0951MediumHash TableStringBit ManipulationSliding WindowRolling HashHash FunctionZ AlgorithmBoyer–Moore String-Search Algorithm

Ten-Letter Runs That Come Round Again

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Trains the technique from

LeetCode 187Repeated DNA Sequences

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 strand reads s, a string over the four letters 'A', 'C', 'G' and 'T'.

Return every run of exactly ten letters that appears in the strand more than once, each run listed once, in the order they first appear.

Examples

Example 1

Input: s = "GGTCAGGTCAGGTCATTTGCA"
Output: ["GGTCAGGTCA"]

The run "GGTCAGGTCA" starts at both position 0 and position 5, so it comes round again. No other ten-letter run repeats.

Example 2

Input: s = "GATTACAGATTACAGATTACA"
Output: ["GATTACAGAT", "ATTACAGATT", "TTACAGATTA", "TACAGATTAC", "ACAGATTACA"]

Two runs repeat here, and they are listed in the order they first appear.

Example 3

Input: s = "ACGTACGTAC"
Output: []

The strand is exactly ten letters long, so there is only one run and nothing can repeat.

Constraints

1 <= s.length <= 10^5
s[i] is 'A', 'C', 'G' or 'T'

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 find_repeated_dna_sequences(s: str) -> list[str]:

Java

public List<String> findRepeatedDnaSequences(String s)

The intended approach

Time O(n) for a strand of length n: one pass with a rolling code.Space O(n) for the tally of codes.

There are only four letters, so each fits in two bits and a run of ten fits in twenty. Rolling the window along one letter at a time means shifting the code up two bits, dropping anything above twenty, and adding the new letter, which gives every run's code without ever cutting a fresh string out of the strand. Tally how often each code has been seen. A code reaching two for the first time is a run that repeats, and that is the moment to record it, which puts the answers in order of first repeat. To list them in order of first appearance instead, remember where each code was first seen and sort the recorded runs by that, or equivalently record the run at its first sighting and mark it once it comes round again.

Pythonentry point: find_repeated_dna_sequences

def find_repeated_dna_sequences(s):
    n = len(s)
    if n < 10:
        return []
    value = {"A": 0, "C": 1, "G": 2, "T": 3}
    mask = (1 << 20) - 1
    code = 0
    first_at = {}
    seen_twice = set()
    repeats = []
    for i in range(n):
        code = ((code << 2) | value[s[i]]) & mask
        if i < 9:
            continue
        start = i - 9
        if code in first_at:
            if code not in seen_twice:
                seen_twice.add(code)
                repeats.append((first_at[code], code))
        else:
            first_at[code] = start
    repeats.sort()
    out = []
    for start, code in repeats:
        out.append(s[start:start + 10])
    return out

The obvious version

Cut every run of ten letters out of the strand as its own string, tally those strings, then walk the strand again reporting each run whose tally is above one the first time it turns up. Builds a fresh string per position rather than rolling a code.

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_find_repeated_dna_sequences(s):
    n = len(s)
    tally = {}
    for i in range(n - 9):
        run = s[i:i + 10]
        tally[run] = tally.get(run, 0) + 1
    out = []
    reported = set()
    for i in range(n - 9):
        run = s[i:i + 10]
        if tally[run] > 1 and run not in reported:
            reported.add(run)
            out.append(run)
    return out

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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