### Hepic_Antony_Skarlatos's blog

By Hepic_Antony_Skarlatos, history, 4 years ago, ,

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' ?

• +1

 » 4 years ago, # | ← Rev. 2 →   0 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)
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Could anyone tell me if that will work?
 » 4 years ago, # | ← Rev. 3 →   +2 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.
•  » » 4 years ago, # ^ |   +5 I just know a MlogN algorithm. How I will get that in O(M) ?
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 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.
•  » » » 4 years ago, # ^ |   0 I think you can get O(M) per query using suffix automaton.
•  » » » » 4 years ago, # ^ |   +3 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).
 » 4 years ago, # |   +25 I think the most efficient algorithm for this kind of problems is Aho–Corasick