187. Repeated DNA Sequences

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

Solution 1: Hash Table

We define a hash table $cnt$ to store the occurrence count of all substrings of length $10$.

We iterate through all substrings of length $10$ in the string $s$. For the current substring $t$, we update its count in the hash table. If the count of $t$ is $2$, we add it to the answer.

After the iteration, we return the answer array.

The time complexity is $O(n \times 10)$, and the space complexity is $O(n \times 10)$. Here, $n$ is the length of the string $s$.

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)

Solution 2: Rabin-Karp String Matching Algorithm

This method essentially combines sliding window and hash. Similar to 28 Find-the-Index-of-the-First-Occurrence-in-a-String. Find the Index of the First Occurrence in a String, this problem can use a hash function to reduce the time complexity of counting subsequences to $O(1)$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

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

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        unordered_map<string, int> cnt;
        vector<string> ans;
        for (int i = 0, n = s.size() - 10 + 1; i < n; ++i) {
            auto t = s.substr(i, 10);
            if (++cnt[t] == 2) {
                ans.emplace_back(t);
            }
        }
        return ans;
    }
};

class Solution:
    def findRepeatedDnaSequences(self, s: str) -> List[str]:
        words = set() # for encoded words
        repeated = set() # for encoded words
        result = [] # not encoded words

        # A => binary: 00
        # C => binary: 01
        # G => binary: 10
        # T => binary: 11
        mapping = {'A': 0, 'C': 1, 'G': 2, 'T': 3}

        for i in range(len(s) - 9):
            v = 0
            for j in range(i, i + 10):
                # every time, use the new 2 bits after shifting for current char
                v <<= 2
                v |= mapping[s[j]]
            if v in words and v not in repeated:
                repeated.add(v)
                result.append(s[i:i + 10])

            words.add(v)

        return result

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

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

func findRepeatedDnaSequences(s string) (ans []string) {
	cnt := map[string]int{}
	for i := 0; i < len(s)-10+1; i++ {
		t := s[i : i+10]
		cnt[t]++
		if cnt[t] == 2 {
			ans = append(ans, t)
		}
	}
	return
}

function findRepeatedDnaSequences(s: string): string[] {
    const n = s.length;
    const cnt: Map<string, number> = new Map();
    const ans: string[] = [];
    for (let i = 0; i <= n - 10; ++i) {
        const t = s.slice(i, i + 10);
        cnt.set(t, (cnt.get(t) ?? 0) + 1);
        if (cnt.get(t) === 2) {
            ans.push(t);
        }
    }
    return ans;
}

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

public class Solution {
    public IList<string> FindRepeatedDnaSequences(string s) {
        var cnt = new Dictionary<string, int>();
        var ans = new List<string>();
        for (int i = 0; i < s.Length - 10 + 1; ++i) {
            var t = s.Substring(i, 10);
            if (!cnt.ContainsKey(t)) {
                cnt[t] = 0;
            }
            if (++cnt[t] == 2) {
                ans.Add(t);
            }
        }
        return ans;
    }
}

use std::collections::HashMap;

impl Solution {
    pub fn find_repeated_dna_sequences(s: String) -> Vec<String> {
        if s.len() < 10 {
            return vec![];
        }
        let mut cnt = HashMap::new();
        let mut ans = Vec::new();
        for i in 0..s.len() - 9 {
            let t = &s[i..i + 10];
            let count = cnt.entry(t).or_insert(0);
            *count += 1;
            if *count == 2 {
                ans.push(t.to_string());
            }
        }
        ans
    }
}

187 - Repeated DNA Sequences

187. Repeated DNA Sequences

Description

Solutions

All Problems

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187. Repeated DNA Sequences

Description

Solutions

All Problems

All Solutions