Question

Formatted question description: https://leetcode.ca/all/187.html

 187	Repeated DNA Sequences

 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"]

Algorithm

Using HashSet can have the characteristics of duplicate items.

A => binary: 00
C => binary: 01
G => binary: 10
T => binary: 11

Basically, this questions is more of a compression problem, how to encode the 10-char string to use less disk space.

plain char is 8 bits (2 bytes)
encoding to 2 bits (00,01,10,11)

Code

public class Repeated_DNA_Sequences {

    public static void main(String[] args) {
        Repeated_DNA_Sequences out = new Repeated_DNA_Sequences();
        Solution s = out.new Solution();

        System.out.println(s.findRepeatedDnaSequences("AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"));
    }

    class Solution_optimize {
        public List<String> findRepeatedDnaSequences(String s) {
            Set<Integer> words = new HashSet<>();

            // need this to remove duplicates
            // Input "AAAAAAAAAAAAA", result ["AAAAAAAAAA"]
            // to avoid result ["AAAAAAAAAA","AAAAAAAAAA","AAAAAAAAAA"]
            Set<Integer> repteated = new HashSet<>();
            List<String> result = new ArrayList<>();

            char[] map = new char[26];
            //map['A' - 'A'] = 0; // binary: 00
            map['C' - 'A'] = 1; // binary: 01
            map['G' - 'A'] = 2; // binary: 10
            map['T' - 'A'] = 3; // binary: 11

            for(int i = 0; i < s.length() - 9; i++) {
                int v = 0;
                for(int j = i; j < i + 10; j++) {
                    v <<= 2; // every time, use the new 2 bits after shifting for current char
                    v |= map[s.charAt(j) - 'A'];
                }
                if(words.contains(v) && !repteated.contains(v)) {
                    repteated.add(v);
                    result.add(s.substring(i, i + 10));
                }

                words.add(v); // re-add, if already contains
            }
            return result;
        }
    }

    class Solution {
        HashSet<String> existed = new HashSet<>();
        HashSet<String> result = new HashSet<>();

        public List<String> findRepeatedDnaSequences(String s) {

            if (s.length() <= 10) {
                return new ArrayList<String>();
            }

            for (int i = 0; i < 10; i++) {
                for (int j = i; j + 9 < s.length(); j += 10) {

                    String one = s.substring(j, j + 10);

                    if (existed.contains(one)) {
                        result.add(one);
                    }

                    existed.add(one);
                }
            }

            return new ArrayList<String>(result);
        }

    }
}

############

class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        int n = s.length() - 10;
        Map<String, Integer> cnt = new HashMap<>();
        List<String> ans = new ArrayList<>();
        for (int i = 0; i <= n; ++i) {
            String sub = s.substring(i, i + 10);
            cnt.put(sub, cnt.getOrDefault(sub, 0) + 1);
            if (cnt.get(sub) == 2) {
                ans.add(sub);
            }
        }
        return ans;
    }
}

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        map<string, int> cnt;
        int n = s.size() - 10;
        vector<string> ans;
        for (int i = 0; i <= n; ++i) {
            string sub = s.substr(i, 10);
            if (++cnt[sub] == 2) {
                ans.push_back(sub);
            }
        }
        return ans;
    }
};

class Solution:
    def findRepeatedDnaSequences(self, s: str) -> List[str]:
        n = len(s) - 10
        cnt = Counter()
        ans = []
        for i in range(n + 1):
            sub = s[i : i + 10]
            cnt[sub] += 1
            if cnt[sub] == 2:
                ans.append(sub)
        return ans

############

class Solution(object):
  def findRepeatedDnaSequences(self, s):
    """
    :type s: str
    :rtype: List[str]
    """
    d = {}
    ans = []
    for i in range(len(s) - 9):
      key = s[i:i + 10]
      if key in d:
        d[key] += 1
        if d[key] == 2:
          ans.append(key)
      else:
        d[key] = 1
    return ans

func findRepeatedDnaSequences(s string) []string {
	hashCode := map[byte]int{'A': 0, 'C': 1, 'G': 2, 'T': 3}
	ans, cnt, left, right := []string{}, map[int]int{}, 0, 0

	sha, multi := 0, int(math.Pow(4, 9))
	for ; right < len(s); right++ {
		sha = sha*4 + hashCode[s[right]]
		if right-left+1 < 10 {
			continue
		}
		cnt[sha]++
		if cnt[sha] == 2 {
			ans = append(ans, s[left:right+1])
		}
		sha, left = sha-multi*hashCode[s[left]], left+1
	}
	return ans
}

function findRepeatedDnaSequences(s: string): string[] {
    const n = s.length;
    const map = new Map<string, boolean>();
    const res = [];
    for (let i = 0; i <= n - 10; i++) {
        const key = s.slice(i, i + 10);
        if (map.has(key) && map.get(key)) {
            res.push(key);
        }
        map.set(key, !map.has(key));
    }
    return res;
}

/**
 * @param {string} s
 * @return {string[]}
 */
var findRepeatedDnaSequences = function (s) {
    const n = s.length - 10;
    let cnt = new Map();
    let ans = [];
    for (let i = 0; i <= n; ++i) {
        let sub = s.slice(i, i + 10);
        cnt[sub] = (cnt[sub] || 0) + 1;
        if (cnt[sub] == 2) {
            ans.push(sub);
        }
    }
    return ans;
};

using System.Collections.Generic;

public class Solution {
    public IList<string> FindRepeatedDnaSequences(string s) {
        var once = new HashSet<int>();
        var moreThanOnce = new HashSet<int>();
        int bits = 0;
        for (var i = 0; i < s.Length; ++i)
        {
            bits <<= 2;
            switch (s[i])
            {
                case 'A':
                    break;
                case 'C':
                    bits |= 1;
                    break;
                case 'G':
                    bits |= 2;
                    break;
                case 'T':
                    bits |= 3;
                    break;
            }
            if (i >= 10)
            {
                bits &= 0xFFFFF;
            }
            if (i >= 9 && !once.Add(bits))
            {
                moreThanOnce.Add(bits);
            }
        }
        
        var results = new List<string>();
        foreach (var item in moreThanOnce)
        {
            var itemCopy = item;
            var charArray = new char[10];
            for (var i = 9; i >= 0; --i)
            {
                switch (itemCopy & 3)
                {
                    case 0:
                        charArray[i] = 'A';
                        break;
                    case 1:
                        charArray[i] = 'C';
                        break;
                    case 2:
                        charArray[i] = 'G';
                        break;
                    case 3:
                        charArray[i] = 'T';
                        break;
                }
                itemCopy >>= 2;
            }
            results.Add(new string(charArray));
        }
        return results;
    }
}

use std::collections::HashMap;

impl Solution {
    pub fn find_repeated_dna_sequences(s: String) -> Vec<String> {
        let n = s.len();
        let mut res = vec![];
        if n < 10 {
            return res;
        }
        let mut map = HashMap::new();
        for i in 0..=n - 10 {
            let key = &s[i..i + 10];
            if map.contains_key(&key) && *map.get(&key).unwrap() {
                res.push(key.to_string());
            }
            map.insert(key, !map.contains_key(&key));
        }
        res
    }
}

187 - Repeated DNA Sequences

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