Search a sequence in a string. DNA - c++

Question

I need to do a program that separate from 3 to the size of a string and compare to the others sequences of 3 in the same string given. I'm going to explain it.
User introduce this DNA string = "ACTGCGACGGTACGCTTCGACGTAG" For example.
We start with n = 3, this is, we take the first three caracters for comparing in the DNA.
The first characters are: "ACT", and we need to compare it with the other sequences of three, like, [CTG,TGC,GCA... until the end].
If we find another sequence equal to "ACT", we save the position.
Here is another example:
DNA: "ACTGCGACGGTACGCTTCGACGTAG" and we find this sequences in his positions:
ACG: 7 - 12 - 20
CGA: 5 - 18
GAC: 6 - 19
GTA: 10 - 22
CGAC: 5 - 18
GACG: 6 - 19
CGACG: 5 - 18
The number is the position of the start of the sequence:
ACTGCGACGGTACGCTTCGACGTAG
You can see that the n = 3, increment in 1 when the we end to find by n = 3, the variable pass to n=4, until n = DNA.size().
My problem is that i have one function for divide the string in a little sequences of the DNA, and I do a push_back() for saving in the vector, and then I can see if there is more sequences or not, but i don't know how can i get the position.
I can use the library algorithm, and for sure, in this library there is a function that do this but i don't know so much this library.
Here is my code:
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
const string DNA = "ACTGCGACGGTACGCTTCGACGTAG";
const size_t taille = DNA.size();
size_t m = 3;
vector<string> v;
/*
struct DNA{
const string dna; // chaine saisie pour l'utilisateur
size_t taille; // Taille de la chaine
string chaine; // Chaine à chercher
};
*/
// what kind of structs can i create? for me it's stupid to make any struct in this program.
bool checkDNA(string &s);
string takeStrings(const string &s,size_t i, size_t m);
void FindSequenceDNA(vector<string>&s,string sq);
size_t incrementValue(size_t &m);
int main(){
string DNAuser;
cout << "Introduce the DNA: ";
cin >> DNAuser;
bool request;
cout << boolalpha;
request = DNAuser.find_first_not_of("AGCT");
cout << request << endl;
vector<string> vectorSq;
size_t auxiliar = 0;
string r;
size_t ocurrencies = DNA.size()-2;
cout << "DNA: " << DNA << endl;
while(auxiliar<ocurrencies){ // This gonna be works with the ocurriences, from 1 to end.
r = takeStrings(DNA,auxiliar,auxiliar+m);
auxiliar++;
if(r.size()==m){
vectorSq.push_back(r);
}
}
// string res = takeStrings(DNA,0,3);
// cout << "res: " << res << endl;
// cout << "Printing vector: " << endl;
// I just need to find the other, the practice is almost done.
for(size_t i = 0; i< vectorSq.size(); i++){
cout << vectorSq[i] << endl;
}
return 0;
}
string takeStrings(const string &s,size_t i, size_t m){
string result;
size_t aux=i;
if(s.size()==0){
cout << "String is empty." << endl;
}
else{
for(;i<s.size()&&i!=m;i++){
result+=s[i];
aux++;
}
}
return result;
}
void FindSequenceDNA(vector<string>&s,string sq){
if(s.size()==0){
cout << "DNA invalid." << endl;
}
else{
for(size_t i=0;i<s.size();i++){
if(sq==s[i]){
cout << "function: " << endl;
cout << s[i] << endl; // I need to calculate the real position in the string, not in the vector
}
}
}
}
bool checkDNA(string &s){
bool res;
if(s.size()==0 || s.size()<3){
cout << "DNA invalid" << endl;
}
else{
for(size_t i=0;i<s.size();i++){
if(s[i]=='A' || s[i]=='C' || s[i]=='G' || s[i]=='T')
{
res = true;
}
else{
res= false;
}
}
}
return res;
}
size_t incrementValue(size_t &m){
if(m<DNA.size()){
m++;
}
return m;
}

Answer 1

Based on Mohit's answer but re-uses pointers to possibly, get better performance (vs string.substr)
#include <iostream>
#include <cstring>
#include <vector>
#include <string>
using namespace std;
static const char* DNAdata = "ACTGCGACGGTACGCTTCGACGTAG";
static const size_t len = strlen(DNAdata);
vector< vector< string > > uniqueKeys(len);
vector< vector< vector<size_t> > > locations(len);
void saveInfo(const char* str, size_t n, size_t loc) {
vector<string>& keys = uniqueKeys[n-1];
vector<vector<size_t> >& locs = locations[n-1];
bool found = false;
for (size_t i=0; i<keys.size(); ++i) {
if (keys[i] == str) {
locs[i].push_back(loc);
found = true;
break;
}
}
if (!found) {
vector<size_t> newcont;
newcont.push_back(loc);
keys.push_back(str);
locs.push_back(newcont);
}
}
void printInfo(const char* str) {
cout << str << endl;
size_t len = strlen(str);
vector<string>& keys = uniqueKeys[len-1];
vector<vector<size_t> >& locs = locations[len-1];
for (size_t i=0; i<keys.size(); ++i) {
if (keys[i] == str) {
vector<size_t>& l = locs[i];
vector<size_t>::iterator iter = l.begin();
for (; iter != l.end(); ++iter) {
cout << *iter << endl;
}
break;
}
}
}
int main() {
char* DNA = new char[len+1];
strcpy(DNA, DNAdata);
char* end = DNA+len;
char* start = DNA;
for (size_t n =3; n<=len; ++n) {
size_t loc = 0;
char* p = start;
char* e = p+n;
while (e <= end) {
char save = *e;
*e = 0;
saveInfo(p++, n, loc++);
*e = save;
++e;
}
}
delete[] DNA;
printInfo("GTA");
printInfo("ACTGCGACGGTACGCTTCGACGTA");
return 0;
}
To print all:
void printAll() {
for (size_t n=3; n<=len; ++n) {
cout << "--> " << n << " <--" << endl;
vector<string>& keys = uniqueKeys[n-1];
vector<vector<size_t> >& locs = locations[n-1];
for (size_t i=0; i<keys.size(); ++i) {
cout << keys[i] << endl;
vector<size_t>& l = locs[i];
vector<size_t>::iterator iter = l.begin();
for (; iter != l.end(); ++iter) {
cout << *iter << endl;
}
}
}
}

Answer 2

How about:
std::map< std::string, std::vectpr<int> > msvi;
std::size_t len = dna.size();
for(size_t from = 0; from < len; ++from) {
for(size_t sz = 3; sz < len; ++sz) {
msvi[ dna.substr(from, sz ].push_back(from);
}
}
This creates all strings of size 3 and saves there position in a map.
Live demo link
Print only the items with 2 or more instances
As you don't want to use std::map, you can construct a trie as shown on this page written in C. Change your tree node to:
struct tree_node {
vector<int> starts;
struct tree_node *children[26]; /* A to Z */
};