*187. Repeated DNA Sequences (hashmap, one for loop)(difference between subsequence & substring)

Posted 2020-11-10 wz30

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.

Example:

Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"

Output: ["AAAAACCCCC", "CCCCCAAAAA"]

 Solution: count the frequency of 10 letter words

 

class Solution {
    //find all the 10-letter-long sequences that occur more than once in a DNA molecule
    public List<String> findRepeatedDnaSequences(String s) {
        //substring -- subset n +n-1+...+1: n-k+1
        List<String> res = new ArrayList<String>();
        Map<String,Integer> map = new HashMap<String,Integer>();
        int n = s.length();
        int k  =10;
        if(n < k) return res;
        for(int i = 0; i<=n-k; i++){//11-10 1
            String sub = s.substring(i, i+k);
            if(map.containsKey(sub)){
                map.put(sub, map.get(sub)+1);
            }else {
                map.put(sub, 1);
            }
        }
        for(Map.Entry<String, Integer> entry : map.entrySet()){
            if(entry.getValue() >1){
                res.add(entry.getKey());
            }
        }
        return res;
    }
}

 

Solution 2: two HashSet with a non-duplicate feature.

public List<String> findRepeatedDnaSequences(String s) {
    Set seen = new HashSet(), repeated = new HashSet();
    for (int i = 0; i + 9 < s.length(); i++) {
        String ten = s.substring(i, i + 10);
        if (!seen.add(ten))//if add then first time, else add it
            repeated.add(ten);
    }
    return new ArrayList(repeated);
}

 

subsequence & substring

subsequence: subset 2^n

substring: continous string : n+n-1+n-2+...+1