java代码训练基因组范围查询

Question

任务是：

给出了一个非空的零索引字符串S。字符串S由N个大写英文字母A，C，G，t组成。

这个字符串实际上代表一个DNA序列，大写字母代表单个核苷酸。

您还将获得由 M 个整数组成的非空零索引数组 P 和 Q。这些数组表示有关最小核苷酸的查询。我们将字符串 S 的字母表示为数组 P 和 Q 中的整数 1、2、3、4，其中 A = 1、C = 2、G = 3、T = 4，我们假设 A

查询K要求您从范围（P[K]，Q[K]），0中找到最小核苷酸≤ P【i】≤ 问题[i]

例如，考虑字符串S=GACACCATA和数组P、Q，这样：

P[0] = 0    Q[0] = 8
P[1] = 0    Q[1] = 2
P[2] = 4    Q[2] = 5
P[3] = 7    Q[3] = 7

这些范围内的最小核苷酸如下:

    (0, 8) is A identified by 1,
    (0, 2) is A identified by 1,
    (4, 5) is C identified by 2,
    (7, 7) is T identified by 4.

写一个函数：

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

给定一个由N个字符组成的非空零索引字符串S和两个由M个整数组成的非空零索引数组P和Q，返回一个由M个字符组成的数组，指定所有查询的连续答案。

序列应返回为：

    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).

例如，给定字符串 S = GACACCATA 和数组 P、Q，使得：

P[0] = 0    Q[0] = 8
P[1] = 0    Q[1] = 2
P[2] = 4    Q[2] = 5
P[3] = 7    Q[3] = 7

如上所述，该函数应返回值 [1， 1， 2， 4]。

假设:

    N is an integer within the range [1..100,000];
    M is an integer within the range [1..50,000];
    each element of array P, Q is an integer within the range [0..N − 1];
    P[i] ≤ Q[i];
    string S consists only of upper-case English letters A, C, G, T.

复杂性：

    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).

输入数组的元素可以修改。

我的解决方案是：

class Solution {
    public int[] solution(String S, int[] P, int[] Q) {
        final  char c[] = S.toCharArray();
        final int answer[] = new int[P.length];
        int tempAnswer;
        char tempC;

        for (int iii = 0; iii < P.length; iii++) {
            tempAnswer = 4;
            for (int zzz = P[iii]; zzz <= Q[iii]; zzz++) {
                tempC = c[zzz];
                if (tempC == 'A') {
                    tempAnswer = 1;
                    break;
                } else if (tempC == 'C') {
                    if (tempAnswer > 2) {
                        tempAnswer = 2;
                    }
                } else if (tempC == 'G') {
                    if (tempAnswer > 3) {
                        tempAnswer = 3;
                    }

                }
            }
            answer[iii] = tempAnswer;
        }

        return answer;
    }
}

这不是最佳的，我相信它应该在一个循环内完成，任何提示，我如何实现它？

您可以在此处检查解决方案的质量https://codility.com/train/测试名称是Genomic-range-query。

Answer 1

Java，100/100，但是没有累积/前缀和！我在一个数组“地图”中隐藏了低3个核苷酸的最后出现索引。稍后我检查最后一个索引是否在P-Q之间。如果是，它返回nuclotide，如果没有找到，它是顶部的一个(T):

class Solution {

int[][] lastOccurrencesMap;

public int[] solution(String S, int[] P, int[] Q) {
    int N = S.length();
    int M = P.length;

    int[] result = new int[M];
    lastOccurrencesMap = new int[3][N];
    int lastA = -1;
    int lastC = -1;
    int lastG = -1;

    for (int i = 0; i < N; i++) {
        char c = S.charAt(i);

        if (c == 'A') {
            lastA = i;
        } else if (c == 'C') {
            lastC = i;
        } else if (c == 'G') {
            lastG = i;
        }

        lastOccurrencesMap[0][i] = lastA;
        lastOccurrencesMap[1][i] = lastC;
        lastOccurrencesMap[2][i] = lastG;
    }

    for (int i = 0; i < M; i++) {
        int startIndex = P[i];
        int endIndex = Q[i];

        int minimum = 4;
        for (int n = 0; n < 3; n++) {
            int lastOccurence = getLastNucleotideOccurrence(startIndex, endIndex, n);
            if (lastOccurence != 0) {
                minimum = n + 1; 
                break;
            }
        }

        result[i] = minimum;
    }
    return result;
}

int getLastNucleotideOccurrence(int startIndex, int endIndex, int nucleotideIndex) {
    int[] lastOccurrences = lastOccurrencesMap[nucleotideIndex];
    int endValueLastOccurenceIndex = lastOccurrences[endIndex];
    if (endValueLastOccurenceIndex >= startIndex) {
        return nucleotideIndex + 1;
    } else {
        return 0;
    }
}
}

Answer 2

简单，优雅，领域特定，100/100 JS解决方案，带注释！

function solution(S, P, Q) {
    var N = S.length, M = P.length;

    // dictionary to map nucleotide to impact factor
    var impact = {A : 1, C : 2, G : 3, T : 4};

    // nucleotide total count in DNA
    var currCounter = {A : 0, C : 0, G : 0, T : 0};

    // how many times nucleotide repeats at the moment we reach S[i]
    var counters = [];

    // result
    var minImpact = [];

    var i;

    // count nucleotides
    for(i = 0; i <= N; i++) {
        counters.push({A: currCounter.A, C: currCounter.C, G: currCounter.G});
        currCounter[S[i]]++;
    }

    // for every query
    for(i = 0; i < M; i++) {
        var from = P[i], to = Q[i] + 1;

        // compare count of A at the start of query with count at the end of equry
        // if counter was changed then query contains A
        if(counters[to].A - counters[from].A > 0) {
            minImpact.push(impact.A);
        }
        // same things for C and others nucleotides with higher impact factor
        else if(counters[to].C - counters[from].C > 0) {
            minImpact.push(impact.C);
        }
        else if(counters[to].G - counters[from].G > 0) {
            minImpact.push(impact.G);
        }
        else { // one of the counters MUST be changed, so its T
            minImpact.push(impact.T);
        }
    }

    return minImpact;
}

Answer 3

这是 codility.com 年100个中的100个的解决方案。请阅读前缀和以了解解决方案：

public static int[] solveGenomicRange(String S, int[] P, int[] Q) {
        //used jagged array to hold the prefix sums of each A, C and G genoms
        //we don't need to get prefix sums of T, you will see why.
        int[][] genoms = new int[3][S.length()+1];
        //if the char is found in the index i, then we set it to be 1 else they are 0
        //3 short values are needed for this reason
        short a, c, g;
        for (int i=0; i<S.length(); i++) {
            a = 0; c = 0; g = 0;
            if ('A' == (S.charAt(i))) {
                a=1;
            }
            if ('C' == (S.charAt(i))) {
                c=1;
            }
            if ('G' == (S.charAt(i))) {
                g=1;
            }
            //here we calculate prefix sums. To learn what's prefix sums look at here https://codility.com/media/train/3-PrefixSums.pdf
            genoms[0][i+1] = genoms[0][i] + a;
            genoms[1][i+1] = genoms[1][i] + c;
            genoms[2][i+1] = genoms[2][i] + g;
        }

        int[] result = new int[P.length];
        //here we go through the provided P[] and Q[] arrays as intervals
        for (int i=0; i<P.length; i++) {
            int fromIndex = P[i];
            //we need to add 1 to Q[i], 
            //because our genoms[0][0], genoms[1][0] and genoms[2][0]
            //have 0 values by default, look above genoms[0][i+1] = genoms[0][i] + a; 
            int toIndex = Q[i]+1;
            if (genoms[0][toIndex] - genoms[0][fromIndex] > 0) {
                result[i] = 1;
            } else if (genoms[1][toIndex] - genoms[1][fromIndex] > 0) {
                result[i] = 2;
            } else if (genoms[2][toIndex] - genoms[2][fromIndex] > 0) {
                result[i] = 3;
            } else {
                result[i] = 4;
            }
        }

        return result;
    }