zoukankan      html  css  js  c++  java
  • GenomicRangeQuery Find the minimal nucleotide from a range of sequence DNA.

    Task description

     

    A DNA sequence can be represented as a string consisting of the letters ACG 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 ACG 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.

    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
  • 相关阅读:
    JavaScript之美读书笔记一
    关于form与表单操作
    正则表达式中的字符转义
    [codeforces 391D2]Supercollider
    [AMPPZ 2013]Bytehattan
    [那些你所不知道的鬼畜写法]平衡树学习笔记
    [AHOI 2006][BZOJ 1269]文本编辑器editor
    [BZOJ 3622]已经没有什么好害怕的了
    [POI 2008]Mafia
    [POI 2008][BZOJ 1132]Tro
  • 原文地址:https://www.cnblogs.com/iter-li/p/13493135.html
Copyright © 2011-2022 走看看