Introduction

In this blog post, we will explore how to solve the LeetCode problem "187" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

The DNA sequence is composed of a series of nucleotides abbreviated as 'A', 'C', 'G', and 'T'. For example, "ACGAATTCCG" is a DNA sequence. When studying DNA, it is useful to identify repeated sequences within the DNA. Given a string s that represents a DNA sequence, return all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in any order. Example 1: Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT" Output: ["AAAAACCCCC","CCCCCAAAAA"] Example 2: Input: s = "AAAAAAAAAAAAA" Output: ["AAAAAAAAAA"] Constraints: 1 <= s.length <= 105 s[i] is either 'A', 'C', 'G', or 'T'.

Explanation

Here's a breakdown of the approach and the code:

• Hashing for Efficiency: Use a hash table (dictionary in Python) to store the count of each 10-letter substring encountered in the DNA sequence. This allows for O(1) average time complexity lookups to determine if a substring has already been seen.

• Sliding Window: Iterate through the DNA sequence using a sliding window of size 10. Extract each 10-letter substring and update its count in the hash table.

• Result Collection: Maintain a set to store the repeated sequences. Using a set avoids duplicate entries in the final output.

• Runtime Complexity: O(N), where N is the length of the DNA sequence s.

• Storage Complexity: O(N), in the worst-case scenario where most 10-letter substrings are unique.

Code

    def findRepeatedDnaSequences(s: str) -> list[str]:
    """
    Finds all 10-letter-long sequences that occur more than once in a DNA molecule.

    Args:
        s: The DNA sequence string.

    Returns:
        A list of the repeated 10-letter sequences.
    """
    if len(s) < 10:
        return []

    seen = {}
    repeated = set()

    for i in range(len(s) - 9):
        subsequence = s[i:i + 10]
        if subsequence in seen:
            repeated.add(subsequence)
        else:
            seen[subsequence] = 1

    return list(repeated)