0381MediumArrayHash TableStringTrie

Unique Fragment Per Code

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Trains the technique from

LeetCode 3076Shortest Uncommon Substring in an Array

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 warehouse labels every bin with a lowercase part code. The picking sheet wants, for each code, a short fragment that identifies it unambiguously.

You are given codes, an array of n part codes. A fragment of a code is any contiguous run of its characters. For each index i, report the shortest fragment of codes[i] that appears in no other code of the array -- appearing anywhere inside another code disqualifies it. When several fragments tie for shortest, report the alphabetically smallest of them. When no fragment of codes[i] qualifies, report the empty string for that index.

Return an array answer where answer[i] is the fragment reported for codes[i].

Examples

Example 1

Input: codes = ["cab", "xyz"]
Output: ["a", "x"]

For "cab" the fragments "a", "b" and "c" each fail to appear in "xyz", and "a" is the alphabetically smallest of them. For "xyz" the single characters all qualify, and "x" is alphabetically smallest.

Example 2

Input: codes = ["ab", "ax"]
Output: ["b", "x"]

The fragment "a" appears in both codes, so it qualifies for neither. "b" appears only in the first code and "x" only in the second.

Example 3

Input: codes = ["abc", "abc"]
Output: ["", ""]

The two codes are identical, so every fragment of one appears in the other and both indices report the empty string.

Example 4

Input: codes = ["zz", "z"]
Output: ["zz", ""]

For the first code, "z" appears in the second code but "zz" does not. The second code has only the fragment "z", which appears in the first code.

Constraints

n == codes.length
2 <= n <= 100
1 <= codes[i].length <= 20
codes[i] 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 shortest_substrings(codes: list[str]) -> list[str]:

Java

public String[] shortestSubstrings(String[] codes)

The intended approach

Time O(n * m^3) for n codes of length at most mSpace O(n * m^3) for the fragment table, plus the returned array

Build one table mapping every fragment to the number of codes that contain it, collecting each code's fragments into a set first so a fragment repeated inside a single code still counts that code once. A fragment of `codes[i]` is then reportable exactly when its table entry is 1. Scan the fragment lengths of each code in increasing order and keep the alphabetically smallest reportable fragment of the first length that has one.

Pythonentry point: shortest_substrings

def shortest_substrings(codes):
    # Time O(n * m^3) -- m^2 fragments per code, each costing O(m) to cut and hash.
    # Space O(n * m^3) for the ownership table of fragments.
    owners = {}
    for code in codes:
        here = set()
        for i in range(len(code)):
            for j in range(i + 1, len(code) + 1):
                here.add(code[i:j])
        for piece in here:
            owners[piece] = owners.get(piece, 0) + 1
    out = []
    for code in codes:
        pick = ""
        for length in range(1, len(code) + 1):
            best = None
            for i in range(len(code) - length + 1):
                piece = code[i:i + length]
                if owners[piece] == 1 and (best is None or piece < best):
                    best = piece
            if best is not None:
                pick = best
                break
        out.append(pick)
    return out

The obvious version

For each code, list every fragment, order the list by length and then alphabetically, and return the first fragment that a direct containment test rejects from all the other codes. No shared table 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_shortest_substrings(codes):
    out = []
    for idx, code in enumerate(codes):
        pieces = []
        for i in range(len(code)):
            for j in range(i + 1, len(code) + 1):
                pieces.append(code[i:j])
        pieces.sort(key=lambda p: (len(p), p))
        answer = ""
        for piece in pieces:
            clean = True
            for jdx, other in enumerate(codes):
                if jdx != idx and piece in other:
                    clean = False
                    break
            if clean:
                answer = piece
                break
        out.append(answer)
    return out

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
4 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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