1 авг. 2024 г. · Can't understand time complexity. Can someone explain it? Some people say it's O(n * n^2), others say it's O(n^2)

19 янв. 2017 г. · If you calculate wordBreak("test") , and then calculate wordBreak("hand") , it will return the result from test because memo[4] contains the ...

5 мая 2023 г. · Time complexity worst case is O(N^M), N-string length, M number of words in the dictionary. Well, maybe I am wrong about the time complexity ...

4 окт. 2024 г. · If there is a more optimal time complexity, I'm not aware. You can technically beat the space complexity a bit by making it not recursive, but ...

22 сент. 2017 г. · To return a list of all possible paths would be O(2n ) even if you use dynamic programming I believe? For word break, it would be O(n2 ) if we' ...

Time complexity is O(n^3). The first O(n) factor is due to enumerating s . The second O(n) factor is because eow can have size up to n for test cases like s = " ...

4 июл. 2024 г. · The helper function has constant time complexity and the whole solution is O(nlogn) from the initial sorting. But the leetcode time complexity feature says it ...

6 нояб. 2020 г. · Then do word break 2 algorithm, dp + trie. Still, this seems like a pain to implement. It will be O(n2) algorithm.

6 дек. 2022 г. · There is an easy, naive solution with (time) complexity O(nm2). With some thought you can improve that to O(nm). In fact, you can find a ...

4 янв. 2024 г. · A better way would be to iterate from the end, if there are spaces in the end ignore them till you encounter the first letter, then move towards ...