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 !

# | User | Rating |
---|---|---|

1 | tourist | 4009 |

2 | jiangly | 3831 |

3 | Radewoosh | 3646 |

4 | jqdai0815 | 3620 |

4 | Benq | 3620 |

6 | orzdevinwang | 3529 |

7 | ecnerwala | 3446 |

8 | Um_nik | 3396 |

9 | gamegame | 3386 |

10 | ksun48 | 3373 |

# | User | Contrib. |
---|---|---|

1 | cry | 164 |

1 | maomao90 | 164 |

3 | Um_nik | 163 |

4 | atcoder_official | 160 |

5 | -is-this-fft- | 158 |

6 | awoo | 157 |

7 | adamant | 156 |

8 | TheScrasse | 154 |

8 | nor | 154 |

10 | Dominater069 | 153 |

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 !

↑

↓

Codeforces (c) Copyright 2010-2024 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Nov/08/2024 21:15:44 (k2).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|

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*_{log}N), making the algorithmO(Q*M*_{log}N). 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