Codility - Lesson 5 Prefix Sums - 4. GenomicRangeQuery

Source Link:
https://codility.com/programmers/lessons/5-prefix_sums/genomic_range_query/

Question:

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:
class Solution { public int[] solution(String S, int[] P, int[] 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.
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 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.

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.

My Solution:
Notes:
1. result: the minimal impact of each query
int[] result = new int[P.length];
2. to count "A"、"C"、"G"、"T"
// A[i] means: num of 'a' from 0 to i-1
// note: we use "S.length()+1" 
// which will let A[0]=0, C[0]=0, G[0]=0, T[0]=0
// becasue we will compute number of 'a' by "A[Q+1] - A[P]"
// we actually shift to right by one, and assume the biginning is a dummy '0' 
int A[] = new int[S.length()+1];
...
int T[] = new int[S.length()+1];   
3. counting ( note: A[0]=0, C[0]=0, G[0]=0, T[0]=0 )
for (int i = 0; i < S.length(); i++) {
if(S.charAt(i) == 'A')
{
A[i+1] = A[i]+1;
C[i+1] = C[i];
G[i+1] = G[i];
T[i+1] = T[i];
}
...
}
4. to handle the queries
for (int i = 0; i < num_of_query; i++) {
int a = A[ Q[i] + 1] - A[ P[i] ]; // num of 'a' between P and Q
int c = C[ Q[i] + 1] - C[ P[i] ]; // num of 'c' between P and Q
int g = G[ Q[i] + 1] - G[ P[i] ]; // num of 'g' between P and Q

if(a > 0){ // there is 'a'
result[i] = 1;
}
...
}  
5.
return result;