Word Abbreviation

Problem Statement

The goal is to generate the minimal possible abbreviations for an array of unique strings. Each string must be uniquely identifiable through its abbreviation. The process of abbreviation encompasses two main rules:

1. Begin with a format where each word is abbreviated to its first letter, a count of how many characters are in between the first and the last letter, and its last letter.
2. In cases where multiple words resolve to the same abbreviated form, incrementally increase the part of the abbreviation that lists actual letters from the word, from the beginning, until all abbreviations are unique.

This continues until any word that cannot be shortened appropriately without losing its unique identity retains its original form. This problem highlights the challenge of succinctly representing each string while maintaining distinguishability among all others.

Examples

Example 1

Input:

words = ["like","god","internal","me","internet","interval","intension","face","intrusion"]

Output:

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

Example 2

Input:

words = ["aa","aaa"]

Output:

["aa","aaa"]

Constraints

• 1 <= words.length <= 400
• 2 <= words[i].length <= 400
• words[i] consists of lowercase English letters.
• All the strings of words are unique.

Approach and Intuition

The task involves determining minimal abbreviations for a list of words ensuring each abbreviation is unique. Here’s a step-by-step breakdown based on the examples provided:

1. Initialize Basic Abbreviations:

   • For each word, compute its initial abbreviation as the first character, the count of intermediate characters, and the last character. For example, "internal" becomes "i6l".

2. Handle Conflicts:

   • When two or more words yield the same abbreviation, start increasing the number of characters taken from the beginning of each word to make distinct abbreviations.
   • Iteratively adjust conflicting abbreviations by including more characters from the start until all abbreviations are unique. e.g., from "a4f" to "abc2f" and "abn2f".

3. Maintain Word Length Check:

   • After resolving conflicts, ensure that the abbreviation isn't lengthier than the original word. If the abbreviation process offers no shortening, revert to the original word. For example, short words like "aa" or "me" remain unchanged because their abbreviations would not shorten the word.

Using the given examples:

• For "like", "god", "face", etc., where no initial conflicts in basic abbreviations exist, the process stops after step 1.
• For more complex sets like those containing "internet" and "internal", repeated applications of step 2 are necessary until a unique identifier for each word is found.

This systematic approach ensures that each word is represented as briefly as possible while maintaining clear distinctions between all others in the list.

Solutions

class Abbreviator {
    public List<String> abbreviateListOfWords(List<String> dict) {
        Map<String, List<PositionedWord>> groupedWords = new HashMap<>();
        String[] abbreviations = new String[dict.size()];
    
        int pos = 0;
        for (String word : dict) {
            String shortForm = getAbbreviation(word, 0);
            if (!groupedWords.containsKey(shortForm))
                groupedWords.put(shortForm, new ArrayList<>());
            groupedWords.get(shortForm).add(new PositionedWord(word, pos));
            pos++;
        }
    
        for (List<PositionedWord> group : groupedWords.values()) {
            Trie trieRoot = new Trie();
            for (PositionedWord pw : group) {
                Trie current = trieRoot;
                for (char ch : pw.word.substring(1).toCharArray()) {
                    if (current.subNodes[ch - 'a'] == null)
                        current.subNodes[ch - 'a'] = new Trie();
                    current.frequency++;
                    current = current.subNodes[ch - 'a'];
                }
            }
    
            for (PositionedWord pw : group) {
                Trie current = trieRoot;
                int idx = 1;
                for (char ch : pw.word.substring(1).toCharArray()) {
                    if (current.frequency == 1) break;
                    current = current.subNodes[ch - 'a'];
                    idx++;
                }
                abbreviations[pw.index] = getAbbreviation(pw.word, idx - 1);
            }
        }
    
        return Arrays.asList(abbreviations);
    }
    
    private String getAbbreviation(String word, int position) {
        int len = word.length();
        if (len - position <= 3) return word;
        return word.substring(0, position + 1) + (len - position - 2) + word.charAt(len - 1);
    }
}
    
class Trie {
    Trie[] subNodes;
    int frequency;
    Trie() {
        subNodes = new Trie[26];
        frequency = 0;
    }
}
    
class PositionedWord {
    String word;
    int index;
    PositionedWord(String str, int idx) {
        word = str;
        index = idx;
    }
}