Algorithm


Problem Name: 966. Vowel Spellchecker

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"]

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.

Code Examples

#1 Code Example with Java Programming

Code - Java Programming


class Solution {
  public String[] spellchecker(String[] wordlist, String[] queries) {
    Map capsMap = new HashMap<>();
    Map < String, String> vowMap = new HashMap<>();
    Set perfect = new HashSet<>();
    for (String word : wordlist) {
      perfect.add(word);
      capsMap.putIfAbsent(word.toLowerCase(), word);
      vowMap.putIfAbsent(getVowelKey(word.toLowerCase()), word);
    }
    String[] ans = new String[queries.length];
    for (int i = 0; i  <  queries.length; i++) {
      if (perfect.contains(queries[i])) {
        ans[i] = queries[i];
      }
      else if (capsMap.containsKey(queries[i].toLowerCase())) {
        ans[i] = capsMap.get(queries[i].toLowerCase());
      }
      else {
        ans[i] = vowMap.getOrDefault(getVowelKey(queries[i].toLowerCase()), "");
      }
    }
    return ans;
  }
  
  private String getVowelKey(String word) {
    StringBuilder sb = new StringBuilder();
    for (char c : word.toCharArray()) {
      if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
        sb.append('*');
      }
      else {
        sb.append(c);
      }
    }
    return sb.toString();
  }
}
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Input

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cmd
wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"]

Output

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["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"]

#2 Code Example with Python Programming

Code - Python Programming


class Solution:
    def spellchecker(self, wordlist, queries):
        """
        :type wordlist: List[str]
        :type queries: List[str]
        :rtype: List[str]
        """
        st, cap, vow = set(wordlist), {}, {}
        for w in wordlist:
            newC = w.lower()
            newW = "".join(c if c not in "aeiou" else "*" for c in newC)
            if newC not in cap:
                cap[newC] = w
            if newW not in vow:
                vow[newW] = w
        for i, w in enumerate(queries):
            if w in st:
                pass
            elif w.lower() in cap:
                queries[i] = cap[w.lower()]
            else:
                new = "".join(c if c not in "aeiou" else "*" for c in w.lower())
                if new in vow:
                    queries[i] = vow[new]
                else:
                    queries[i] = ""
        return queries
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Input

x
+
cmd
wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"]

Output

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["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"]
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