Question

966. Vowel Spellchecker

中文文档

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"
• 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.

Solutions

Solution 1

class Solution:
  def spellchecker(self, wordlist: List[str], queries: List[str]) -> List[str]:
    def f(w):
      t = []
      for c in w:
        t.append("*" if c in "aeiou" else c)
      return "".join(t)

    s = set(wordlist)
    low, pat = {}, {}
    for w in wordlist:
      t = w.lower()
      low.setdefault(t, w)
      pat.setdefault(f(t), w)

    ans = []
    for q in queries:
      if q in s:
        ans.append(q)
        continue
      q = q.lower()
      if q in low:
        ans.append(low[q])
        continue
      q = f(q)
      if q in pat:
        ans.append(pat[q])
        continue
      ans.append("")
    return ans

class Solution {
  public String[] spellchecker(String[] wordlist, String[] queries) {
    Set<String> s = new HashSet<>();
    Map<String, String> low = new HashMap<>();
    Map<String, String> pat = new HashMap<>();
    for (String w : wordlist) {
      s.add(w);
      String t = w.toLowerCase();
      low.putIfAbsent(t, w);
      pat.putIfAbsent(f(t), w);
    }
    int m = queries.length;
    String[] ans = new String[m];
    for (int i = 0; i < m; ++i) {
      String q = queries[i];
      if (s.contains(q)) {
        ans[i] = q;
        continue;
      }
      q = q.toLowerCase();
      if (low.containsKey(q)) {
        ans[i] = low.get(q);
        continue;
      }
      q = f(q);
      if (pat.containsKey(q)) {
        ans[i] = pat.get(q);
        continue;
      }
      ans[i] = "";
    }
    return ans;
  }

  private String f(String w) {
    char[] cs = w.toCharArray();
    for (int i = 0; i < cs.length; ++i) {
      char c = cs[i];
      if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
        cs[i] = '*';
      }
    }
    return String.valueOf(cs);
  }
}

class Solution {
public:
  vector<string> spellchecker(vector<string>& wordlist, vector<string>& queries) {
    unordered_set<string> s(wordlist.begin(), wordlist.end());
    unordered_map<string, string> low;
    unordered_map<string, string> pat;
    auto f = [](string& w) {
      string res;
      for (char& c : w) {
        if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
          res += '*';
        } else {
          res += c;
        }
      }
      return res;
    };
    for (auto& w : wordlist) {
      string t = w;
      transform(t.begin(), t.end(), t.begin(), ::tolower);
      if (!low.count(t)) {
        low[t] = w;
      }
      t = f(t);
      if (!pat.count(t)) {
        pat[t] = w;
      }
    }
    vector<string> ans;
    for (auto& q : queries) {
      if (s.count(q)) {
        ans.emplace_back(q);
        continue;
      }
      transform(q.begin(), q.end(), q.begin(), ::tolower);
      if (low.count(q)) {
        ans.emplace_back(low[q]);
        continue;
      }
      q = f(q);
      if (pat.count(q)) {
        ans.emplace_back(pat[q]);
        continue;
      }
      ans.emplace_back("");
    }
    return ans;
  }
};

func spellchecker(wordlist []string, queries []string) (ans []string) {
  s := map[string]bool{}
  low := map[string]string{}
  pat := map[string]string{}
  f := func(w string) string {
    res := []byte(w)
    for i := range res {
      if res[i] == 'a' || res[i] == 'e' || res[i] == 'i' || res[i] == 'o' || res[i] == 'u' {
        res[i] = '*'
      }
    }
    return string(res)
  }
  for _, w := range wordlist {
    s[w] = true
    t := strings.ToLower(w)
    if _, ok := low[t]; !ok {
      low[t] = w
    }
    if _, ok := pat[f(t)]; !ok {
      pat[f(t)] = w
    }
  }
  for _, q := range queries {
    if s[q] {
      ans = append(ans, q)
      continue
    }
    q = strings.ToLower(q)
    if s, ok := low[q]; ok {
      ans = append(ans, s)
      continue
    }
    q = f(q)
    if s, ok := pat[q]; ok {
      ans = append(ans, s)
      continue
    }
    ans = append(ans, "")
  }
  return
}