notes/OJ notes/pages/Leetcode Repeated-DNA-Sequences.md

# Leetcode Repeated-DNA-Sequences

2022-09-06 19:58

> ##### Data structures:
>
> #DS #hash_table #string
>
> ##### Difficulty:
>
> #coding_problem #difficulty_medium
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Links:

- [Link to problem](https://leetcode.com/problems/repeated-dna-sequences/)

---

### Problem

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**.

#### Examples

**Example 1:**

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

**Example 2:**

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

#### Constraints

### Thoughts

> [!summary]
> This is a #hash_table problem.

The question ask for an answer, and the substrings can
overlap. So, using a map is prefered(Why?)

Two reasons:

- Easy way to know if a array is a duplicate (set, map can
  suffice.)
- Keep information on how many duplicates found, so we only
  append it to the answer the first time we meet it.

One trip-over hole: in the for loop, upper bound should be:

```cpp
for (int i = 0, top = s.size() - 9; i < top; i++)
                                ^^^
```

Minus 9, because 9 is the extended length for an subarray starting with i.

```
1234567890
^        ^
|--------|
i       i+9

i + 9 - i + 1 = 10.
```

With these edge-cases taken care of, we can proceed to the
solution:

### Solution

```cpp
class Solution {
public:
  vector<string> findRepeatedDnaSequences(string s) {
    unordered_map<string, int> used;
    vector<string> ans = {};
    for (int i = 0, size = s.size() - 9; i < size; i++) {
      string tmp = s.substr(i, 10);

      if ((used[tmp]++) == 1) {
        ans.push_back(tmp);
      }
    }

    return ans;
  }
};
```