Codility 5.2 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 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 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:

def solution(S, P, 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.
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这道题是典型的用空间换时间的例子。下面的解法并非最优，还需找到一种优化获得最小值的方法。

def solution(S, P, Q):
    # write your code in Python 3.6
    l = {"A":1,"C":2,"G":3,"T":4}
    prefix_sum=[]
    m =[]
    for i in S:
        prefix_sum.append(l[i])
    for i in range(len(P)):
        m.append(min(prefix_sum[P[i]:Q[i]+1]))
    return m

用四个Prefix Sums数组分别用来统计ACGT从m到n的个数，如果A个数为0就看C，如此类推。
下面这个方法是打印出每个核苷酸在每个位置的下一次出现的位置信息。（来源于CODEsays）

P = [2,5,0]
Q = [4,5,6]
S ='CAGCCTA'
result = []
DNA_len = len(S)
mapping = {"A":1, "C":2, "G":3, "T":4}
# next_nucl is used to store the position information
# next_nucl[0] is about the "A" nucleotides, [1] about "C"
#    [2] about "G", and [3] about "T"
# next_nucl[i][j] = k means: for the corresponding nucleotides i,
#    at position j, the next corresponding nucleotides appears
#    at position k (including j)
# k == -1 means: the next corresponding nucleotides does not exist
next_nucl = [[-1]*DNA_len, [-1]*DNA_len, [-1]*DNA_len, [-1]*DNA_len]

# Scan the whole DNA sequence, and retrieve the position information
next_nucl[mapping[S[-1]] - 1][-1] = DNA_len-1

for index in range(DNA_len-2,-1,-1):
    next_nucl[0][index] = next_nucl[0][index+1]
    next_nucl[1][index] = next_nucl[1][index+1]
    next_nucl[2][index] = next_nucl[2][index+1]
    next_nucl[3][index] = next_nucl[3][index+1]
    next_nucl[mapping[S[index]] - 1][index] = index
print(next_nucl)
for index in range(0,len(P)):
    if next_nucl[0][P[index]] != -1 and next_nucl[0][P[index]] <= Q[index]:
        result.append(1)
    elif next_nucl[1][P[index]] != -1 and next_nucl[1][P[index]] <= Q[index]:
        result.append(2)
    elif next_nucl[2][P[index]] != -1 and next_nucl[2][P[index]] <= Q[index]:
        result.append(3)
    else:
        result.append(4)
result
next_nucl[1][P[0]]