Lesson 5 - GenomicRangeQuery

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 QK.

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 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[0] = 2    Q[0] = 4
P[1] = 5    Q[1] = 5
P[2] = 0    Q[2] = 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.

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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) {
        int[] answers = new int[P.length];

        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);
                
            }

            answers[i] = minimulImpactFactor;
        }

        return answers;
    }

    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;
    }
}

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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[] answer = new int[P.length];
 
    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[0]++;
        }
        else if (c == 'C') {
            C[0]++;
        }
        else if (c == 'G') {
            G[0]++;
        }
    }
 
    // 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]) {
                answer[i] = 1;
            }
            else if (C[Q[i]] > C[P[i] - 1]) {
                answer[i] = 2;
            }
            else if (G[Q[i]] > G[P[i] - 1]) {
                answer[i] = 3;
            }
            else {
                answer[i] = 4;
            }
        }
        else {
            if (A[Q[i]] > 0) {
                answer[i] = 1;
            }
            else if (C[Q[i]] > 0) {
                answer[i] = 2;
            }
            else if (G[Q[i]] > 0) {
                answer[i] = 3;
            }
            else {
                answer[i] = 4;
            }
        }
    }
 
    return answer;
}