[LeetCode#187]Repeated DNA Sequences

Problem:

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"].

Analysis:

This problem has a genius solution.
If you have not encounter it before, you may never be able to solve it out.

Idea:
Since we only have four characters "A", "C", "G", "T", We can map each character with a sole 2 bits. (Note: not the ASCII code)
And each sub sequence is 10 characters long, after mapping, which would only take up 20 bits. (Since an Integer in Java takes up 32 bits, a subsequence could be represented into an Integer, or we call this as an Integer hash code)

Another benefits of this mapping is that, as long we add new character, we can update on related hash code through bit movement operation.

1. prepare the HashMap for the mapping.

HashMap<Character, Integer> map = new HashMap<Character, Integer> ();
map.put('A', 0);
map.put('C', 1);
map.put('G', 2);
map.put('T', 3);

2. move the subsequence window, and get realted Hashcode.
int hash = 0;
for (int i = 0; i < s.length(); i++) {
    if (i < 9) {
        hash = (hash << 2) + map.get(s.charAt(i));
    } else{
        hash = (hash << 2) + map.get(s.charAt(i));
        hash = hash & ((1 << 20) - 1);
        ...

    }
}
Note: once the slide window's size meet 10 characters, we should get the hash code for the window. The skill here is to use '&' with a 20 bits "1" to get those bits.
2.1  get 20 bits '1'.
((1 << 20) - 1)
The idea is not hard: like 4 - 1 = 100 - 1 = 011
2.2  use '&'' operator to get the bits.
hash = hash & ((1 << 20) - 1);

Errors:
When you put a <key, value> pair into hashmap, and the value based on the existing in the HashMap, you must test if the pair exist or not.
if (counted.containsKey(hash))
    counted.put(hash, counted.get(hash)+1);
else
    counted.put(hash, 1);

Solution:

public class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        ArrayList<String> ret = new ArrayList<String> ();
        if (s.length() < 10)
            return ret;
        HashMap<Character, Integer> map = new HashMap<Character, Integer> ();
        map.put('A', 0);
        map.put('C', 1);
        map.put('G', 2);
        map.put('T', 3);

        HashMap<Integer, Integer> counted = new HashMap<Integer, Integer> ();
        int hash = 0;
        for (int i = 0; i < s.length(); i++) {
            if (i < 9) {
                hash = (hash << 2) + map.get(s.charAt(i));
            } else{
                hash = (hash << 2) + map.get(s.charAt(i));
                hash = hash & ((1 << 20) - 1);
                if (counted.containsKey(hash) && counted.get(hash) == 1) {
                    ret.add(s.substring(i-9, i+1));
                    counted.put(hash, 2);
                } else{
                    if (counted.containsKey(hash))
                        counted.put(hash, counted.get(hash)+1);
                    else
                        counted.put(hash, 1);
                }
            }
        }
        return ret;
    }
}

Actually, since we only care about if a subsequence has appeared twice, we could use two HashSet to avoid the above ugly code.

public class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        ArrayList<String> ret = new ArrayList<String> ();
        if (s.length() < 10)
            return ret;
        HashMap<Character, Integer> map = new HashMap<Character, Integer> ();
        map.put('A', 0);
        map.put('C', 1);
        map.put('G', 2);
        map.put('T', 3);
        HashSet<Integer> appeared = new HashSet<Integer> ();
        HashSet<Integer> counted = new HashSet<Integer> ();

        int hash = 0;
        for (int i = 0; i < s.length(); i++) {
            if (i < 9) {
                hash = (hash << 2) + map.get(s.charAt(i));
            } else{
                hash = (hash << 2) + map.get(s.charAt(i));
                hash = hash & ((1 << 20) - 1);
                if (appeared.contains(hash) && !counted.contains(hash)) {
                    ret.add(s.substring(i-9, i+1));
                    counted.add(hash);
                } else{
                    appeared.add(hash);
                }
            }
        }
        return ret;
    }
}