1019MediumArrayHash TableString

Labels That Cover Every Pattern

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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.

Two lists of lowercase strings are given, labels and patterns, and no two labels are alike.

A label covers a pattern when, for every letter, the label holds that letter at least as many times as the pattern does.

Return the labels that cover every pattern in the list, in the order they appear in labels.

Examples

Example 1

Input: labels = ["carrot", "cabbage", "radish"], patterns = ["ca", "ar"]
Output: ["carrot"]

Between them the patterns demand at least one `a`, one `c` and one `r`. Only the first label holds all three: the second has no `r` and the third no `c`.

Example 2

Input: labels = ["tin", "tint"], patterns = ["tt"]
Output: ["tint"]

The pattern demands two `t`s, so holding a `t` is not enough. The first label has one and falls short; the second has two.

Example 3

Input: labels = ["a"], patterns = ["b"]
Output: []

The only label has no `b` at all, so nothing covers the pattern.

Constraints

1 <= labels.length <= 10^4
1 <= patterns.length <= 10^4
1 <= labels[i].length <= 10
1 <= patterns[i].length <= 10
Every string is made of lowercase English letters.
No two labels are alike.

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 word_subsets(labels: list[str], patterns: list[str]) -> list[str]:

Java

public List<String> wordSubsets(String[] labels, String[] patterns)

The intended approach

Time O(P * L + N * L) for N labels, P patterns and strings of length at most L.Space O(1) beyond the answer, since the tallies are a fixed size.

Covering all the patterns at once is the same as covering a single combined demand: for each letter, the most times any one pattern asks for it. Meeting the largest demand for a letter automatically meets every smaller demand for that letter, and the demands never interfere, since each letter is counted on its own. So sweep the patterns once, tallying each one's letters and keeping the running largest per letter. Then tally each label's letters and keep the label when no letter falls short of the combined demand. That is one pass over each list with a fixed twenty-six-letter tally, instead of comparing every label against every pattern, which would cost their two lengths multiplied together.

Pythonentry point: word_subsets

def word_subsets(labels, patterns):
    need = [0] * 26
    for word in patterns:
        here = [0] * 26
        for ch in word:
            here[ord(ch) - 97] += 1
        for at in range(26):
            if here[at] > need[at]:
                need[at] = here[at]
    out = []
    for word in labels:
        have = [0] * 26
        for ch in word:
            have[ord(ch) - 97] += 1
        ok = True
        for at in range(26):
            if have[at] < need[at]:
                ok = False
                break
        if ok:
            out.append(word)
    return out

The obvious version

Take each label and test it against each pattern in turn, counting for every letter of the pattern how many times it appears in the label and in the pattern, and dropping the label the moment one falls short. Costs the labels times the patterns rather than a single combined demand.

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_word_subsets(labels, patterns):
    out = []
    for word in labels:
        ok = True
        for pattern in patterns:
            for ch in set(pattern):
                if word.count(ch) < pattern.count(ch):
                    ok = False
                    break
            if not ok:
                break
        if ok:
            out.append(word)
    return out

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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