[LeetCode] Repeated DNA Sequences

This link has a great discussion about this problem. You may refer to it if you like. In fact, the idea and code in this passage is from the former link.

Well, there is a very intuitive solution to this problem. That is, starting from the first letter of the string, extract a substring of length 10, check whether it has occurred and not been added to the result. If so, add it to the result; otherwise, visit the next letter and repeat the above process. However, a naive implementation of this idea will give the MLE error, and this is the real obstacle of the problem.

Then we need to save spaces. Instead of keeping the whole substring, can be convert it to other formats? Well, you have noticed that there are only 4 letters A, T, C, G in the substring. If we assign each letter 2 bits, then a 10-letter substring will only cost 20 bits and can thus be accommodated by a 32-bit integer, greatly lowering the space complexity.

Then you may put this idea into code and get an simple Accepted solution as follows. Congratulations!

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        unordered_map<int, int> mp;
        vector<string> res;
        int i = 0, code = 0;
        while (i < 9)
            code = ((code << 2) | mapping(s[i++]));
        for (; i < (int)s.length(); i++) {
            code = (((code << 2) & 0xfffff) | mapping(s[i]));
            if (mp[code]++ == 1)
                res.push_back(s.substr(i - 9, 10));
        }
        return res;
    }
private:
    int mapping(char s) {
        if (s == 'A') return 0;
        if (s == 'C') return 1;
        if (s == 'G') return 2;
        if (s == 'T') return 3;
    }
};

Do you see the logic in the above code? Well, we first merge 9 letters into code. Then, each time we meet a new letter, we merge it to code by | mapping(s[i]) and mask the leftmost letter by & 0xfffff (20 bits take 5 hexadecimal digits). Thus we have a code for the current 10-letter substring. We check whether it has occurred exactly for once to decide whether to push it to the result or not.

The above code can still be shorten using tricks from the above link. In fact, if we code A, T, C, G using 3 bits, the code will be as short as 10 lines! Refer to the above link to learn more!