Leetcode - Camelcase Matching Solution
1 min read

Leetcode - Camelcase Matching Solution

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.

Solution in Python

class Solution:
    def camelMatch(self, queries: List[str], pattern: str) -> List[bool]:
        for query in queries:
            pattern_stack = deque(pattern)
            query_stack = deque(query)
            while pattern_stack and query_stack:
                if pattern_stack[0] == query_stack[0]:
                    pattern_stack.popleft()
                    query_stack.popleft()
                elif query_stack[0].islower():
                    query_stack.popleft()
                else:
                    yield False
                    break
            else:
                if pattern_stack or any((i.isupper() for i in query_stack)):
                    yield False
                else:
                    yield True

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