Question

Formatted question description: https://leetcode.ca/all/288.html

 288	Unique Word Abbreviation

 An abbreviation of a word follows the form
    <first letter><number><last letter>.

 Below are some examples of word abbreviations:

 a) it                      --> it    (no abbreviation)

      1
 b) d|o|g                   --> d1g

               1    1  1
      1---5----0----5--8
 c) i|nternationalizatio|n  --> i18n

               1
      1---5----0
 d) l|ocalizatio|n          --> l10n


 Assume you have a dictionary and given a word, find whether its abbreviation is unique in the dictionary.
 A word's abbreviation is unique if no other word from the dictionary has the same abbreviation.

 Example:

 Given dictionary = [ "deer", "door", "cake", "card" ]

 isUnique("dear") -> false
 isUnique("cart") -> true
 isUnique("cane") -> false
 isUnique("make") -> true

 @tag-design

Algorithm

Code

Java

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

public class Unique_Word_Abbreviation {

    /*
        more test cases

        1. dictionary = {"dear"},  isUnique("door") -> false

        2. dictionary = {"door", "door"}, isUnique("door") -> true

        3. dictionary = {"dear", "door"}, isUnique("door") -> false

     */
    public static void main(String[] args) {
        Unique_Word_Abbreviation out = new Unique_Word_Abbreviation();
        ValidWordAbbr s = out.new ValidWordAbbr(new String[]{"deer", "door", "cake", "card"});

        System.out.println(s.isUnique("cart"));
    }

    public class ValidWordAbbr {

        Map<String, Set<String>> map;

        public boolean isUnique(String word) {
            String abbr = getAbbr(word);
            // @note: same word, and only this same word
            if (!map.containsKey(abbr) || (map.get(abbr).contains(word) && map.get(abbr).size() == 1)) {
                return true;
            }

            return false;
        }

        public ValidWordAbbr(String[] dictionary) {
            map = new HashMap<>();
            for (String s : dictionary) {
                String abbr = getAbbr(s);
                if (!map.containsKey(abbr)) {
                    map.put(abbr, new HashSet<String>());
                }
                map.get(abbr).add(s);
            }
        }

        private String getAbbr(String s) {
            if (s.length() < 3) {
                return s;
            }
            int len = s.length();

            return s.substring(0, 1) + (len - 2) + s.substring(len - 1);
        }
    }
}

```
Todo
```

class ValidWordAbbr(object):
  def __init__(self, dictionary):
    """
    initialize your data structure here.
    :type dictionary: List[str]
    """
    self.d = {}
    self.dict = dictionary = set(dictionary)
    for word in dictionary:
      wordLen = len(word)
      if wordLen > 2:
        key = word[0] + str(wordLen - 2) + word[-1]
        self.d[key] = self.d.get(key, 0) + 1
      else:
        self.d[word] = self.d.get(word, 0) + 1

  def isUnique(self, word):
    """
    check if a word is unique.
    :type word: str
    :rtype: bool
    """
    wordLen = len(word)
    key = None
    if wordLen > 2:
      key = word[0] + str(wordLen - 2) + word[-1]
    else:
      key = word
    if key in self.d:
      if self.d[key] == 1 and word in self.dict:
        return True
      return False
    else:
      return True

# Your ValidWordAbbr object will be instantiated and called as such:
# vwa = ValidWordAbbr(dictionary)
# vwa.isUnique("word")
# vwa.isUnique("anotherWord")

288 - Unique Word Abbreviation

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