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 Thave 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] = 6
The 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:
class Solution { public int[] solution(string S, int[] P, int[] Q); }
that, given a non-empty zero-indexed string S consisting of N characters and two non-empty zero-indexed arrays P and Q consisting of M integers, returns an array consisting of M integers specifying the consecutive answers to all queries.
SIMPLE SOLUTION - 100% correctness, 0% performance:
using System;
using System.Collections.Generic;
using System.Linq;
class Solution {
public int[] solution(string S, int[] P, int[] Q) {
string dna = "0ACGT";
int m = P.Length;
int [] result = new int [m];
for (int i=0; i<m; i++)
{
string subS= S.Substring(P[i], (Q[i]-P[i]+1));
char [] charSubS= subS.ToCharArray();
Array.Sort(charSubS);
result[i]=dna.IndexOf(charSubS[0]);
}
return result;
}
}
100% SCORE SOLUTION:
class Solution {
public int[] solution(string S, int[] P, int[] Q) {
var nucleo = new int[S.Length + 1, 4];
for (var count = 0; count < S.Length; count++)
{
if (count > 0)
{
for (var index = 0; index < 4; index++)
{
nucleo[count + 1, index] += nucleo[count, index];
}
}
switch (S[count])
{
case 'A':
nucleo[count + 1, 0]++;
break;
case 'C':
nucleo[count + 1, 1]++;
break;
case 'G':
nucleo[count + 1, 2]++;
break;
case 'T':
nucleo[count + 1, 3]++;
break;
}
}
var result = new int[P.Length];
for (var count = 0; count < P.Length; count++) {
if(P[count] == Q[count])
{
switch(S[P[count]]) {
case 'A':
result[count] = 1;
break;
case 'C':
result[count] = 2;
break;
case 'G':
result[count] = 3;
break;
case 'T':
result[count] = 4;
break;
}
} else {
for(var index = 0; index < 4; index++) {
if((nucleo[Q[count] + 1, index] - nucleo[P[count], index]) > 0) {
result[count] = index + 1;
break;
}
}
}
}
return result;
}
}
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