
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:
- 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.
- 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:
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".
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".
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
- Java
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;
}
}
The solution provided is a Java class named Abbreviator
designed to abbreviate a list of words. The aim is to create shortened forms of the words such that they remain unique amongst each other if possible. A detailed outline of the approach and functionalities within the code are as follows:
Class
Abbreviator
:- Method
abbreviateListOfWords(List<String> dict)
: This method takes a list of words and returns their abbreviated forms as a list.- A
HashMap
namedgroupedWords
categorizes words by their initial abbreviation which is derived using helper methodgetAbbreviation(word, 0)
. - For each word, this initial abbreviation serves as a key, and the value is a list of
PositionedWord
objects containing the word and its original position. - Further, the method processes these groups to potentially shorten the abbreviations while ensuring they remain unique within their group.
- A
Trie
data structure used here helps in counting the occurrences of each character sequence, aiding in determining up to which character an abbreviation needs to be unique. - Upon identifying the minimum unique abbreviation length for each word, the final abbreviation is recalculated and stored.
- A
- Method
Helper Methods:
getAbbreviation(String word, int position)
: Constructs the abbreviation based on the specified starting position, adhering to a formula that includes the number of characters between the first and the last character of the abbreviation.
Supporting Classes:
Trie
: Facilitates the uniqueness check for the abbreviations by holding the frequency of character sequences as paths are defined through the Trie nodes (subNodes
).PositionedWord
: Utilized to associate a word with its position within the original list, helping maintain order in the final output list.
Characters and Indices:
- The solution manages character indices and mapping through use of arrays (
subNodes
in the Trie) and positions in thePositionedWord
objects.
- The solution manages character indices and mapping through use of arrays (
Output:
- Finally, the solution returns the list of unique abbreviations maintaining the order they were provided.
This Java solution effectively handles the problem of abbreviating words uniquely by grouping similar abbreviations and using a combination of Trie data structures and indices, ensuring efficient processing and accurate mapping from input to output.
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