codility - GenomicRangeQuery ( java )

traning center [Genomic Range Query]

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 Gand 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.

The sequence should be returned as:

• a Results structure (in C), or
• a vector of integers (in C++), or
• a Results record (in Pascal), or
• an array of integers (in any other programming language).

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.

Assume that:

• 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.

Complexity:

• expected worst-case time complexity is O(N+M);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

처음에는 contain을 통해 배열 안의 내용을 확인 하는 것이 시간복잡도에 영향을 줄 것이라고 생각을 못했었다. 그래서 contain을 사용해서 너무 쉬운데?? 하면서 제출을 했는데 결과는 timeout이 나오면서 66%가 되었다.

다시 코드로 돌아와 처음부터 String을 숫자로 바꾸어 계산해야겠다고 생각을 했다. 2차원 배열을 만들까 하다가, 2차원 배열은 왠지 공간복잡도 N*M으로 나올 것 같아서 한참을 헤메다 보니, 2차원 배열의 공간 복잡도는 N이었다. 그래서 배열을 통해 해당 index까지 몇 개가 나왔는지 세어주는 index라는 배열을 만들어 주었고, 연산을 해서 출력했다.

처음에는 String을 변환 시키는 과정에서 switch case를 사용하다가 enum을 사용해 보고 싶어 enum을 만들어서 작성했다.

public enum Index {
    A(1), C(2), G(3), T(4);
    int value;
    Index(int value ) {
        this.value = value;
    }
 
    int getValue() {
        return this.value;
    }
 
    static int getValue(char value) {
        return valueOf(String.valueOf(value)).getValue();
    }
}
 
public int[] solution(String S, int[] P, int[] Q) {
    int[][] index = new int[S.length()][4];
 
    for ( int i = 0; i < S.length() ; i ++) {
        index[i][Index.getValue(S.charAt(i)) - 1] = 1;
    }
 
    for ( int i = 1; i  < S.length(); i ++) {
        for ( int j = 0; j < 4; j ++) {
            index[i][j] += index[i - 1][j];
        }
    }
 
    int[] result = new int[P.length];
 
    for ( int i = 0; i < P.length ; i ++) {
        if (P[i] == 0) {
            for ( int j = 0; j < 4; j ++) {
                if (index[Q[i]][j] > 0) {
                    result[i] = j + 1;
                    break;
                }
            }
        } else {
            for ( int j = 0; j < 4; j ++) {
                if (index[Q[i]][j] - index[P[i] - 1][j] > 0) {
                    result[i] = j + 1;
                    break;
                }
            }
        }
    }
    return result;
}

Test Result Report