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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] = 6The 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] = 6the 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.
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int[] solution(String S, int[] P, int[] Q) {
// write your code in Java SE 8
int[] res = new int[P.length];
int strLength = S.length();
if(P.length != Q.length)
return res;
int[] A = new int[strLength];
int[] C = new int[strLength];
int[] G = new int[strLength];
int[] T = new int[strLength];
switch(S.charAt(0)) {
case ‘A‘:
A[0] = 1;
break;
case ‘C‘:
C[0] = 2;
break;
case ‘G‘:
G[0] = 3;
break;
case ‘T‘:
T[0] = 4;
break;
default:
break;
}
for(int i=1; i<strLength;i++) {
switch(S.charAt(i)) {
case ‘A‘:
A[i] = A[i-1]+1;
C[i] = C[i-1];
G[i] = G[i-1];
T[i] = T[i-1];
break;
case ‘C‘:
A[i] = A[i-1];
C[i] = C[i-1] + 2;
G[i] = G[i-1];
T[i] = T[i-1];
break;
case ‘G‘:
A[i] = A[i-1];
C[i] = C[i-1];
G[i] = G[i-1] + 3;
T[i] = T[i-1];
break;
case ‘T‘:
A[i] = A[i-1];
C[i] = C[i-1];
G[i] = G[i-1];
T[i] = T[i-1] + 4;
break;
default:
break;
}
}
for(int i=0; i<P.length; i++) {
if(P[i] == 0) {
if(A[Q[i]] > 0)
res[i] =1;
else if(C[Q[i]] > 0)
res[i] =2;
else if(G[Q[i]] > 0)
res[i] =3;
else if(T[Q[i]] > 0)
res[i] =4;
}else {
if(A[Q[i]] - A[P[i]-1] > 0)
res[i] = 1;
else if(C[Q[i]] - C[P[i]-1] > 0)
res[i] = 2;
else if(G[Q[i]] - G[P[i]-1] > 0)
res[i] = 3;
else if(T[Q[i]] - T[P[i]-1] > 0)
res[i] = 4;
}
}