Problem Description

In the given LeetCode problem, we are tasked with analyzing a string that represents a DNA sequence. A DNA sequence consists of a series of nucleotides, which are denoted by the characters 'A', 'C', 'G', and 'T'. Our goal is to find all the unique substrings of length 10 that repeat more than once within this DNA sequence. We need to return these substrings in any order, without accounting for their frequency beyond the first occurrence.

Intuition

Encountering this problem, the first intuition is to track the frequency of every possible 10-letter-long substring within the larger DNA string. To efficiently look up these sequences, a HashTable (Python's Counter dictionary can be used here for convenience) is a natural choice because it allows insertions and lookups in constant time, assuming a good hash distribution.

The thinking process goes something like this:

1. We want to iterate over the sequence only once and keep track of every 10-letter sequence as we go. This is a sliding window problem.

2. Each time we encounter a new 10-letter sequence, we add it to the HashTable. If the sequence already exists in the table, we increment its count.

3. Instead of waiting until the end to find out which sequences have occurred more than once, we can check the count as we go. If a sequence's count reaches 2, we add it to the answer list. We only need to do this once per sequence because the problem asks for sequences that occur 'more than once,' not the specific number of times they occur.

4. When a sequence's count exceeds 2, we don't need to do anything more with it, so we can ignore it. This avoids adding duplicates to our results list and keeps the operation more efficient.

5. Finally, we return the list of repeating 10-letter sequences.

By using this approach, we optimize for both time and space by not keeping track of sequences that don't matter (those with counts less than 2 after processing) and by not duplicating entries in our results list.

Learn more about Sliding Window patterns.

Solution Approach

The implementation leverages a HashTable to track 10-letter sequences, which provides an efficient way to check for duplicate sequences as we iterate through the DNA string. Here's how the solution unfolds:

1. Define a variable n which represents the range we need to check within the string. Since we're looking for sequences of length 10, we set n to len(s) - 10. This ensures we don't go over the string length when checking for the substrings at the end of the DNA sequence.

2. Initialize a Counter (a special HashTable from Python's collections library) to keep track of the counts of all 10-letter sequences we find.

3. Define an empty list ans, which will store the sequences that repeat more than once.

4. Iterate over the DNA string using a range that goes from 0 to n+1. This range represents the starting index of each 10-letter sequence we want to examine.

5. Inside the loop, extract a substring sub of length 10 starting from the current index i.

6. Increment the count of this substring in the Counter. The Counter automatically handles the creation of new keys if the substring hasn't been seen before.

7. Check if the count of the current substring has reached 2. If it has, it means this sequence has been encountered once before and now is the second occurrence. It is now qualified as a repeating sequence and should be appended to the ans list.

8. Continue this process until we've checked all possible starting indices of 10-letter sequences in the DNA string.

9. Once the loop completes, return the ans list holding all the sequences that repeated more than once.

The application of this method ensures we only store and return unique sequences that fulfill the problem's conditions. The usage of the Counter allows for both efficient insertion and lookup operations, where the worst-case time complexity is O(n * 10) accounting for the substring operation in each iteration. The space complexity is O(n), as we potentially store information about each 10-letter sequence present in the DNA string.

Example Walkthrough

Let’s consider a small example to illustrate the solution approach using a given DNA string: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT".

1. We initialize n as len(s) - 10, which equates to 20 since the string length is 30. This signifies that we will check for 10-letter sequences starting at indices 0 to 20.

2. We then create a Counter to keep track of the sequences.

3. We initialize an empty list ans to store our answers.

4. We begin iterating from index 0 to 20 (inclusive) to check all possible 10-letter-long sequences.

   • At index 0, we get the substring sub = "AAAAACCCCC". We increment its count in the Counter. It's the first time we've seen it, so its count is now 1.

   • Moving to index 1, the substring is sub = "AAAACCCCCA". We increment its count in the Counter.

   • This process continues with new substrings as the index increases.

   • At index 10, we reach the substring sub = "AAAAACCCCC", which was previously seen at index 0. Its count in the Counter becomes 2. Since the count is 2, we add this sequence to the ans list.

   • At index 15, the substring sub = "CCCCCAAAAA" appears. This is also the second time we've encountered it, so its count becomes 2, and it is added to the ans list.

5. After iterating through all starting indices from 0 to 20, we finish with ans containing the sequences that appeared more than once. In this case, ans = ["AAAAACCCCC", "CCCCCAAAAA"].

6. Finally, we return the list ans as our answer, which contains the repeated sequences, satisfying the problem's conditions.

This example demonstrates the process of using the sliding window technique, combined with a Counter, to efficiently determine sequences that repeat more than once within a given DNA string. By checking and updating counts as we iterate, we're able to avoid unnecessary work and produce our answer without redundancies.

Solution Implementation

Time and Space Complexity

The provided code snippet finds all the 10-letter-long sequences in the string s that appear more than once. To analyze the time and space complexity of this Python code, we will review each significant portion of the code.

Time Complexity:

The code loops through the string s, starting from index 0 to index n (which is len(s) - 10). In each iteration, it creates a substring of length 10 and increases the corresponding counter for this substring. Since the loop runs for (len(s) - 10) + 1 times and each operation within the loop (creating a substring and updating the counter) can be considered as O(1), the overall time complexity is O(n - 10 + 1) which simplifies to O(n) where n is the length of the input string s.

Space Complexity:

The space complexity is determined by the additional space required by the algorithm which is primarily dependent on the cnt and ans variables.

• The Counter object will at most store each unique 10-letter-long sequence present in s. In the worst case, this will be n - 9 unique sequences if every sequence is unique. This gives O(n) as the space complexity contribution from cnt.
• The ans list stores each 10-letter sequence that appears exactly twice. In the worst case, if every possible 10-letter sequence appeared exactly twice, this would yield O(n/10) space complexity which simplifies to O(n).

By combining the above considerations, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.