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(核苷).
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.
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.
思路：
开四个Prefix Sums数组分别用来统计ACGT从m到n的个数，如果A个数为0就看C，如此类推。
如果不是事先知道这题应该用Prefix Sums，可能没那么容易想到。
代码：
1 vector<int> solution(string &S, vector<int> &P, vector<int> &Q) { 2 int n = S.length(); 3 vector<vector<int> > vACGT(4, vector<int>(1,0)); 4 int count[4] = {0,0,0,0}; 5 for(int i = 0; i < n; i++){ 6 switch(S[i]){ 7 case 'A': 8 count[0] += 1; 9 break; 10 case 'C': 11 count[1] += 1; 12 break; 13 case 'G': 14 count[2] += 1; 15 break; 16 case 'T': 17 count[3] += 1; 18 break; 19 } 20 for(int k = 0; k < 4; k++){ 21 vACGT[k].push_back(count[k]); 22 } 23 } 24 25 vector<int> vres; 26 for(int i = 0; i < P.size(); i++){ 27 for(int k = 0; k < 4; k++){ 28 if(vACGT[k][Q[i]+1]-vACGT[k][P[i]] > 0){ 29 vres.push_back(k+1); 30 break; 31 } 32 } 33 } 34 return vres; 35 }