Find the minimal nucleotide from a range of sequence DNA.
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 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 integers 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 = 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.
方法一：用 mapping 先記錄各字元最早出現的位置
Correctness：100%、Performance：100%
```python mapping = {"A":100001, "C":100001, "G":100001} def solutionByMapping(S, P, Q): length = len(S) matrix = [([0] * length) for i in range(len(mapping))] for i in range(length-1, -1, -1): mapping[S[i]] = i matrix[0][i] = mapping['A'] matrix[1][i] = mapping['C'] matrix[2][i] = mapping['G'] length = len(P) result = [0] * length for i in range(length): if matrix[0][P[i]] <= Q[i]: result[i] = 1 elif matrix[1][P[i]] <= Q[i]: result[i] = 2 elif matrix[2][P[i]] <= Q[i]: result[i] = 3 else: result[i] = 4 return result ```
方法二：使用 slice 概念的簡易方法（效能O(N*M)被扣分）
Correctness：100%、Performance：33%
```python
def solutionBySlice(S, P, Q):
result = []
length = len(P)
for i in range(length):
temp = (S[P[i]:Q[i]+1])
if "A" in temp:
result.append(1)
elif "C" in temp:
result.append(2)
elif "G" in temp:
result.append(3)
elif "T" in temp:
result.append(4)
return result
```
完整練習題 source code 請參閱：github
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