0037MediumStringStackRecursion

Expand the Weave Pattern

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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 loom is driven by a shorthand pattern. Each lowercase letter stands for one pass of a particular yarn, and a repeated stretch is written compactly as a count immediately followed by the stretch it applies to, wrapped in square brackets: 4[xy] weaves the stretch xy four times in a row.

A bracketed stretch may itself contain counts and brackets, in which case the inner shorthand is expanded first and the whole result of the outer bracket is then repeated. Counts may have more than one digit. Every digit in pattern belongs to a count that is attached to a bracketed stretch, so no digit is ever part of the woven output.

Return the fully written-out sequence of yarn passes. You may assume pattern is well formed: brackets are balanced and each opening bracket is preceded by a count.

Examples

Example 1

Input: pattern = "4[xy]z"
Output: "xyxyxyxyz"

The stretch xy is woven four times and the trailing z is woven once.

Example 2

Input: pattern = "2[b3[ca]]"
Output: "bcacacabcacaca"

The inner bracket becomes cacaca, so the outer stretch is bcacaca, and that whole stretch is woven twice.

Example 3

Input: pattern = "12[q]"
Output: "qqqqqqqqqqqq"

The count spans two digits, so q is woven twelve times rather than once then twice.

Constraints

1 <= pattern.length <= 30
pattern contains lowercase English letters, digits and the characters '[' and ']' only
pattern is well formed
Every count in pattern is between 1 and 300

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_weave(pattern: str) -> str:

Java

public String expandWeave(String pattern)

The intended approach

Time O(L) where L is the length of the returned sequenceSpace O(L)

Scan once, building the text of the stretch currently open. Digits accumulate into a running count. An opening bracket parks the text built so far together with its count on a stack and starts a fresh stretch; a closing bracket pops that pair, repeats the finished stretch and appends it to the parked text. Nesting works because each level keeps its own prefix on the stack.

Pythonentry point: expand_weave

def expand_weave(pattern):
    stack = []
    current = []
    count = 0
    for ch in pattern:
        if ch.isdigit():
            count = count * 10 + int(ch)
        elif ch == '[':
            stack.append((current, count))
            current = []
            count = 0
        elif ch == ']':
            prefix, times = stack.pop()
            prefix.append(''.join(current) * times)
            current = prefix
        else:
            current.append(ch)
    return ''.join(current)

The obvious version

Repeatedly find any innermost bracket, one with only letters inside, and splice in its expansion, until no brackets are left. Quadratic in the output length because the string is rebuilt every time.

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_weave(pattern):
    import re
    innermost = re.compile(r'(\d+)\[([a-z]*)\]')
    text = pattern
    while True:
        found = innermost.search(text)
        if not found:
            return text
        piece = found.group(2) * int(found.group(1))
        text = text[:found.start()] + piece + text[found.end():]

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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