Task description

A DNA sequence can be represented as a string consisting of the lettersA,C,GandT, which correspond to the types of successive nucleotides in the sequence. Each nucleotide has animpact factor, which is an integer. Nucleotides of typesA,C,GandThave 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 =CAGCCTAand arrays P, Q such that:

The answers to these M = 3 queries are as follows:

• The part of the DNA between positions 2 and 4 contains nucleotidesGandC(twice), whose impact factors are 3 and 2 respectively, so the answer is 2.
• The part between positions 5 and 5 contains a single nucleotideT, 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 nucleotideAwhose 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 =CAGCCTAand arrays P, Q such that:

the function should return the values [2, 4, 1], as explained above.

def solution(S,P,Q):
    data = []
    result = []
    for item in S:
        if item == 'A':
            data.append(1)
        elif item == 'C':
            data.append(2)
        elif item == 'G':
            data.append(3)
        else:
            data.append(4)
    #print(data)
    for K in range(len(P)):
        count = min(data[P[K]:Q[K]+1])
        result.append(count)
    return result
    pass