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Published: Mar 23, 2021 License: MIT Imports: 0 Imported by: 0

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1023. Camelcase Matching (Medium)

A query word matches a given pattern if we can insert lowercase letters to the pattern word so that it equals the query. (We may insert each character at any position, and may insert 0 characters.)

Given a list of queries, and a pattern, return an answer list of booleans, where answer[i] is true if and only if queries[i] matches the pattern.

 

Example 1:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
Output: [true,false,true,true,false]
Explanation: 
"FooBar" can be generated like this "F" + "oo" + "B" + "ar".
"FootBall" can be generated like this "F" + "oot" + "B" + "all".
"FrameBuffer" can be generated like this "F" + "rame" + "B" + "uffer".

Example 2:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
Output: [true,false,true,false,false]
Explanation: 
"FooBar" can be generated like this "Fo" + "o" + "Ba" + "r".
"FootBall" can be generated like this "Fo" + "ot" + "Ba" + "ll".

Example 3:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
Output: [false,true,false,false,false]
Explanation: 
"FooBarTest" can be generated like this "Fo" + "o" + "Ba" + "r" + "T" + "est".

 

Note:

  1. 1 <= queries.length <= 100
  2. 1 <= queries[i].length <= 100
  3. 1 <= pattern.length <= 100
  4. All strings consists only of lower and upper case English letters.

[Trie] [String]

Hints

Hint 1 Given a single pattern and word, how can we solve it?
Hint 2 One way to do it is using a DP (pos1, pos2) where pos1 is a pointer to the word and pos2 to the pattern and returns true if we can match the pattern with the given word.
Hint 3 We have two scenarios: The first one is when `word[pos1] == pattern[pos2]`, then the transition will be just DP(pos1 + 1, pos2 + 1). The second scenario is when `word[pos1]` is lowercase then we can add this character to the pattern so that the transition is just DP(pos1 + 1, pos2) The case base is `if (pos1 == n && pos2 == m) return true;` Where n and m are the sizes of the strings word and pattern respectively.

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