288 Unique Word Abbreviation
Problem:
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
Solutions:
public class ValidWordAbbr {
HashMap<String, Integer> app = new HashMap<String, Integer>();
HashSet<String> cand = new HashSet();
public ValidWordAbbr(String[] dictionary) {
for (int i = 0; i < dictionary.length; i ++) {
if (cand.contains(dictionary[i])) {
continue;
}
String s = encode(dictionary[i]);
if (!app.containsKey(s)) {
app.put(s, 1);
}
else {
app.put(s, app.get(s) + 1);
}
cand.add(dictionary[i]);
}
}
private String encode(String s) {
if (s.length() <= 2) {
return s;
}
String result = s.charAt(0) + "" + (s.length() - 2) + "" + s.charAt(s.length() - 1);
return result;
}
public boolean isUnique(String word) {
String s = encode(word);
if (app.containsKey(s)) {
if (!cand.contains(word)) {
return false;
}
return app.get(s) == 1;
}
else {
if (!cand.contains(word)) {
return true;
}
return false;
}
}
}
public class ValidWordAbbr {
HashMap<String, String> app = new HashMap<String, String>();
public ValidWordAbbr(String[] dictionary) {
for (int i = 0; i < dictionary.length; i ++) {
String s = encode(dictionary[i]);
if (app.containsKey(s) && !app.get(s).equals(dictionary[i])) {
app.put(s, "");
}
else {
app.put(s, dictionary[i]);
}
}
}
private String encode(String s) {
if (s.length() <= 2) {
return s;
}
String result = s.charAt(0) + "" + (s.length() - 2) + "" + s.charAt(s.length() - 1);
return result;
}
public boolean isUnique(String word) {
String s = encode(word);
if (app.containsKey(s)) {
return app.get(s).equals(word);
}
return true;
}
}