0723EasyArrayString

Codes Printed On The Label Strip

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Trains the technique from

LeetCode 1967Number of Strings That Appear as Substrings in Word

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 label printer has produced one long strip of lowercase letters, given as strip. A picking sheet lists short product codes in codes, each also made of lowercase letters.

A code is printed on the strip when its letters occur in the strip next to each other and in the same order, that is, when the code is exactly some run of consecutive letters of the strip.

Return how many entries of codes are printed on the strip. The sheet may list the same code more than once, and each entry counts on its own.

Examples

Example 1

Input: codes = ["ab", "ab", "zz"], strip = "cabd"
Output: 2

The letters `ab` sit next to each other in the strip at offset 1, and the sheet lists that code twice, so it contributes two. The strip holds no `z` at all.

Example 2

Input: codes = ["kite", "ki", "ite", "kit", "tie"], strip = "spikite"
Output: 4

`kite`, `ki`, `ite` and `kit` each appear as a run of consecutive letters of the strip. `tie` does not: the strip has a `t`, an `i` and an `e`, but never in that order side by side.

Example 3

Input: codes = ["ac"], strip = "abc"
Output: 0

The strip holds both an `a` and a `c`, but a `b` stands between them, so the code is not a run of consecutive letters of the strip.

Constraints

1 <= codes.length <= 100
1 <= codes[i].length <= 100
1 <= strip.length <= 100
codes[i] and strip hold 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 codes_on_strip(codes: list[str], strip: str) -> int:

Java

public int codesOnStrip(String[] codes, String strip)

The intended approach

Time O(C * n * m) for C codes, a strip of n letters and codes of up to m letters.Space O(m) for the block of the strip being compared.

Handle the codes one at a time. For a code of `m` letters there are only `n - m + 1` offsets in a strip of `n` letters where it could sit, so compare the code against the block of letters at each offset and stop at the first offset that matches, adding one to the tally. Because every entry of the sheet is handled separately, a code listed twice adds two.

Pythonentry point: codes_on_strip

def codes_on_strip(codes, strip):
    # Slide each code along the strip and compare it against the block of letters
    # starting at each offset, stopping at the first place it lines up.
    # Time O(C * n * m) for C codes, a strip of n letters and codes of up to m
    # letters.  Space O(m) for the block being compared.
    found = 0
    n = len(strip)
    for code in codes:
        m = len(code)
        for start in range(n - m + 1):
            if strip[start:start + m] == code:
                found += 1
                break
    return found

The obvious version

Cut the strip into every possible run of consecutive letters and drop them all into a set, then walk the sheet and count the entries that turn up in that set.

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_codes_on_strip(codes, strip):
    blocks = set()
    n = len(strip)
    for i in range(n):
        for j in range(i + 1, n + 1):
            blocks.add(strip[i:j])
    total = 0
    for code in codes:
        if code in blocks:
            total += 1
    return total

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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