0288. Unique Word Abbreviation

https://leetcode.com/problems/unique-word-abbreviation

Description

The abbreviation of a word is a concatenation of its first letter, the number of characters between the first and last letter, and its last letter. If a word has only two characters, then it is an abbreviation of itself.

For example:

• dog --> d1g because there is one letter between the first letter 'd' and the last letter 'g'.

• internationalization --> i18n because there are 18 letters between the first letter 'i' and the last letter 'n'.

• it --> it because any word with only two characters is an abbreviation of itself.

Implement the ValidWordAbbr class:

• ValidWordAbbr(String[] dictionary) Initializes the object with a dictionary of words.

• boolean isUnique(string word) Returns true if either of the following conditions are met (otherwise returns false):

  • There is no word in dictionary whose abbreviation is equal to word's abbreviation.

  • For any word in dictionary whose abbreviation is equal to word's abbreviation, that word and word are the same.

Example 1:

**Input**
["ValidWordAbbr", "isUnique", "isUnique", "isUnique", "isUnique", "isUnique"]
[[["deer", "door", "cake", "card"]], ["dear"], ["cart"], ["cane"], ["make"], ["cake"]]
**Output**
[null, false, true, false, true, true]
**Explanation**
ValidWordAbbr validWordAbbr = new ValidWordAbbr(["deer", "door", "cake", "card"]);
validWordAbbr.isUnique("dear"); // return false, dictionary word "deer" and word "dear" have the same abbreviation "d2r" but are not the same.
validWordAbbr.isUnique("cart"); // return true, no words in the dictionary have the abbreviation "c2t".
validWordAbbr.isUnique("cane"); // return false, dictionary word "cake" and word "cane" have the same abbreviation  "c2e" but are not the same.
validWordAbbr.isUnique("make"); // return true, no words in the dictionary have the abbreviation "m2e".
validWordAbbr.isUnique("cake"); // return true, because "cake" is already in the dictionary and no other word in the dictionary has "c2e" abbreviation.

Constraints:

• 1 <= dictionary.length <= 3 * 104

• 1 <= dictionary[i].length <= 20

• dictionary[i] consists of lowercase English letters.

• 1 <= word.length <= 20

• word consists of lowercase English letters.

• At most 5000 calls will be made to isUnique.

ac1

use 2 map 1 list, time consuming

class ValidWordAbbr {
    Map<Integer, Map<String, List<String>>> map;

    public ValidWordAbbr(String[] dictionary) {
        map = new HashMap<Integer, Map<String, List<String>>>();

        for (String s : dictionary) {
            int len = s.length();
            String abbr = len != 0 ? s.substring(0,1) + s.substring(len-1) : "";
            if (!map.containsKey(len)) map.put(len, new HashMap<String, List<String>>());
            if (!map.get(len).containsKey(abbr)) map.get(len).put(abbr, new ArrayList<String>());
            map.get(len).get(abbr).add(s);
        }
    }

    public boolean isUnique(String word) {
        // edge case
        int len = word.length();
        if (len == 0) return true;

        String abbr = word.substring(0,1) + word.substring(len-1);

        if (map.containsKey(len) && map.get(len).containsKey(abbr)) {
            for (String s : map.get(len).get(abbr)) {
                if (!word.equals(s)) return false;
            }
        }

        return true;
    }
}

/**
 * Your ValidWordAbbr object will be instantiated and called as such:
 * ValidWordAbbr obj = new ValidWordAbbr(dictionary);
 * boolean param_1 = obj.isUnique(word);
 */