0533MediumArrayStringBinary SearchTrieSortingHeap (Priority Queue)

Prefix Hints for the Parts Catalogue

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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 stores terminal holds codes, the part codes it stocks, all of them different and written in lowercase letters. A clerk looks a part up by typing typed one letter at a time, left to right.

After each letter the terminal offers hints: the part codes that begin with everything typed so far. At most three are offered, and when more than three codes qualify the terminal offers the three that come first in dictionary order. The hints in a single offer are listed in dictionary order too, and an offer with no qualifying code is empty.

Return one list of hints per letter typed, in the order the letters were typed, so the result holds exactly typed.length lists.

Examples

Example 1

Input: codes = ["bolt", "boltcap", "bo", "brace", "band"], typed = "bol"
Output: [["band", "bo", "bolt"], ["bo", "bolt", "boltcap"], ["bolt", "boltcap"]]

After "b" four codes qualify, so the first three in dictionary order are offered. After "bo" the qualifying codes are "bo", "bolt" and "boltcap". After "bol" only "bolt" and "boltcap" qualify.

Example 2

Input: codes = ["nut"], typed = "nuts"
Output: [["nut"], ["nut"], ["nut"], []]

The single code qualifies for "n", "nu" and "nut", and nothing in the catalogue begins with "nuts", so the last offer is empty.

Example 3

Input: codes = ["washer", "wedge", "well"], typed = "we"
Output: [["washer", "wedge", "well"], ["wedge", "well"]]

After "w" all three codes qualify and are listed in dictionary order. After "we" the code "washer" no longer qualifies.

Constraints

1 <= codes.length <= 1000
1 <= codes[i].length <= 3000
1 <= sum of codes[i].length <= 2 * 10^4
The part codes are all different.
codes[i] holds only lowercase English letters.
1 <= typed.length <= 1000
typed holds only lowercase 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 prefix_hints(codes: list[str], typed: str) -> list[list[str]]:

Java

public List<List<String>> prefixHints(String[] codes, String typed)

The intended approach

Time O(k * m log k + L * (log k + m)) for k codes of length up to m and L letters typed.Space O(k) for the ordered catalogue, plus the returned hints.

Put the catalogue in dictionary order once. In that order every code sharing a given prefix sits in one unbroken block, and the block begins exactly where the prefix itself would be inserted, which a binary search finds. From there walk forward at most three times, stopping early if a code no longer opens with the prefix, and that walk already produces the hints in dictionary order.

Pythonentry point: prefix_hints

def prefix_hints(codes, typed):
    # Sort once, then every prefix is a contiguous block of that order, so a
    # binary search finds where the block starts and three steps finish the job.
    from bisect import bisect_left
    ordered = sorted(codes)
    out = []
    prefix = ""
    for ch in typed:
        prefix += ch
        i = bisect_left(ordered, prefix)
        block = []
        while i < len(ordered) and len(block) < 3 and ordered[i].startswith(prefix):
            block.append(ordered[i])
            i += 1
        out.append(block)
    return out

The obvious version

For each prefix of the typed text, scan the whole catalogue comparing the opening letters of every code, sort the codes that matched and keep the first three.

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_prefix_hints(codes, typed):
    out = []
    for size in range(1, len(typed) + 1):
        prefix = typed[:size]
        hits = []
        for code in codes:
            if len(code) >= size and code[:size] == prefix:
                hits.append(code)
        hits.sort()
        out.append(hits[:3])
    return out

How this problem was checked

Reference solution executed against 11 test cases — 11 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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