# Repeated DNA Sequences in C++

Suppose we have a DNA sequence. As we know, all DNA is composed of a series of nucleotides abbreviated such as A, C, G, and T, for example: "ACGAATTCCG". When we are studying DNA, it is sometimes useful to identify repeated sequences within the DNA.

We have to write one method to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.

So if the input is like “AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT”, then the output will be ["AAAAACCCCC", "CCCCCAAAAA"].

To solve this, we will follow these steps −

• Define an array ret, n := size of s, create two sets called visited and visited2

• define a map called bitVal.

• Store corresponding values for ACGT like 0123 into butVal.

• for i in range 0 to n – 1

• mask := mast OR bitVal[s[i]]

• if i < 9, then just continue to the next iteration

• insert substring form index i – 9 to 9, into ret

• insert mark into visited2.

• return ret

## Example(C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
typedef long long int lli;
class Solution {
public:
vector<string>findRepeatedDnaSequences(string s) {
vector <string> ret;
int n = s.size();
set <int> visited;
set <int> visited2;
map <char, int> bitVal;
bitVal['A'] = 0;
bitVal['C'] = 1;
bitVal['G'] = 2;
bitVal['T'] = 3;
for(int i = 0; i < n; i++){
if(i < 9) continue;
ret.push_back(s.substr(i - 9, 10));
}
}
return ret;
}
};
main(){
Solution ob;
print_vector(ob.findRepeatedDnaSequences("AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"));
}

## Input

"AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"

## Output

[AAAAACCCCC, CCCCCAAAAA, ]