Repeated DNA Sequences 解答

Question

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.

Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.

For example,

Given s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT", Return: ["AAAAACCCCC", "CCCCCAAAAA"].

Solution -- Bit Manipulation

Original idea is to use a set to store each substring. Time complexity is O(n) and space cost is O(n). But for details of space cost, a char is 2 bytes, so we need 20 bytes to store a substring and therefore (20n) space.

If we represent DNA substring by integer, the space is cut down to  (4n).

public List<String> findRepeatedDnaSequences(String s) {
    List<String> result = new ArrayList<String>();
 
    int len = s.length();
    if (len < 10) {
        return result;
    }
 
    Map<Character, Integer> map = new HashMap<Character, Integer>();
    map.put('A', 0);
    map.put('C', 1);
    map.put('G', 2);
    map.put('T', 3);
 
    Set<Integer> temp = new HashSet<Integer>();
    Set<Integer> added = new HashSet<Integer>();
 
    int hash = 0;
    for (int i = 0; i < len; i++) {
        if (i < 9) {
            //each ACGT fit 2 bits, so left shift 2
            hash = (hash << 2) + map.get(s.charAt(i)); 
        } else {
            hash = (hash << 2) + map.get(s.charAt(i));
            //make length of hash to be 20
            hash = hash &  (1 << 20) - 1; 
 
            if (temp.contains(hash) && !added.contains(hash)) {
                result.add(s.substring(i - 9, i + 1));
                added.add(hash); //track added
            } else {
                temp.add(hash);
            }
        }
 
    }
 
    return result;
}