This is the final challenge of JetBrains Quest.
The second round of challenges began the day after the first challenge. Let’s finish it.
In the v2ex forum, I saw someone mention that JetBrains launched a challenge on Twitter and I’m very interested in it.
The following content records the problem-solving process in the past few days. There are three rounds. The difficulty is not high and it is very interesting.
In the previous article, I mentioned that when doing fuzzy matching, Levenshtein distance was selected as the reference value. At the time, I directly used the Python library to implement it, which did not run fast in actual use.
In recent days, I have stayed at home and took the time to write several implementations of Levenshtein Distance in Kotlin.
The question is as follows: There is an encrypted string
s[1....n], and there is a codebook that records the original-ciphertext data of the key-value type. If the time of the table is Invariant, the result of the lookup table is treated as
w is a string, if it matches, it returns
f(w), if it does not match, it returns
null, try dynamic programming to determine the string. Whether
s[1...n] can be decrypted.