Colidity--GenomicRangeQuery
2014-09-15 21:58
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思路:统计每一个字符前面的四个字符的个数(利用前缀和数组),这样就能在O(1)时间得到某个区间某个字符的个数
开始想到的是O(n^2)的空间,这个思路比较好
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:
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 Awhose impact factor is 1, so the answer is 1.
Write a function:
vector<int> solution(string &S, vector<int> &P, vector<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:
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).
开始想到的是O(n^2)的空间,这个思路比较好
// you can use includes, for example: // #include <algorithm> // you can write to stdout for debugging purposes, e.g. // cout << "this is a debug message" << endl; vector<int> solution(string &S, vector<int> &P, vector<int> &Q) { // write your code in C++11 int len = S.length(); vector<int> res; vector<vector<int> > arr(len); for(int k = 0 ; k < len ; ++k) { arr[k].resize(4); } int i,j; for(i = 0; i < len; i++) { char c = S[i]; if(c == 'A') arr[i][0] = 1; if(c == 'C') arr[i][1] = 1; if(c == 'G') arr[i][2] = 1; if(c == 'T') arr[i][3] = 1; } for(i = 1 ; i < len ; ++i) { for(j = 0 ; j < 4 ; ++j) { arr[i][j] += arr[i-1][j]; } } for(i = 0 ; i < P.size() ; ++i) { int x = P[i]; int y = Q[i]; for(j = 0 ; j < 4 ; ++j) { int sub = 0; if(x-1>=0) { sub = arr[x-1][j]; } if(arr[y][j] - sub > 0) { res.push_back(j+1); break; } } } return res; }
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 Awhose impact factor is 1, so the answer is 1.
Write a function:
vector<int> solution(string &S, vector<int> &P, vector<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).
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