0537EasyTwo PointersString

Checking a Shorthand Archive Label

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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 box is titled with a lowercase word label. To save ink a clerk writes a shorthand short, made of lowercase letters and digits, and it is read left to right like this: a letter stands for itself and has to line up with that same letter of the label, while a run of digits gives how many letters of the label were left out at that point.

Two rules keep the reading unambiguous. A run of digits is read as one whole number, so "12" asks for twelve letters to be dropped rather than one and then two. And a run of digits never starts with 0, which makes a shorthand such as "012" invalid whatever the label is.

Return true when reading short this way spells out label exactly, using up every letter of the label and every character of the shorthand, and false otherwise.

Examples

Example 1

Input: label = "internal", short = "5nal"
Output: true

Dropping the first five letters of "internal" leaves "nal", which is exactly what the rest of the shorthand spells.

Example 2

Input: label = "banana", short = "3an"
Output: false

Dropping three letters lands on the "a" at index 3, and the shorthand then spells "an", which covers only indices 3 and 4. Index 5 of the label is left over, so the shorthand does not spell the whole label.

Example 3

Input: label = "gadget", short = "012"
Output: false

The digit run "012" opens with a zero, so this shorthand is invalid.

Constraints

1 <= label.length <= 20
label holds only lowercase English letters.
1 <= short.length <= 10
short holds only lowercase English letters and digits.
Every number written in short fits in a 32-bit signed integer.

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 matches_shorthand(label: str, short: str) -> bool:

Java

public boolean matchesShorthand(String label, String short_)

The intended approach

Time O(len(label) + len(short)): neither finger ever moves backwards.Space O(1).

Keep a finger on the label and a finger on the shorthand and advance them together. A letter under the shorthand finger has to equal the letter under the label finger, and then both move on. A digit under the shorthand finger is rejected outright when it is a zero; otherwise the whole run of digits is gathered into a number and the label finger jumps that far. The answer is yes only when both fingers reach the end at the same time, which is what rules out a jump that runs off the label or a label with letters to spare.

Pythonentry point: matches_shorthand

def matches_shorthand(label, short):
    # O(len(label) + len(short)): one finger on each string, never moving back.
    i = 0
    j = 0
    while i < len(label) and j < len(short):
        ch = short[j]
        if ch.isdigit():
            if ch == "0":
                return False
            count = 0
            while j < len(short) and short[j].isdigit():
                count = count * 10 + (ord(short[j]) - 48)
                j += 1
            i += count
        else:
            if label[i] != ch:
                return False
            i += 1
            j += 1
    return i == len(label) and j == len(short)

The obvious version

Split the shorthand into runs of letters and runs of digits, refusing any digit run that opens with a zero, build a pattern that spells the letter runs out and allows exactly the stated number of any characters for each gap, and test the whole label against that pattern.

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_matches_shorthand(label, short):
    import re
    pieces = []
    i = 0
    while i < len(short):
        if short[i].isdigit():
            if short[i] == "0":
                return False
            j = i
            while j < len(short) and short[j].isdigit():
                j += 1
            pieces.append(("gap", int(short[i:j])))
            i = j
        else:
            j = i
            while j < len(short) and not short[j].isdigit():
                j += 1
            pieces.append(("text", short[i:j]))
            i = j
    pattern = "^"
    for kind, value in pieces:
        if kind == "gap":
            pattern += ".{" + str(value) + "}"
        else:
            pattern += re.escape(value)
    pattern += "$"
    return re.match(pattern, label) is not None

How this problem was checked

Reference solution executed against 18 test cases — 18 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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