Group Anagrams II in Java with O(n * L * log L) Time Complexity

Given an array of strings, group the anagrams together. You can return the answer in any order.

An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once.

Example:

Input:   strings = ["eat","tea","tan","ate","nat","bat"]

Output:  [["bat"],["nat","tan"],["ate","eat","tea"]]

Note:

Your algorithm should run in O(n * L * log L) time and use O(n * L) extra space.

Understanding the Problem

The core challenge of this problem is to identify and group words that are anagrams of each other. An anagram is a rearrangement of the letters of a word to form another word. For example, "eat" and "tea" are anagrams because they contain the same letters.

This problem has practical applications in text analysis, cryptography, and data clustering.

Potential pitfalls include not correctly identifying anagrams or not efficiently grouping them.

Approach

To solve this problem, we need to group words that are anagrams. A naive solution might involve comparing each word with every other word, but this would be inefficient with a time complexity of O(n^2 * L). Instead, we can use sorting and a hash map to achieve a more optimal solution.

Naive Solution

The naive approach involves comparing each word with every other word to check if they are anagrams. This is not optimal due to its high time complexity.

Optimized Solution

We can use a hash map to group anagrams efficiently. The key idea is to sort each word and use the sorted word as a key in the hash map. All anagrams will have the same sorted key.

Steps:

Initialize an empty hash map.
For each word in the input array:
- Sort the word.
- Use the sorted word as a key in the hash map.
- Append the original word to the list of words for that key.
Return the values of the hash map as the result.

Algorithm

Here is a step-by-step breakdown of the algorithm:

Create a hash map where the key is the sorted word and the value is a list of anagrams.
Iterate through each word in the input array.
Sort the word and use it as the key in the hash map.
Add the original word to the list corresponding to the sorted key.
After processing all words, return the values of the hash map.

Code Implementation

import java.util.*;

public class GroupAnagrams {
    public List<List<String>> groupAnagrams(String[] strs) {
        // Create a hash map to store the grouped anagrams
        Map<String, List<String>> map = new HashMap<>();
        
        // Iterate through each word in the input array
        for (String str : strs) {
            // Convert the word to a character array and sort it
            char[] charArray = str.toCharArray();
            Arrays.sort(charArray);
            // Convert the sorted character array back to a string
            String sortedStr = new String(charArray);
            
            // If the sorted string is not in the map, add it with an empty list
            if (!map.containsKey(sortedStr)) {
                map.put(sortedStr, new ArrayList<>());
            }
            // Add the original word to the list corresponding to the sorted string
            map.get(sortedStr).add(str);
        }
        
        // Return the values of the map as the result
        return new ArrayList<>(map.values());
    }
    
    public static void main(String[] args) {
        GroupAnagrams ga = new GroupAnagrams();
        String[] input = {"eat", "tea", "tan", "ate", "nat", "bat"};
        System.out.println(ga.groupAnagrams(input));
    }
}

Complexity Analysis

The time complexity of this solution is O(n * L * log L), where n is the number of words and L is the maximum length of a word. This is because sorting each word takes O(L * log L) time, and we do this for each of the n words.

The space complexity is O(n * L) because we store each word in the hash map.

Edge Cases

Potential edge cases include:

Empty input array: The output should be an empty list.
Single word: The output should be a list containing a single list with that word.
All words are anagrams: All words should be grouped together.

These edge cases are handled by the algorithm as it processes each word independently and groups them based on their sorted form.

Testing

To test the solution comprehensively, consider the following test cases:

Simple cases with a few words.
Cases with no anagrams.
Cases with all words being anagrams.
Edge cases like empty input or single word input.

JUnit or any other testing framework can be used to automate these tests.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Understand the problem requirements and constraints.
Think about different ways to represent the problem (e.g., sorting, hash maps).
Start with a simple solution and then optimize it.
Consider edge cases and how your solution handles them.

Practice solving similar problems to improve your problem-solving skills.

Conclusion

Grouping anagrams is a common problem that can be efficiently solved using sorting and hash maps. Understanding the problem, considering different approaches, and optimizing the solution are key steps in solving such problems.

Practice and exploration of similar problems will help in mastering these concepts.

Additional Resources

For further reading and practice, consider the following resources: