# Lesson 5 - GenomicRangeQuery

xeropise1·2023년 5월 22일
0

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 = SS...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 QK.

For example, consider string S = CAGCCTA and arrays P, Q such that:

P = 2    Q = 4
P = 5    Q = 5
P = 0    Q = 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 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 an array of integers.

For example, given the string S = CAGCCTA and arrays P, Q such that:

P = 2    Q = 4
P = 5    Q = 5
P = 0    Q = 6

the 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 and 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.

### 내 답안

답은 다 도출되었다. 퍼포먼스에서 모두 fail이 났다.
아마 알고리즘이 필요한듯?

import java.lang.Math;

class Solution {
public int[] solution(String S, int[] P, int[] Q) {

char[] chars = S.toCharArray();

for(int i=0; i<P.length; i++) {
int p = P[i];
int q = Q[i];

int minimulImpactFactor = 5;
for(int j=p; j<=q; j++) {
String s = String.valueOf(chars[j]);
int impactFactor = getImpactFactor(s);
minimulImpactFactor = Math.min(minimulImpactFactor, impactFactor);

}

}

}

private int getImpactFactor(String s) {
int impactFactor = 0;

if (s.equals("A")) {
impactFactor = 1;
}

if (s.equals("C")) {
impactFactor = 2;
}

if (s.equals("G")) {
impactFactor = 3;
}

if (s.equals("T")) {
impactFactor = 4;
}

return impactFactor;
}
}

### 베스트 답안

부분합이라는 알고리즘 개념이 필요하다.

무슨말인지는 알겠는데 숫자의 흐름을 전부 이해하기가 좀 어렵다. 문제 풀이를 많이 해봐야겠다.

public int[] solution(String S, int[] P, int[] Q) {

int[] A = new int[S.length()];
int[] C = new int[S.length()];
int[] G = new int[S.length()];

// index == 0
{
char c = S.charAt(0);

if (c == 'A') {
A++;
}
else if (c == 'C') {
C++;
}
else if (c == 'G') {
G++;
}
}

// index > 0
for (int i = 1; i < S.length(); i++) {
char c = S.charAt(i);

if (c == 'A') {
A[i] = A[i - 1] + 1;
C[i] = C[i - 1];
G[i] = G[i - 1];
}
else if (c == 'C') {
A[i] = A[i - 1];
C[i] = C[i - 1] + 1;
G[i] = G[i - 1];
}
else if (c == 'G') {
A[i] = A[i - 1];
C[i] = C[i - 1];
G[i] = G[i - 1] + 1;
}
else {
A[i] = A[i - 1];
C[i] = C[i - 1];
G[i] = G[i - 1];
}
}

for (int i = 0; i < P.length; i++) {
if (P[i] > 0) {
if (A[Q[i]] > A[P[i] - 1]) {
}
else if (C[Q[i]] > C[P[i] - 1]) {
}
else if (G[Q[i]] > G[P[i] - 1]) {
}
else {
}
}
else {
if (A[Q[i]] > 0) {
}
else if (C[Q[i]] > 0) {
}
else if (G[Q[i]] > 0) {