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Every Code A Part Template Allows

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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 catalogue writes a family of part codes as one template s, built from lowercase letters and from groups wrapped in curly brackets. A group such as {a,c,f} lists its options separated by commas; the options inside one group are single letters, all different from each other, and a group is never written inside another group. A letter outside every bracket appears in the code as it stands, and each group supplies exactly one of its letters.

Return every part code the template allows, in lexicographic order, that is the order a dictionary would list them in.

Examples

Example 1

Input: s = "{a,b}p{c,d}"
Output: ["apc", "apd", "bpc", "bpd"]

The first group supplies a or b, the letter p is fixed, and the second group supplies c or d. All four codes are listed in dictionary order.

Example 2

Input: s = "{b,a}"
Output: ["a", "b"]

The template is one group offering b or a, so the codes are the single letters a and b, listed in dictionary order rather than in the order the template wrote them.

Example 3

Input: s = "{q}"
Output: ["q"]

The group offers one option, so the template allows the single code q.

Constraints

1 <= s.length <= 50
s consists of curly brackets, commas and lowercase English letters.
s is guaranteed to be a valid template.
No group appears inside another group.
The letters listed inside one group are all different.

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 expand(s: str) -> list[str]:

Java

public String[] expand(String s)

The intended approach

Time O(len(s) + K * T) — building the slots is linear in the template, then T codes of length K are assembled.Space O(K * T) for the codes returned.

Read the template once into a list of slots, one slot per output character: a bare letter becomes a slot with that single option, and a braced group becomes a slot holding its letters put into order. Then start from the empty code and extend the collection slot by slot, left to right, trying the options of each slot in order. Because each group supplies exactly one letter, every code is the same length, and extending in this order means the codes come out sorted with no separate sorting step.

Pythonentry point: expand

def expand(s):
    # O(len(s) + K * T) time and space, with T the number of codes returned and K
    # the length of each. Turn every template position into the sorted list of
    # letters allowed there: a plain letter gives a one-item list, a braced group
    # gives its letters in order. Then grow the codes slot by slot from left to
    # right. Every group contributes exactly one letter, so all codes share the
    # same length K, and growing with sorted choices in this order emits them
    # already sorted.
    slots = []
    i = 0
    n = len(s)
    while i < n:
        if s[i] == "{":
            j = i + 1
            letters = []
            while s[j] != "}":
                if s[j] != ",":
                    letters.append(s[j])
                j += 1
            letters.sort()
            slots.append(letters)
            i = j + 1
        else:
            slots.append([s[i]])
            i += 1
    codes = [""]
    for letters in slots:
        grown = []
        for prefix in codes:
            for ch in letters:
                grown.append(prefix + ch)
        codes = grown
    return codes

The obvious version

Split the template into slots by locating each closing bracket and cutting the group body on commas. Explore every combination with an explicit stack, collecting a finished code each time all slots have been chosen, then sort the collection before returning it.

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_expand(s):
    parts = []
    i = 0
    while i < len(s):
        if s[i] == "{":
            j = s.index("}", i)
            body = s[i + 1:j]
            parts.append([piece for piece in body.split(",") if piece])
            i = j + 1
        else:
            parts.append([s[i]])
            i += 1
    out = []
    stack = [("", 0)]
    while stack:
        cur, idx = stack.pop()
        if idx == len(parts):
            out.append(cur)
            continue
        for ch in parts[idx]:
            stack.append((cur + ch, idx + 1))
    out.sort()
    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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