An abbreviation of a word follows the form
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
Solution
public class ValidWordAbbr {
Map<String, Map<Integer, Integer>> map = new HashMap<String, Map<Integer, Integer>>();
Map<String, Integer> occur = new HashMap<String, Integer>();
public ValidWordAbbr(String[] dictionary) {
for(String s: dictionary) {
StringBuilder sb = new StringBuilder();
if(s.length() <= 1) sb.append(s);
else {sb.append(s.charAt(0)); sb.append(s.charAt(s.length()-1));}
String key = sb.toString();
if(!map.containsKey(key)) map.put(key, new HashMap<Integer, Integer>());
if(!map.get(key).containsKey(s.length())) map.get(key).put(s.length(), 0);
map.get(key).put(s.length(), map.get(key).get(s.length()) + 1);
occur.put(s, occur.getOrDefault(s, 0) + 1);
}
}
public boolean isUnique(String word) {
StringBuilder sb = new StringBuilder();
if(word.length() <= 1) sb.append(word);
else {sb.append(word.charAt(0)); sb.append(word.charAt(word.length()-1));}
String key = sb.toString();
if(!map.containsKey(key)) return true;
if(!map.get(key).containsKey(word.length())) return true;
return occur.containsKey(word) && map.get(key).get(word.length()) == occur.get(word);
}
}
// Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");