Mastering Unique Elements and Anagram Detection with Java HashSets

Mastering Unique Elements and Anagram Detection with Java HashSets | CodeSignal Learn

Introduction

Welcome to our focused exploration of Java's HashSet and its remarkable applications in solving algorithmic challenges. In this lesson, " Mastering Unique Elements and Anagram Detection with Java HashSets," we'll explore how this powerful data structure can be used to approach and solve certain types of problems commonly encountered in technical interviews.

Problem 1: Unique Echo

Picture this: you're given a vast list of words, and you must identify the final word that stands proudly solitary — the last word that is not repeated. Imagine sorting through a database of unique identifiers and finding one identifier towards the end of the list that is unlike any others.

Problem 1: Naive Approach

O(n^2)

Problem 1: Efficient Approach

O(n)

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We can use two HashSet instances: wordsSet to maintain unique words and duplicatesSet to keep track of duplicate words. By the end, we can remove all duplicated words from wordsSet to achieve our goal. Here is how to use HashSet to solve the problem:

Create a HashSet instance to store unique words:

Java

HashSet<String> wordsSet = new HashSet<>();

Initialize another HashSet to monitor duplicates:

Java

HashSet<String> duplicatesSet = new HashSet<>();

Iterate the word array, filling wordsSet and duplicatesSet:

Java

for (String word : words) {
    if (wordsSet.contains(word)) {
        duplicatesSet.add(word);
    } else {
        wordsSet.add(word);
    }
}

for (String word : words) {
    if (wordsSet.contains(word)) {
        duplicatesSet.add(word);
    } else {
        wordsSet.add(word);
    }
}

Use the removeAll method from the HashSet API to remove all duplicated words from wordsSet:

Java

wordsSet.removeAll(duplicatesSet);

wordsSet.removeAll(duplicatesSet);

Now, wordsSet only contains unique words. Find the last unique word by iterating through the original word list from the end:

Java

String lastUniqueWord = "";
for (int i = words.length - 1; i >= 0; i--) {
   if (wordsSet.contains(words[i])){
       lastUniqueWord = words[i];
       break;
   }
}

String lastUniqueWord = "";
for (int i = words.length - 1; i >= 0; i--) {
   if (wordsSet.contains(words[i])){
       lastUniqueWord = words[i];
       break;
   }
}

And finally, return the last unique word:

Java

return lastUniqueWord;

return lastUniqueWord;

This efficient approach, with a time complexity closer to $O(n)$ , is far superior to the naive method and showcases your proficiency at solving algorithmic problems with Java's HashSet.

Problem 2: Solution Building

O(m\log m)

Let's break down the anagram matcher:

Construct a method to create sorted character signatures from the input string:

Java

private static String sortCharacters(String input) {
   char[] chars = input.toCharArray();
   Arrays.sort(chars);
   return new String(chars);
}

Store these sorted characters from array2 in a HashSet for fast lookup:

Java

HashSet<String> sortedWordsInArray2 = new HashSet<>();
for (String word : array2) {
   sortedWordsInArray2.add(sortCharacters(word));
}

For each word in array1, check for its sorted signature in the HashSet and track the found anagrams:

Java

HashSet<String> anagramsMatched = new HashSet<>();
ArrayList<String> result = new ArrayList<>();
for (String word : array1) {
    if (sortedWordsInArray2.contains(sortCharacters(word))) {
        if (!anagramsMatched.contains(word)) {
            result.add(word);
            anagramsMatched.add(word);
        }
    }
}

HashSet<String> anagramsMatched = new HashSet<>();
ArrayList<String> result = new ArrayList<>();
for (String word : array1) {
    if (sortedWordsInArray2.contains(sortCharacters(word))) {
        if (!anagramsMatched.contains(word)) {
            result.add(word);
            anagramsMatched.add(word);
        }
    }
}

The ArrayList result stores the matches, ensuring that we return unique anagrams, while the HashSet anagramsMatched prevents duplication in our result.

Our final step is to return the list of anagrams found:

Java

return result;

By utilizing HashSets in this manner, we achieve efficient anagram checking with reduced complexity, considering both the $O(m\log m)$ character sorting for each word and the $O(n)$ comparison for $n$ words.