1023. Camelcase Matching

Problem Link

Description

---

Given an array of strings queries and a string pattern, return a boolean array answer where answer[i] is true if queries[i] matches pattern, and false otherwise.

A query word queries[i] matches pattern if you can insert lowercase English letters into the pattern so that it equals the query. You may insert a character at any position in pattern or you may choose not to insert any characters at all.

 

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".

 

Constraints:

• 1 <= pattern.length, queries.length <= 100
• 1 <= queries[i].length <= 100
• queries[i] and pattern consist of English letters.

Solution

---

Python3

class Solution:
    def camelMatch(self, queries: List[str], pattern: str) -> List[bool]:
        m, n = len(queries), len(pattern)
        res = [False] * m
        
        for i, words in enumerate(queries):
            index = 0
            valid = True
            
            for word in words:
                if ord("A") <= ord(word) <= ord("Z") and (index == n or word != pattern[index]):
                    valid = False
                    break
                    
                if index < n and word == pattern[index]:
                    index += 1
            
            res[i] = valid and index == n
        
        return res