966. Vowel Spellchecker
Description
Given a wordlist
, we want to implement a spellchecker that converts a query word into a correct word.
For a given query
word, the spell checker handles two categories of spelling mistakes:
- Capitalization: If the query matches a word in the wordlist (case-insensitive), then the query word is returned with the same case as the case in the wordlist.
- Example:
wordlist = ["yellow"]
,query = "YellOw"
:correct = "yellow"
- Example:
wordlist = ["Yellow"]
,query = "yellow"
:correct = "Yellow"
- Example:
wordlist = ["yellow"]
,query = "yellow"
:correct = "yellow"
- Example:
- Vowel Errors: If after replacing the vowels
('a', 'e', 'i', 'o', 'u')
of the query word with any vowel individually, it matches a word in the wordlist (case-insensitive), then the query word is returned with the same case as the match in the wordlist.- Example:
wordlist = ["YellOw"]
,query = "yollow"
:correct = "YellOw"
- Example:
wordlist = ["YellOw"]
,query = "yeellow"
:correct = ""
(no match) - Example:
wordlist = ["YellOw"]
,query = "yllw"
:correct = ""
(no match)
- Example:
In addition, the spell checker operates under the following precedence rules:
- When the query exactly matches a word in the wordlist (case-sensitive), you should return the same word back.
- When the query matches a word up to capitlization, you should return the first such match in the wordlist.
- When the query matches a word up to vowel errors, you should return the first such match in the wordlist.
- If the query has no matches in the wordlist, you should return the empty string.
Given some queries
, return a list of words answer
, where answer[i]
is the correct word for query = queries[i]
.
Example 1:
Input: wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"] Output: ["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"]
Example 2:
Input: wordlist = ["yellow"], queries = ["YellOw"] Output: ["yellow"]
Constraints:
1 <= wordlist.length, queries.length <= 5000
1 <= wordlist[i].length, queries[i].length <= 7
wordlist[i]
andqueries[i]
consist only of only English letters.
Solutions
Solution 1: Hash Table
We traverse the $\textit{wordlist}$ and store the words in hash tables $\textit{low}$ and $\textit{pat}$ according to case-insensitive and vowel-insensitive rules, respectively. The key of $\textit{low}$ is the lowercase form of the word, and the key of $\textit{pat}$ is the string obtained by replacing the vowels of the word with *
, with the value being the word itself. We use the hash table $\textit{s}$ to store the words in $\textit{wordlist}$.
We traverse $\textit{queries}$, for each word $\textit{q}$, if $\textit{q}$ is in $\textit{s}$, it means $\textit{q}$ is in $\textit{wordlist}$, and we directly add $\textit{q}$ to the answer array $\textit{ans}$; otherwise, if the lowercase form of $\textit{q}$ is in $\textit{low}$, it means $\textit{q}$ is in $\textit{wordlist}$ and is case-insensitive, and we add $\textit{low}[q.\text{lower}()]$ to the answer array $\textit{ans}$; otherwise, if the string obtained by replacing the vowels of $\textit{q}$ with *
is in $\textit{pat}$, it means $\textit{q}$ is in $\textit{wordlist}$ and is vowel-insensitive, and we add $\textit{pat}[f(q)]$ to the answer array $\textit{ans}$; otherwise, it means $\textit{q}$ is not in $\textit{wordlist}$, and we add an empty string to the answer array $\textit{ans}$.
Finally, we return the answer array $\textit{ans}$.
The time complexity is $O(n + m)$, and the space complexity is $O(n)$, where $n$ and $m$ are the lengths of $\textit{wordlist}$ and $\textit{queries}$, respectively.
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