Find Repetition

You are given a short string short_s and a long string long_s. Return the largest integer k such that short_s repeated k times appears as one contiguous substring of long_s.
The copies in a repetition must be adjacent and non-overlapping. Return 0 if either string is empty or if short_s does not occur.

Function
findRepetitions(short_s: String, long_s: String) → int

Examples
Example 1
short_s = "AB"
long_s = "ABBAC"
return = 1
AB occurs once in ABBAC, so the maximum consecutive repetition count is 1.
Example 2
short_s = "AB"
long_s = "ABCABCABAB"
return = 2
AB starts at indices 0, 3, 6, and 8. Only the occurrences at indices 6 and 8 are adjacent, forming ABAB, so the answer is 2.

Constraints
0 <= short_s.length < 10
0 <= long_s.length < 1000000

The trick: find every index where short_s matches in long_s, then look for the longest chain where each next match starts exactly len(short_s) later. Walk i from 0 to n-m. At each i, if long_s[i:i+m] equals short_s, count how many consecutive copies follow by jumping m at a time. Track the max. The common pitfall is the overlap case. With short_s like AA and long_s AAA, matches at 0 and 1 overlap, so they don't chain. Only jump by m, never by 1, inside a run. Another pitfall is forgetting the empty-string case, which must return 0 before you loop. Since m is under 10, direct comparison per index costs about 10n, which is fine for a million characters. Avoid building a repeated string and calling find in a loop, that gets slow. StealthCoder is your hedge if the index math slips under pressure during the live OA.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.