Codility. Train. Genomic-range-query ★★★

A non-empty zero-indexed string S is given. String S consists of N characters from the set of upper-case English letters A, C, G, T.

This string actually represents a DNA sequence, and the upper-case letters represent single nucleotides.

You are also given non-empty zero-indexed arrays P and Q consisting of M integers. These arrays represent queries about minimal nucleotides. We represent the letters of string S as integers 1, 2, 3, 4 in arrays P and Q, where A = 1, C = 2, G = 3, T = 4, and we assume that A < C < G < T.

Query K requires you to find the minimal nucleotide from the range (P[K], Q[K]), 0 ≤ P[i] ≤ Q[i] < N.

For example, consider string S = GACACCATA and arrays P, Q such that:

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

The minimal nucleotides from these ranges are as follows:

• (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.

Write a function:

def solution(S, P, Q)

that, given a non-empty zero-indexed string S consisting of N characters and two non-empty zero-indexed arrays P and Q consisting of M integers, returns an array consisting of M characters specifying the consecutive answers to all queries.

The sequence should be returned as:

• 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).

For example, given the string S = GACACCATA and arrays P, Q such that:

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

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

Assume that:

• 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.

Complexity:

• 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).

Elements of input arrays can be modified.

Solution:

def solution(S, P, Q):
  vMap = dict(A = 1, C = 2, G = 3, T = 4)
  vLst = dict(A =[], C =[], G =[], T =[])
  for i in xrange(len(S)):
    vLst[S[i]].append(i)
  for i in xrange(len(P)):
    for j in sorted(vLst):
      iF, iL = 0, len(vLst[j])
      if iL == iF or \
         vLst[j][iF] > Q[i] or \
         vLst[j][iL - 1] < P[i]:
        continue
      while iF < iL:
        iM = iF + (iL - iF) / 2
        if P[i] <= vLst[j][iM]:
          if vLst[j][iM] <= Q[i]:
            P[i] = vMap[j]
            break
          iL = iM
        else:
          iF = iM + 1
      else:
        continue
      break
  return P