What is the most efficient algorithm when the problem gives you a string of N length and asks you to answer in Q queries if the ith word of length M (where M is much lesser than N -> M << N) is contained into 'N length word' ?
Thanks in advance !
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3839 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3612 |
7 | Geothermal | 3569 |
8 | ecnerwala | 3494 |
9 | Um_nik | 3396 |
10 | gamegame | 3386 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | Um_nik | 164 |
2 | -is-this-fft- | 162 |
3 | maomao90 | 159 |
3 | atcoder_official | 159 |
5 | cry | 158 |
5 | awoo | 158 |
7 | adamant | 155 |
8 | nor | 154 |
9 | TheScrasse | 153 |
10 | Dominater069 | 152 |
What is the most efficient algorithm when the problem gives you a string of N length and asks you to answer in Q queries if the ith word of length M (where M is much lesser than N -> M << N) is contained into 'N length word' ?
Thanks in advance !
Название |
---|
Well, if M is really small, you can compute hash for all possible words of size <= M inside the string N. Then, just compute the hash for the ith word and check if the same hash was found inside the string N.
Edit: this would be O(n*m + q*m)
Could anyone tell me if that will work?
You can compute the suffix array for the string of length N and then answer each query in O(M * logN), making the algorithm O(Q * M * logN). If M is small, it should run in time.
I just know a MlogN algorithm. How I will get that in O(M) ?
Yes, you're right. I fixed the typo. I'd need to know the actual constraints, but I guess this solution should be fast enough.
I think you can get O(M) per query using suffix automaton.
Yep. Just build a suffix automaton on the string of length N and after that for each query run a dfs from the start node of the automaton. If you can do all M transitions between the automaton states then the small string is contained in the big one. The time complexity is O(M)*O(Q)=O(M*Q).
I think the most efficient algorithm for this kind of problems is Aho–Corasick