Can anybody help me with optimization of my LONGEST COMMON SUBSTRING problem? I must read really big files (up to 2 Gb), but i cant figure out which structure to use... In c++ there is no hash maps.. There is concurrent hash map in TBB but it is very complicated to use with this algorithm. I have this problem solved with **L matrix but it is greedy and cannot be used for large inputs. Matrix is full of zeros, and that can be eliminated by i.e. using map> and store only non-zeros but that is really slow and practicaly unusable. Speed is very important. Here is the code :
// L[i][j] will contain length of the longest substring
// ending by positions i in refSeq and j in otherSeq
size_t **L = new size_t*[refSeq.length()];
for(size_t i=0; i<refSeq.length();++i)
L[i] = new size_t[otherSeq.length()];
// iteration over the characters of the reference sequence
for(size_t i=0; i<refSeq.length();i++){
// iteration over the characters of the sequence to compare
for(size_t j=0; j<otherSeq.length();j++){
// if the characters are the same,
// increase the consecutive matching score from the previous cell
if(refSeq[i]==otherSeq[j]){
if(i==0 || j==0)
L[i][j]=1;
else
L[i][j] = L[i-1][j-1] + 1;
}
// or reset the matching score to 0
else
L[i][j]=0;
}
}
// output the matches for this sequence
// length must be at least minMatchLength
// and the longest possible.
for(size_t i=0; i<refSeq.length();i++){
for(size_t j=0; j<otherSeq.length();j++){
if(L[i][j]>=minMatchLength) {
//this sequence is part of a longer one
if(i+1<refSeq.length() && j+1<otherSeq.length() && L[i][j]<=L[i+1][j+1])
continue;
//this sequence is part of a longer one
if(i<refSeq.length() && j+1<otherSeq.length() && L[i][j]<=L[i][j+1])
continue;
//this sequence is part of a longer one
if(i+1<refSeq.length() && j<otherSeq.length() && L[i][j]<=L[i+1][j])
continue;
cout << i-L[i][j]+2 << " " << i+1 << " " << j-L[i][j]+2 << " " << j+1 << "\n";
// output the matching sequences for debugging :
//cout << refSeq.substr(i-L[i][j]+1,L[i][j]) << "\n";
//cout << otherSeq.substr(j-L[i][j]+1,L[i][j]) << "\n";
}
}
}
