966. Vowel Spellchecker

966. Vowel Spellchecker

难度: Medium

刷题内容

原题连接

内容描述

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"
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)
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"]


Note:

1 <= wordlist.length <= 5000
1 <= queries.length <= 5000
1 <= wordlist[i].length <= 7
1 <= queries[i].length <= 7
All strings in wordlist and queries consist only of english letters.

解题方案

思路 1
*- 时间复杂度: O(len(wordlist) * len(queries) * 5^len(query))- 空间复杂度: O(len(wordlist))*

开始对于一个query的vowel变化还写了个dfs求所有的转变可能性,明摆着超时

```
class Solution:
def spellchecker(self, wordlist, queries):
"""
:type wordlist: List[str]
:type queries: List[str]
:rtype: List[str]
"""
def dfs(q, idx):
res = set()
if idx >= len(q):
return res
if q[idx] not in 'AEIOUaeiou':
res |= dfs(q, idx + 1)
return res
for c in 'aeiou':
tmp = q[:idx] + c + q[idx + 1:]
res.add(tmp.lower())
res |= dfs(tmp, idx + 1)
return res

    lookup = set(wordlist)
    lower_lookup = collections.defaultdict(list)
    for idx, word in enumerate(wordlist):
        lower_lookup[word.lower()].append(idx)

    def helper(w, q):
        if q in lookup:
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