A prefix of a string S is any leading contiguous part of S. For example, "c" and "cod" are prefixes of the string "codility". For simplicity, we require prefixes to be non-empty.
The product of prefix P of string S is the number of occurrences of P multiplied by the length of P. More precisely, if prefix P consists of K characters and P occurs exactly T times in S, then the product equals K * T.
For example, S = "abababa" has the following prefixes:
- "a", whose product equals 1 * 4 = 4,
- "ab", whose product equals 2 * 3 = 6,
- "aba", whose product equals 3 * 3 = 9,
- "abab", whose product equals 4 * 2 = 8,
- "ababa", whose product equals 5 * 2 = 10,
- "ababab", whose product equals 6 * 1 = 6,
- "abababa", whose product equals 7 * 1 = 7.
The longest prefix is identical to the original string. The goal is to choose such a prefix as maximizes the value of the product. In above example the maximal product is 10.
In this problem we consider only strings that consist of lower-case English letters (a−z).
Write a function
int solution(string &S);
that, given a string S consisting of N characters, returns the maximal product of any prefix of the given string. If the product is greater than 1,000,000,000 the function should return 1,000,000,000.
For example, for a string:
- S = "abababa" the function should return 10, as explained above,
- S = "aaa" the function should return 4, as the product of the prefix "aa" is maximal.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..300,000];
- string S is made only of lowercase letters (a−z).
#include <string>
#include <vector>
using namespace std;
int solution(string &S) { // I couldn't make it O(N)
vector<size_t> repetitions(S.size(), 0);
size_t max_rep = 0;
for(size_t i = 0, imax = S.size(); i < imax; ++i) {
for(size_t j = 0, jmax = imax - i; j < jmax && S[i + j] == S[j]; ++j) {
if(repetitions[j] < 1000000000) {
repetitions[j] += j + 1;
}
if(repetitions[j] > 1000000000) {
repetitions[j] = 1000000000;
}
if(repetitions[j] > max_rep) {
max_rep = repetitions[j];
}
}
}
return max_rep;
}
#include <string>
#include <vector>
using namespace std;
int solution(string &S) { // I couldn't make it O(N)
vector<size_t> repetitions(S.size(), 0);
size_t max_rep = 0;
for(size_t i = 0, imax = S.size(); i < imax; ++i) {
for(size_t j = 0, jmax = imax - i; j < jmax && S[i + j] == S[j]; ++j) {
repetitions[j] += j + 1;
if(repetitions[j] >= 1000000000) {
return 1000000000;
}
if(repetitions[j] > max_rep) {
max_rep = repetitions[j];
}
}
}
return max_rep;
}
#include <string>
#include <vector>
using namespace std;
int solution(string &S) { // I couldn't make it O(N)
vector<size_t> repetitions(S.size(), 0);
size_t max_rep = 0;
for(size_t i = 0, imax = S.size(); i < imax; ++i) {
for(size_t j = 0, jmax = imax - i; j < jmax && S[i + j] == S[j]; ++j) {
repetitions[j] += j + 1;
if(repetitions[j] >= 1000000000) {
return 1000000000;
}
if(repetitions[j] > max_rep) {
max_rep = repetitions[j];
}
}
}
return max_rep;
}
#include <string>
#include <vector>
using namespace std;
int solution(string &S) { // I couldn't make it O(N)
vector<size_t> repetitions(S.size(), 0);
size_t max_rep = 0;
for(size_t i = 0, imax = S.size(); i < imax; ++i) {
for(size_t j = 0, jmax = imax - i; j < jmax && S[i + j] == S[j]; ++j) {
repetitions[j] += j + 1;
if(repetitions[j] >= 1000000000) {
return 1000000000;
}
if(repetitions[j] > max_rep) {
max_rep = repetitions[j];
}
}
}
return max_rep;
}
The following issues have been detected: timeout errors.
running time: >0.40 sec., time limit: 0.10 sec.
running time: >1.02 sec., time limit: 0.16 sec.