1058MediumArrayHash TableString

Matching Lookups Against a Word List

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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 word list is given as entries and a run of lookups as asks, both made of English letters in either case.

Answer each lookup by the first of these rules that finds a match, and where a rule finds several matches, the earliest such entry in entries wins:

the lookup matches an entry exactly, letter for letter and case for case: answer that entry;
the lookup matches an entry once case is ignored: answer that entry;
the lookup matches an entry once case is ignored and every vowel is treated as interchangeable with every other vowel, the vowels being a, e, i, o and u: answer that entry;
nothing matches at all: answer the empty string.

Return the answers in the order the lookups are given.

Examples

Example 1

Input: entries = ["Book", "book", "cat"], asks = ["Book", "BOOK", "beek", "cot", "dog"]
Output: ["Book", "Book", "Book", "cat", ""]

The first lookup matches an entry exactly. The second matches once case is ignored, and the earliest entry doing so is Book. The third differs only in its vowels, again answering Book. The fourth differs only in its vowels from cat. The last matches nothing at all.

Example 2

Input: entries = ["yellow"], asks = ["YellOw"]
Output: ["yellow"]

No entry matches case for case, but folding case away leaves the one entry, so it is the answer with its own spelling kept.

Example 3

Input: entries = ["cat"], asks = ["dog"]
Output: [""]

The lookup differs from the only entry in a consonant, which no rule forgives, so nothing matches.

Constraints

1 <= entries.length <= 5000
1 <= asks.length <= 5000
1 <= entries[i].length <= 7
1 <= asks[i].length <= 7
Every string is made of English letters.

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 spellchecker(entries: list[str], asks: list[str]) -> list[str]:

Java

public String[] spellchecker(String[] entries, String[] asks)

The intended approach

Time O((E + Q) * L) for E entries, Q lookups and strings of length at most L.Space O(E * L) for the three tables.

Each rule is a matter of matching on some flattening of the string, so build one lookup table per rule and then answer every lookup in constant time. The first table holds the entries as they stand. The second is keyed on the entry folded to lowercase. The third is keyed on the entry folded to lowercase with every vowel replaced by one stand-in mark, which makes exactly the strings that differ only in their vowels share a key. Filling the second and third tables, keep only the first entry to produce each key, which is what makes the earliest matching entry win without any comparing later on. Then each lookup tries the three tables in order and falls back on the empty string. One pass over the entries and one over the lookups, with the strings short enough that folding one costs a fixed amount.

Pythonentry point: spellchecker

def spellchecker(entries, asks):
    vowels = "aeiou"
    def blur(word):
        out = []
        for ch in word:
            if ch in vowels:
                out.append('*')
            else:
                out.append(ch)
        return "".join(out)
    exact = set()
    folded = {}
    blurred = {}
    for word in entries:
        exact.add(word)
        low = word.lower()
        if low not in folded:
            folded[low] = word
        key = blur(low)
        if key not in blurred:
            blurred[key] = word
    out = []
    for ask in asks:
        if ask in exact:
            out.append(ask)
            continue
        low = ask.lower()
        if low in folded:
            out.append(folded[low])
            continue
        key = blur(low)
        if key in blurred:
            out.append(blurred[key])
            continue
        out.append("")
    return out

The obvious version

Answer each lookup by scanning the word list three times over: once for an exact match, once comparing with case folded away, and once comparing with case folded away and every vowel treated alike, taking the first entry to match on each pass. Costs the entries times the lookups rather than three tables built once.

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_spellchecker(entries, asks):
    vowels = "aeiou"
    def blur(word):
        out = []
        for ch in word.lower():
            if ch in vowels:
                out.append('*')
            else:
                out.append(ch)
        return "".join(out)
    out = []
    for ask in asks:
        answer = None
        for word in entries:
            if word == ask:
                answer = word
                break
        if answer is None:
            for word in entries:
                if word.lower() == ask.lower():
                    answer = word
                    break
        if answer is None:
            target = blur(ask)
            for word in entries:
                if blur(word) == target:
                    answer = word
                    break
        out.append(answer if answer is not None else "")
    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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