0266EasyTwo PointersStringDynamic Programming

Shorthand Fits the 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 archivist writes a short code on a box and a longer label on the shelf card it belongs to. Both are made of lowercase English letters.

The code fits the label when every letter of the code can be matched to a position of the label so that the matched positions increase from the first letter of the code to the last, and no position of the label is used twice. Matched positions do not have to sit next to each other, and letters of the label may be left unmatched.

Return true when the code fits the label and false when it does not. A code with no letters fits every label.

Examples

Example 1

Input: code = "fig", label = "flight"
Output: true

Matching `f` to position 0, `i` to position 2 and `g` to position 3 uses increasing positions of the label for the three letters in order.

Example 2

Input: code = "gif", label = "flight"
Output: false

The label carries `g` only at position 3, and it holds no `i` or `f` after that position, so the letters cannot be matched in the order the code gives them.

Example 3

Input: code = "tt", label = "tap"
Output: false

The label holds one `t` and a position cannot be used twice, so the second `t` of the code has nothing left to match.

Constraints

0 <= code.length <= 100
0 <= label.length <= 10^4
code and label consist 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 code_fits_label(code: str, label: str) -> bool:

Java

public boolean codeFitsLabel(String code, String label)

The intended approach

Time O(len(label)), one pass over the label with constant work per positionSpace O(1)

Read the label once with a finger resting on the letter of the code that still needs matching. Whenever the label letter under the reading position equals the letter under the finger, move the finger on, because taking the earliest label position that matches leaves the widest possible remainder of the label for the rest of the code. The code fits exactly when the finger runs off the end of the code before the label runs out.

Pythonentry point: code_fits_label

def code_fits_label(code, label):
    if not code:
        return True
    j = 0
    for letter in label:
        if letter == code[j]:
            j += 1
            if j == len(code):
                return True
    return False

The obvious version

Fill a table where `fit[i][j]` says whether the first `i` letters of the code can be matched inside the first `j` positions of the label, either by ignoring position `j` or by matching it against the i-th letter. O(len(code) * len(label)).

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_fits_label(code, label):
    n = len(code)
    m = len(label)
    fit = [[False] * (m + 1) for _ in range(n + 1)]
    for j in range(m + 1):
        fit[0][j] = True
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if fit[i][j - 1]:
                fit[i][j] = True
            elif fit[i - 1][j - 1] and code[i - 1] == label[j - 1]:
                fit[i][j] = True
    return fit[n][m]

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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