgrammingKnapsack Problem0-1 Knapsack

Stamping the Most Part Codes

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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 workshop marks parts with codes written only from the digits '0' and '1'. The list codes holds the codes waiting to be marked; the same code may appear more than once, and each entry in the list may be stamped at most once.

The tray holds zeros zero-stamps and ones one-stamps. Stamping a code consumes one stamp per digit of that code, of the matching kind, and every stamp is single use. A code counts only when all of its digits are stamped.

Return the largest number of entries from codes that can be stamped in full with the stamps in the tray.

Examples

Example 1

Input: codes = ["01", "0", "1"], zeros = 1, ones = 1
Output: 2

Stamping `"0"` uses the zero-stamp and stamping `"1"` uses the one-stamp, which finishes two entries. The tray is then empty.

Example 2

Input: codes = ["11"], zeros = 1, ones = 1
Output: 0

The only code needs two one-stamps and the tray holds one, so no entry can be finished.

Example 3

Input: codes = ["0000", "1111"], zeros = 3, ones = 4
Output: 1

The four one-stamps cover `"1111"` exactly. `"0000"` needs four zero-stamps and only three are in the tray.

Constraints

1 <= codes.length <= 600
1 <= codes[i].length <= 100
1 <= zeros, ones <= 100
Each code is made only of the digits '0' and '1'.

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 most_codes(codes: list[str], zeros: int, ones: int) -> int:

Java

public int mostCodes(String[] codes, int zeros, int ones)

The intended approach

Time O(L * zeros * ones + total code length), for L codes.Space O(zeros * ones).

Track, for every pair of remaining stamp counts, the most codes that could be finished using at most that many stamps of each kind, and fold the codes in one at a time. A code with `p` zeros and `q` ones improves the cell `(z, o)` to one more than the cell `(z - p, o - q)` held before that code was folded in, so both budget loops run downwards and each list entry contributes at most once. Codes that cannot fit in a full tray are skipped outright.

Pythonentry point: most_codes

def most_codes(codes, zeros, ones):
    # best[z][o] = most codes stampable from those considered so far when z
    # zero-stamps and o one-stamps are on hand
    best = [[0] * (ones + 1) for _ in range(zeros + 1)]
    for code in codes:
        need_z = code.count('0')
        need_o = len(code) - need_z
        if need_z > zeros or need_o > ones:
            continue
        # both budgets run downwards so this code is offered to each cell at most
        # once, which keeps a single tray entry from being stamped twice
        for z in range(zeros, need_z - 1, -1):
            row = best[z]
            below = best[z - need_z]
            for o in range(ones, need_o - 1, -1):
                cand = below[o - need_o] + 1
                if cand > row[o]:
                    row[o] = cand
    return best[zeros][ones]

The obvious version

Enumerate every subset of the list, add up the zeros and ones it would need, and keep the largest subset that fits in the tray. Exponential in the number of codes.

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_most_codes(codes, zeros, ones):
    n = len(codes)
    best = 0
    for mask in range(1 << n):
        z = 0
        o = 0
        picked = 0
        for i in range(n):
            if mask >> i & 1:
                z += codes[i].count('0')
                o += codes[i].count('1')
                picked += 1
        if z <= zeros and o <= ones and picked > best:
            best = picked
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 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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