Codility 11번 GenomicRangeQuery

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 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 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, 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.

 

 

import java.util.ArrayList;

class Solution {
    public int[] solution(String S, int[] P, int[] Q) {
        // write your code in Java SE 8
        ArrayList<Integer> aArray = new ArrayList<>();
        ArrayList<Integer> cArray = new ArrayList<>();
        ArrayList<Integer> gArray = new ArrayList<>();
        ArrayList<Integer> tArray = new ArrayList<>();
        for (int i = 0; i < S.length(); i++) {
            switch (S.charAt(i)) {
                case 'A':
                    aArray.add(i);
                    break;
                case 'C':
                    cArray.add(i);
                    break;
                case 'G':
                    gArray.add(i);
                    break;
                case 'T':
                    tArray.add(i);
            }
        }
        int[] ans = new int[P.length];
        System.out.println(ans[0]);
        for (int i = 0; i < P.length; i++) {

            for (int j = 0; j < aArray.size(); j++) {
                if (aArray.get(j) >= P[i] && aArray.get(j) <= Q[i]) {
                    ans[i] = 1;
                    break;
                }
            }

            if (ans[i] == 0) {
                for (int j = 0; j < cArray.size(); j++) {
                    if (cArray.get(j) >= P[i] && cArray.get(j) <= Q[i]) {
                        ans[i] = 2;
                        break;
                    }
                }
            }
            if (ans[i] == 0) {
                for (int j = 0; j < gArray.size(); j++) {
                    if (gArray.get(j) >= P[i] && gArray.get(j) <= Q[i]) {
                        ans[i] = 3;
                        break;
                    }
                }
            }
            if (ans[i] == 0) {
                for (int j = 0; j < tArray.size(); j++) {
                    if (tArray.get(j) >= P[i] && tArray.get(j) <= Q[i]) {
                        ans[i] = 4;
                        break;
                    }
                }
            }
        }

        return ans;
    }
}

class Solution {
    public int[] solution(String S, int[] P, int[] Q) {
        int[] aArray = new int[S.length()];
        int[] cArray = new int[S.length()];
        int[] gArray = new int[S.length()];

        switch (S.charAt(0)) {
            case 'A':
                aArray[0]++;
                break;
            case 'C':
                cArray[0]++;
                break;
            case 'G':
                gArray[0]++;
                break;
        }

        for (int i = 1; i <S.length(); i++) {
            switch (S.charAt(i)) {
                case 'A':
                    aArray[i] = aArray[i-1] + 1;
                    cArray[i] = cArray[i-1];
                    gArray[i] = gArray[i-1];
                    break;
                case 'C':
                    aArray[i] = aArray[i-1];
                    cArray[i] = cArray[i-1] + 1;
                    gArray[i] = gArray[i-1];
                    break;
                case 'G':
                    aArray[i] = aArray[i-1];
                    cArray[i] = cArray[i-1];
                    gArray[i] = gArray[i-1] + 1;
                    break;
                case 'T':
                    aArray[i] = aArray[i-1];
                    cArray[i] = cArray[i-1];
                    gArray[i] = gArray[i-1];
                    break;
            }
        }
        int[] ans = new int[P.length];
        for (int i = 0; i < P.length; i++) {
            
                if(P[i] == 0){
                    if(aArray[Q[i]] > 0){
                        ans[i] = 1;
                    } else if(cArray[Q[i]] > 0){
                        ans[i] = 2;
                    } else if(gArray[Q[i]] > 0){
                        ans[i] = 3;
                    } else{
                        ans[i] = 4;
                    }
                } else{
                    if(aArray[Q[i]] > aArray[P[i]-1]){
                        ans[i] = 1;
                    } else if(cArray[Q[i]] > cArray[P[i]-1]){
                        ans[i] = 2;
                    } else if(gArray[Q[i]] > gArray[P[i]-1]){
                        ans[i] = 3;
                    } else{
                        ans[i] = 4;
                    }
                }
        
        }

        return ans;

    }
}

 

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