527 Word Abbreviation

527. Word Abbreviation

1. Question

Given an array of n distinct non-empty strings, you need to generate minimal possible abbreviations for every word following rules below.

1. Begin with the first character and then the number of characters abbreviated, which followed by the last character.

2. If there are any conflict, that is more than one words share the same abbreviation, a longer prefix is used instead of only the first character until making the map from word to abbreviation become unique. In other words, a final abbreviation cannot map to more than one original words.

3. If the abbreviation doesn't make the word shorter, then keep it as original.

Example:

Input:
["like", "god", "internal", "me", "internet", "interval", "intension", "face", "intrusion"]

Output:
["l2e","god","internal","me","i6t","interval","inte4n","f2e","intr4n"]

Note:

1. Both n and the length of each word will not exceed 400.

2. The length of each word is greater than 1.

3. The words consist of lowercase English letters only.

4. The return answers should be in the same order as the original array.

2. Implementation

(1) Trie

思路:

class Solution {
    class TrieNode {
        Word word;
        TrieNode[] childNode;

        public TrieNode() {
            childNode = new TrieNode[26];
        }
    }

    class Word {
        String word;
        int index;

        public Word(String word, int index) {
            this.word = word;
            this.index = index;
        }
    }

    public List<String> wordsAbbreviation(List<String> dict) {
        Map<String, TrieNode> map = new HashMap();
        String[] res = new String[dict.size()];

        for (int i = 0; i < dict.size(); i++) {
            String curWord = dict.get(i);
            String abbr = abbreviate(curWord, 1);

            if (!map.containsKey(abbr)) {
                TrieNode node = new TrieNode();
                node.childNode[curWord.charAt(0) - 'a'] = new TrieNode();
                node.childNode[curWord.charAt(0) - 'a'].word = new Word(curWord, i);
                map.put(abbr, node);
                res[i] = abbr;
            }
            else {
                TrieNode node = map.get(abbr);

                for (int j = 0; j < curWord.length(); j++) {
                    char c = curWord.charAt(j);

                    if (node.childNode[c - 'a'] == null) {
                        node.childNode[c - 'a'] = new TrieNode();
                        node.childNode[c - 'a'].word = new Word(curWord, i);
                        res[i] = abbreviate(curWord, j + 1);
                        break;
                    }
                    node = node.childNode[c - 'a'];

                    if (node.word != null) {
                        Word conflictWord = node.word;
                        node.word = null;
                        int k = j + 1;

                        while (conflictWord.word.charAt(k) == curWord.charAt(k)) {
                            node.childNode[curWord.charAt(k) - 'a'] = new TrieNode();
                            node = node.childNode[curWord.charAt(k) - 'a'];
                            ++k;
                        }

                        node.childNode[conflictWord.word.charAt(k) - 'a'] = new TrieNode();
                        node.childNode[conflictWord.word.charAt(k) - 'a'].word = conflictWord;
                        res[conflictWord.index] = abbreviate(conflictWord.word, k + 1);
                        node.childNode[curWord.charAt(k) - 'a'] = new TrieNode();
                        node.childNode[curWord.charAt(k) - 'a'].word = new Word(curWord, i);
                        res[i] = abbreviate(curWord, k + 1);
                        break;
                    }
                }
            }
        }
        return Arrays.asList(res);
    }

    public String abbreviate(String word, int k) {
        if (k >= word.length() - 2) {
            return word;
        }
        StringBuilder res = new StringBuilder();
        res.append(word.substring(0, k));
        res.append(word.length() - 1 - k);
        res.append(word.charAt(word.length() - 1));
        return res.toString();
    } 
}

(2) Hash Set

class Solution {
    public List<String> wordsAbbreviation(List<String> dict) {
        int len = dict.size();
        int[] prefixLen = new int[len];
        String[] ans = new String[len];

        for (int i = 0; i < len; i++) {
            prefixLen[i] = 1;
            ans[i] = abbreviate(dict.get(i), 1);
        }

        for (int i = 0; i < len - 1; i++) {
            while (true) {
                Set<Integer> set = new HashSet();
                // Put the index of word in the dictionary in the hash set if their abbreviation 
                // are the same
                for (int j = i + 1; j < len; j++) {
                    if (ans[i].equals(ans[j])) {
                        set.add(j);
                    }
                }

                if (set.isEmpty()) break;

                set.add(i);

                for (int k : set) {
                    ans[k] = abbreviate(dict.get(k), ++prefixLen[k]);
                }
            }
        }
        return Arrays.asList(ans);
    }

    public String abbreviate(String word, int k) {
        if (k >= word.length() - 2) {
            return word;
        }
        StringBuilder res = new StringBuilder();
        res.append(word.substring(0, k));
        res.append(word.length() - 1 - k);
        res.append(word.charAt(word.length() - 1));
        return res.toString();
    } 
}

3. Time & Space Complexity