GenomicRangeQuery
Task
A DNA sequence can be represented as a string consisting of the letters A, C, G and T, which correspond to the types of successive nucleotides in the sequence. Each nucleotide has an impact factor, which is an integer. Nucleotides of types A, C, G and T have impact factors of 1, 2, 3 and 4, respectively. You are going to answer several queries of the form: What is the minimal impact factor of nucleotides contained in a particular part of the given DNA sequence?
The DNA sequence is given as a non-empty string S = S[0]S[1]…S[N-1] consisting of N characters. There are M queries, which are given in non-empty arrays P and Q, each consisting of M integers. The K-th query (0 ≤ K < M) requires you to find the minimal impact factor of nucleotides contained in the DNA sequence between positions P[K] and Q[K] (inclusive).
For example, consider string S = CAGCCTA and arrays P, Q such that:
P[0] = 2 Q[0] = 4 P[1] = 5 Q[1] = 5 P[2] = 0 Q[2] = 6The answers to these M = 3 queries are as follows:
- The part of the DNA between positions 2 and 4 contains nucleotides G and C (twice), whose impact factors are 3 and 2 respectively, so the answer is 2.
- The part between positions 5 and 5 contains a single nucleotide T, whose impact factor is 4, so the answer is 4.
- The part between positions 0 and 6 (the whole string) contains all nucleotides, in particular nucleotide A whose impact factor is 1, so the answer is 1.
Write a function:
vector<int> solution(string &S, vector<int> &P, vector<int> &Q);that, given a non-empty string S consisting of N characters and two non-empty arrays P and Q consisting of M integers, returns an array consisting of M integers specifying the consecutive answers to all queries.
Result array should be returned as a vector of integers.
For example, given the string S = CAGCCTA and arrays P, Q such that:
P[0] = 2 Q[0] = 4 P[1] = 5 Q[1] = 5 P[2] = 0 Q[2] = 6the function should return the values [2, 4, 1], as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..100,000];
- M is an integer within the range [1..50,000];
- each element of arrays P, Q is an integer within the range [0..N − 1];
- P[K] ≤ Q[K], where 0 ≤ K < M;
- string S consists only of upper-case English letters A, C, G, T.
Solution
#include <cassert>
#include <math.h>
#include <vector>
vector<int> solution(string &S, vector<int> &P, vector<int> &Q) {
// write your code in C++14 (g++ 6.2.0)
size_t N = S.size();
size_t M = P.size();
assert(Q.size() == M);
//arrays are count of all previous to. so add extra zero items which is zero
vector<int> As_prev(N+1);
vector<int> Cs_prev(N+1);
vector<int> Gs_prev(N+1);
As_prev[0] = 0;
Cs_prev[0] = 0;
Gs_prev[0] = 0;
// build index of count of previous nucleotides of diff types
int A_count = 0, C_count = 0, G_count = 0;
for (std::string::size_type i = 0; i < S.size(); i++) {
char nucleo = S[i];
switch (nucleo) {
case ‘A’:
A_count++;
break;
case ‘C’:
C_count++;
break;
case ‘G’:
G_count++;
break;
case ‘T’:
break;
default:
assert(false);
}
As_prev[i+1] = A_count;
Cs_prev[i+1] = C_count;
Gs_prev[i+1] = G_count;
}
// now calc minimums
vector<int> res(M);
for (size_t i = 0; i < M; i++) {
int mini = 5;
size_t from = P[i];
size_t to = Q[i];
// includes last
int numAs = As_prev[to+1] – As_prev[from];
assert(numAs >= 0);
if (numAs > 0) {
mini = 1;
} else {
int numCs = Cs_prev[to+1] – Cs_prev[from];
assert(numCs >= 0);
if (numCs > 0) {
mini = 2;
} else {
int numGs = Gs_prev[to+1] – Gs_prev[from];
assert(numGs >= 0);
if (numGs > 0) {
mini = 3;
} else {
// only Ts left
mini = 4;
}
}
}// numAs else
// mini calc’d
assert(mini<=4);
res[i] = mini;
}// for
return res;
}Results
| Correctness | 100% |
| Performance | 100% |
| Time Complexity | O(N+M) |