288. Unique Word Abbreviation

An abbreviation of a word follows the form . 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

The tricky part of this question is the case of duplications in the dictionary. if input dictionary is ["a", "a", "a"], and query is isUnique("a"), this should return true; because "a" is unique. another point require attention is to query the exactly word in the dictionary, this also should return true.

public class ValidWordAbbr {
    private Map<String, Set<String>> dict = new HashMap<>();

    public ValidWordAbbr(String[] dictionary) {
        for(String s : dictionary){

            String key = buildKey(s);
            if(dict.containsKey(key)){
                dict.get(key).add(s);
            }else{
                Set<String> t = new HashSet<>();
                t.add(s);
                dict.put(key, t);
            }
        }    
    }
    private String buildKey(String s){
        if(s.length() <= 2 ) return s;
        else return s.charAt(0) + Integer.toString(s.length()-2) + s.charAt(s.length()-1);
    }
    public boolean isUnique(String word) {
        String key = buildKey(word);
        if(!dict.containsKey(key)){
            return true;
        }else{
            return dict.get(key).contains(word) && dict.get(key).size() <=1;
        }
    }
}


// Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");