0816EasyStringDynamic ProgrammingString Matching

Longest Run of a Repeated Tag

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Trains the technique from

LeetCode 1668Maximum Repeating Substring

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 device writes a single line of lowercase letters, given as stream, and a watchdog looks for a marker, given as tag.

The marker is said to run k times when k copies of the whole marker, written one straight after another with nothing in between, appear as a contiguous stretch of stream.

Return the largest k for which that happens. If the marker never appears in the line at all, return 0.

Examples

Example 1

Input: stream = "xxabcabcabcyy", tag = "abc"
Output: 3

Three copies of the marker written together spell "abcabcabc", and that stretch sits in the line starting at position 2.

Example 2

Input: stream = "abxxab", tag = "ab"
Output: 1

The marker appears at position 0 and again at position 4, but "abab" is nowhere in the line, so the longest run is a single copy.

Example 3

Input: stream = "hi", tag = "hihi"
Output: 0

The marker is longer than the line, so not even one copy fits.

Constraints

1 <= stream.length <= 100
1 <= tag.length <= 100
stream and tag 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 longest_tag_run(stream: str, tag: str) -> int:

Java

public int longestTagRun(String stream, String tag)

The intended approach

Time O(n * m) for a line of n characters and a marker of m characters.Space O(n) for the count of copies ending at each position.

Sweep the positions where a copy of the marker could end. When the marker sits there, the number of copies ending back to back at that position is one more than the number that ended a marker-length earlier, so a single left-to-right pass over those counts gives the longest run. A marker longer than the line can never fit, which is handled up front.

Pythonentry point: longest_tag_run

def longest_tag_run(stream, tag):
    # For every position where the marker ends, count how many copies end there
    # back to back: one more than the count that ended one marker-length earlier.
    # Time: O(n * m) for a log of n characters and a marker of m.
    # Space: O(n) for the per-position counts.
    n, m = len(stream), len(tag)
    if m > n:
        return 0
    ending = [0] * (n + 1)
    best = 0
    for end in range(m, n + 1):
        if stream[end - m:end] == tag:
            ending[end] = ending[end - m] + 1
            if ending[end] > best:
                best = ending[end]
    return best

The obvious version

Build the marker repeated once, twice, three times and so on, and ask each time whether the line contains that string, stopping at the first count that does not fit. Leans entirely on substring search and keeps no state.

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_longest_tag_run(stream, tag):
    """Write out k copies of the marker and ask whether the log contains them."""
    k = 0
    while tag * (k + 1) in stream:
        k += 1
    return k

How this problem was checked

Reference solution executed against 16 test cases — 16 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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