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Repeated DNA Sequences

Problem Description​

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.

Examples​

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'.

Approach​

  1. Initialize Sets: Use two hash sets—seen to track all 10-letter-long substrings encountered, and ans to store substrings that appear more than once.

  2. Iterate Through String: Loop through the string, extracting each 10-letter-long substring.

  3. Check and Record: For each substring, check if it is already in seen. If it is, add it to ans. Regardless, add the substring to seen.

  4. Return Results: Convert the ans set to a vector and return it as the result.

Solution​

C++​

class Solution {
 public:
  vector<string> findRepeatedDnaSequences(string s) {
    unordered_set<string> ans;
    unordered_set<string_view> seen;
    const string_view sv(s);

    for (int i = 0; i + 10 <= s.length(); ++i) {
      if (seen.count(sv.substr(i, 10)))
        ans.insert(s.substr(i, 10));
      seen.insert(sv.substr(i, 10));
    }

    return {ans.begin(), ans.end()};
  }
};

PYthon​

class Solution:
  def findRepeatedDnaSequences(self, s: str) -> List[str]:
    ans = set()
    seen = set()

    for i in range(len(s) - 9):
      seq = s[i:i + 10]
      if seq in seen:
        ans.add(seq)
      seen.add(seq)

    return list(ans)

Java​

class Solution {
  public List<String> findRepeatedDnaSequences(String s) {
    Set<String> ans = new HashSet<>();
    Set<String> seen = new HashSet<>();

    for (int i = 0; i + 10 <= s.length(); ++i) {
      final String seq = s.substring(i, i + 10);
      if (seen.contains(seq))
        ans.add(seq);
      seen.add(seq);
    }

    return new ArrayList<>(ans);
  }
}

Complexity analysis​

  • Time Complexity: 𝑂(n)𝑂(n)
  • Space Complexity: O(n)O(n)