0127HardArrayHash TableStringDynamic ProgrammingBacktrackingTrieMemoization

Tape Track Splits

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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.

An archive holds reel tapes whose printed labels ran the track names together with every separator worn away. You are given the label tape and titles, the catalogue of track names the archive recognises.

Restore the label. A restoration is an ordered run of catalogue titles whose letters, written one after another with nothing between them, give tape exactly. A title may be reused as often as it helps, and the catalogue may hold titles that no restoration uses.

Return every restoration as a string, with one blank space between neighbouring titles. The restorations may come back in any order. Return an empty list when the label cannot be restored from the catalogue at all.

Examples

Example 1

Input: tape = "dubdub", titles = ["dub", "du", "b", "bdub"]
Output: ["du b du b", "du b dub", "du bdub", "dub du b", "dub dub"]

Three runs of titles spell the label: dub then dub, du then bdub, and du then b then du then b.

Example 2

Input: tape = "harpsong", titles = ["harp", "song", "harps", "on", "g"]
Output: ["harp song", "harps on g"]

The label splits after harp and after harps, so two restorations exist and both are reported.

Example 3

Input: tape = "mix", titles = ["mi"]
Output: []

No catalogue title covers the trailing x, so nothing can be restored and the list is empty.

Constraints

1 <= tape.length <= 20
1 <= titles.length <= 1000
1 <= titles[i].length <= 10
tape and titles[i] hold lowercase English letters only
The titles are pairwise distinct
The input is chosen so the returned list holds at most 10^5 restorations

The values you return may be in any order.

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 tape_track_splits(tape: str, titles: list[str]) -> list[str]:

Java

public List<String> tapeTrackSplits(String tape, List<String> titles)

The intended approach

Time O(n^2 * L + total output size), where n is the label length and L the longest titleSpace O(n * total output size) for the cached suffix answers

Solve suffixes. For a position in the label, the restorations of everything from that position onward are found by trying each catalogue title that the label carries at that position and prefixing it onto every restoration of the shorter suffix that follows. The empty suffix has one restoration, the empty run. Cache the answer per position so a position shared by many prefixes is expanded once instead of once per route into it, then join each run with single spaces.

Pythonentry point: tape_track_splits

def tape_track_splits(tape, titles):
    known = set(titles)
    longest = max((len(t) for t in known), default=0)
    n = len(tape)
    cache = {}

    def from_pos(start):
        if start == n:
            return [[]]
        if start in cache:
            return cache[start]
        out = []
        for end in range(start + 1, min(n, start + longest) + 1):
            head = tape[start:end]
            if head in known:
                for rest in from_pos(end):
                    out.append([head] + rest)
        cache[start] = out
        return out

    return [" ".join(pieces) for pieces in from_pos(0)]

The obvious version

Enumerate every way to chop the label into pieces by iterating over all subsets of cut positions, then keep a chopping only when each of its pieces is in the catalogue. 2^(n-1) choppings, no caching and no per-position reasoning.

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_tape_track_splits(tape, titles):
    known = set(titles)
    n = len(tape)
    found = []
    for mask in range(1 << (n - 1)):
        cuts = [i + 1 for i in range(n - 1) if (mask >> i) & 1]
        edges = [0] + cuts + [n]
        pieces = [tape[a:b] for a, b in zip(edges, edges[1:])]
        ok = True
        for piece in pieces:
            if piece not in known:
                ok = False
                break
        if ok:
            found.append(" ".join(pieces))
    return found

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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