0909HardArrayStringTrie

Best Match by Shared Ending

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Trains the technique from

LeetCode 3093Longest Common Suffix Queries

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 catalogue is given as wordsContainer and a list of lookups as wordsQuery, both lists of lowercase words.

For each lookup, find the catalogue entry sharing the longest ending with it, where an ending means a run of characters at the very end of both words; sharing no ending at all counts as a shared ending of length zero. Break a tie on that length by taking the shorter catalogue entry, and break a remaining tie by taking the one listed earliest.

Return the position in the catalogue of the entry chosen for each lookup, in the order the lookups are given.

Examples

Example 1

Input: wordsContainer = ["crate", "grate", "plate", "late"], wordsQuery = ["slate", "ornate", "crate"]
Output: [3, 3, 0]

The lookup "slate" shares the ending "late" with both "plate" and "late", and the shorter of those wins. The lookup "ornate" shares only "ate", which the same three entries offer, so the shortest one wins again. The lookup "crate" shares all five characters with the entry at position 0.

Example 2

Input: wordsContainer = ["aa", "bb", "cc"], wordsQuery = ["dd", "aa"]
Output: [0, 0]

The lookup "dd" shares no ending with anything, so the tie-break falls back to the shortest entry and then the earliest, giving position 0. The lookup "aa" matches the entry at position 0 outright.

Example 3

Input: wordsContainer = ["abc", "abc", "abc"], wordsQuery = ["abc"]
Output: [0]

All three entries share the whole word and are the same length, so the earliest one is chosen.

Constraints

1 <= wordsContainer.length <= 10^4
1 <= wordsQuery.length <= 10^4
1 <= wordsContainer[i].length <= 5 * 10^3
1 <= wordsQuery[i].length <= 5 * 10^3
The total length of wordsContainer is at most 5 * 10^5
The total length of wordsQuery is at most 5 * 10^5
Every word consists of lowercase English letters only

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 string_indices(wordsContainer: list[str], wordsQuery: list[str]) -> list[int]:

Java

public int[] stringIndices(String[] wordsContainer, String[] wordsQuery)

The intended approach

Time O(C + Q) over the total catalogue and lookup lengths, each character costing one tree step.Space O(C) for the tree, bounded by the total catalogue length.

Shared endings are shared beginnings read backwards, so build a prefix tree over the catalogue entries reversed. Walking an entry into the tree marks every node on its path, and at each node keep the best entry passing through it under the stated tie-break: shorter wins, and among equals the earlier position wins. Because the rule is settled as entries are inserted, each node already holds its answer by the time any lookup arrives. Answering a lookup is then walking it reversed down the tree as far as the nodes go and taking the best entry recorded at the deepest node reached, which is the longest shared ending available. Falling off the tree straight away lands on the root, whose record covers the whole catalogue and gives the shared-nothing answer.

Pythonentry point: string_indices

def string_indices(wordsContainer, wordsQuery):
    # each node: [children, best index, best length]
    root = [{}, -1, 0]

    def offer(node, idx, length):
        if node[1] < 0 or length < node[2] or (length == node[2] and idx < node[1]):
            node[1] = idx
            node[2] = length

    for idx in range(len(wordsContainer)):
        word = wordsContainer[idx]
        offer(root, idx, len(word))
        node = root
        for at in range(len(word) - 1, -1, -1):
            ch = word[at]
            nxt = node[0].get(ch)
            if nxt is None:
                nxt = [{}, -1, 0]
                node[0][ch] = nxt
            node = nxt
            offer(node, idx, len(word))
    out = []
    for word in wordsQuery:
        node = root
        for at in range(len(word) - 1, -1, -1):
            nxt = node[0].get(word[at])
            if nxt is None:
                break
            node = nxt
        out.append(node[1])
    return out

The obvious version

For each lookup, compare it against every catalogue entry, walking both from the last character backwards to measure the shared ending, and keep the best under the tie-break. Quadratic in the total input and it builds no structure at all.

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_string_indices(wordsContainer, wordsQuery):
    out = []
    for query in wordsQuery:
        best_idx = -1
        best_len = -1
        best_size = 0
        for idx in range(len(wordsContainer)):
            word = wordsContainer[idx]
            shared = 0
            while (shared < len(word) and shared < len(query)
                   and word[len(word) - 1 - shared] == query[len(query) - 1 - shared]):
                shared += 1
            if best_idx < 0:
                take = True
            elif shared > best_len:
                take = True
            elif shared < best_len:
                take = False
            elif len(word) < best_size:
                take = True
            else:
                take = False
            if take:
                best_idx = idx
                best_len = shared
                best_size = len(word)
        out.append(best_idx)
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 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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