leetcode 288: Unique Word Abbreviation

Unique Word Abbreviation

Total Accepted: 351 Total Submissions: 2106 Difficulty: Easy

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

[CODE]

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

    public ValidWordAbbr(String[] dictionary) {
        for(int i=0; i<dictionary.length; i++) {
            String s = dictionary[i];
            if(s.length() > 2 ) {
                s = s.charAt(0) + Integer.toString(s.length()-2) + s.charAt(s.length()-1);
            }
            if(map.containsKey(s) ) {
                map.get(s).add(dictionary[i]);
            } else {
                Set<String> set = new HashSet<String>();
                set.add(dictionary[i]);
                map.put(s, set);
            }
        }
    }

    public boolean isUnique(String word) {
        //input check
        String s = word;
        if(s.length() > 2 ) {
            s = s.charAt(0) + Integer.toString(s.length()-2) + s.charAt(s.length()-1);
        }
        if(!map.containsKey(s)) return true;
        else return map.get(s).contains(word) && map.get(s).size()<=1;
        
    }
}


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