0712MediumStringBacktrackingBit Manipulation

Every Spelling of a Part Code

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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 warehouse scanner reads part codes made of English letters and digits, and it ignores letter case, so the same part code can be keyed in more than one way.

Return every spelling of code that can be produced by choosing, independently for each letter, whether that letter is written in uppercase or in lowercase. Digits are copied over unchanged.

The spellings may be returned in any order, and each distinct spelling must appear exactly once.

Examples

Example 1

Input: code = "p4q"
Output: ["p4q", "p4Q", "P4q", "P4Q"]

Two letters each have two cases, so four spellings come out, and the digit 4 stays put in every one of them.

Example 2

Input: code = "58"
Output: ["58"]

There is no letter to choose a case for, so the code has exactly one spelling, itself.

Example 3

Input: code = "Wz"
Output: ["wz", "wZ", "Wz", "WZ"]

Each of the two letters is written either way, whatever case it arrived in, giving four spellings.

Constraints

1 <= code.length <= 12
code consists of lowercase English letters, uppercase English letters and digits.

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 code_spellings(code: str) -> list[str]:

Java

public List<String> codeSpellings(String code)

The intended approach

Time O(n * 2^k) for a code of n characters with k letters, which is the size of the output.Space O(n * 2^k) for the returned spellings.

Build the answer left to right, holding the list of spellings of the prefix seen so far. Starting from the single empty prefix, a digit is appended to every spelling in the list and leaves its length alone, while a letter replaces each spelling with two, one carrying the lowercase form and one the uppercase form. After the last character the list holds one entry per combination of choices, all different because two entries always disagree at the position where their choices first differed.

Pythonentry point: code_spellings

def code_spellings(code):
    # O(n * 2^k) for n characters of which k are letters: grow the list of
    # partial spellings one character at a time, doubling it at every letter.
    spellings = [""]
    for ch in code:
        if ch.isalpha():
            low = ch.lower()
            high = ch.upper()
            spellings = [pre + c for pre in spellings for c in (low, high)]
        else:
            spellings = [pre + ch for pre in spellings]
    return spellings

The obvious version

Run over every subset of the character positions, uppercase the positions in the subset and lowercase the rest, and collect the results in a set so the duplicates produced by digit positions collapse. Enumerates 2^n candidates regardless of how many are letters.

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_code_spellings(code):
    n = len(code)
    found = set()
    for mask in range(1 << n):
        chars = []
        for i in range(n):
            ch = code[i]
            chars.append(ch.upper() if (mask >> i) & 1 else ch.lower())
        found.add("".join(chars))
    return sorted(found)

How this problem was checked

Reference solution executed against 13 test cases — 13 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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