README_EN.md

187. Repeated DNA Sequences

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Description

The DNA sequence is composed of a series of nucleotides abbreviated as 'A', 'C', 'G', and 'T'.

• For example, "ACGAATTCCG" is a DNA sequence.

When studying DNA, it is useful to identify repeated sequences within the DNA.

Given a string s that represents a DNA sequence, return all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in any order.

 

Example 1:

Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
Output: ["AAAAACCCCC","CCCCCAAAAA"]

Example 2:

Input: s = "AAAAAAAAAAAAA"
Output: ["AAAAAAAAAA"]

 

Constraints:

• 1 <= s.length <= 105
• s[i] is either 'A', 'C', 'G', or 'T'.

Solutions

Approach 1: HashTable

Time complexity $O(n \times 10)$, Space complexity $O(n)$.

Approach 2: Rabin-Karp

Time complexity $O(n)$, Space complexity $O(n)$.

Python3

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

Java

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;
    }
}

JavaScript

/**
 * @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;
};

Go

HashTable:

func findRepeatedDnaSequences(s string) []string {
	ans, cnt := []string{}, map[string]int{}
	for i := 0; i <= len(s)-10; i++ {
		sub := s[i : i+10]
		cnt[sub]++
		if cnt[sub] == 2 {
			ans = append(ans, sub)
		}
	}
	return ans
}

Rabin-Karp:

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
}

C++

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;
    }
};

C#

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;
    }
}

TypeScript

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;
}

Rust

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
    }
}

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