966. Vowel Spellchecker

Problem Link

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.

  <ul>
  	<li>Example: <code>wordlist = [&quot;yellow&quot;]</code>, <code>query = &quot;YellOw&quot;</code>: <code>correct = &quot;yellow&quot;</code></li>
  	<li>Example: <code>wordlist = [&quot;Yellow&quot;]</code>, <code>query = &quot;yellow&quot;</code>: <code>correct = &quot;Yellow&quot;</code></li>
  	<li>Example: <code>wordlist = [&quot;yellow&quot;]</code>, <code>query = &quot;yellow&quot;</code>: <code>correct = &quot;yellow&quot;</code></li>
  </ul>
  </li>
  <li>Vowel Errors: If after replacing the vowels <code>(&#39;a&#39;, &#39;e&#39;, &#39;i&#39;, &#39;o&#39;, &#39;u&#39;)</code> of the query word with any vowel individually, it matches a word in the wordlist (<strong>case-insensitive</strong>), then the query word is returned with the same case as the match in the wordlist.
  <ul>
  	<li>Example: <code>wordlist = [&quot;YellOw&quot;]</code>, <code>query = &quot;yollow&quot;</code>: <code>correct = &quot;YellOw&quot;</code></li>
  	<li>Example: <code>wordlist = [&quot;YellOw&quot;]</code>, <code>query = &quot;yeellow&quot;</code>: <code>correct = &quot;&quot;</code> (no match)</li>
  	<li>Example: <code>wordlist = [&quot;YellOw&quot;]</code>, <code>query = &quot;yllw&quot;</code>: <code>correct = &quot;&quot;</code> (no match)</li>
  </ul>
  </li>
  

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] and queries[i] consist only of only English letters.

Solution

---

Python3

class Solution:
    def spellchecker(self, wordlist: List[str], queries: List[str]) -> List[str]:
        wordList = set(wordlist)
        words = {}
        shuffled = {}
        done_shuffled = set()
        res = []
        
        
        for word in wordlist:
            lower = word.lower()
            
            if lower not in words:
                words[lower] = word
            
            if lower in done_shuffled: continue
                
            # word, current processed index
            queue = deque([(word.lower(), 0)])
            
            while queue:
                currWord, index = queue.popleft()
                
                if index == len(word): continue
                
                if currWord[index] in "aeiou":
                    for vowel in "aeiou":
                        newWord = currWord[:index] + vowel + currWord[index + 1:]
                        if newWord not in shuffled:
                            shuffled[newWord] = word
                        queue.append((newWord, index + 1))
                else:
                    queue.append((currWord, index + 1))
            
            done_shuffled.add(word)
            
        for word in queries:
            if word in wordList:
                res.append(word)
                continue
            
            lower = word.lower()
            
            if lower in words:
                res.append(words[lower])
                continue
            
            if lower in shuffled:
                res.append(shuffled[lower])
                continue
            
            res.append("")
        
        return res