Group Anagrams in Java

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:

For this lesson, your algorithm should run in O(n * log n * L * log L) time and use O(n * L) extra space.
(There are faster solutions which we will discuss in future lessons)

Problem Definition

The problem requires us to group an array of strings into anagrams. 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.

Input:

An array of strings.

Output:

A list of lists, where each sublist contains strings that are anagrams of each other.

Constraints and Assumptions:

All inputs are lowercase letters.
The length of each string is bounded by L.
The number of strings is bounded by n.

Example:

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

Understanding the Problem

The core challenge is to identify and group strings that are anagrams. Anagrams have the same characters in the same frequency but in different orders. This problem is significant in text processing and can be applied in various fields such as cryptography, linguistics, and data analysis.

Potential Pitfalls and Misconceptions:

Assuming that the order of characters in the output matters.
Not considering the efficiency of the solution.

Approach

To solve this problem, we need to identify a way to group strings that are anagrams. A naive solution would involve comparing each string with every other string, which is inefficient.

Naive Solution:

Compare each string with every other string to check if they are anagrams. This approach has a time complexity of O(n^2 * L), which is not optimal.

Optimized Solution:

A more efficient approach involves sorting the characters of each string and using the sorted string as a key in a hash map. Strings that are anagrams will have the same sorted string, thus can be grouped together.

Thought Process:

Sort each string to create a key.
Use a hash map to group strings by their sorted key.
Return the values of the hash map as the result.

Algorithm

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

Initialize a hash map to store groups of anagrams.
Iterate through each string in the input array.
Sort the characters of the string to form a key.
Check if the key exists in the hash map:
- If it exists, append the string to the corresponding list.
- If it does not exist, create a new list with the string and add it to the hash map.
Return the values of the hash map as the result.

Code Implementation

import java.util.*;

public class GroupAnagrams {
    public List<List<String>> groupAnagrams(String[] strs) {
        // HashMap to store the grouped anagrams
        Map<String, List<String>> map = new HashMap<>();
        
        // Iterate through each string in the input array
        for (String s : strs) {
            // Convert the string to a character array and sort it
            char[] charArray = s.toCharArray();
            Arrays.sort(charArray);
            // Convert the sorted character array back to a string
            String sortedString = new String(charArray);
            
            // If the sorted string is not in the map, add it with a new list
            if (!map.containsKey(sortedString)) {
                map.put(sortedString, new ArrayList<>());
            }
            // Add the original string to the list corresponding to the sorted string
            map.get(sortedString).add(s);
        }
        
        // 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 approach is O(n * log n * L * log L), where n is the number of strings and L is the maximum length of a string. Sorting each string takes O(L * log L) time, and we do this for each of the n strings. The space complexity is O(n * L) due to the storage of the hash map and the lists of strings.

Edge Cases

Potential edge cases include:

Empty input array: The output should be an empty list.
Strings with different lengths: The algorithm handles this naturally as sorting and grouping are based on character content.
All strings being anagrams of each other: The output should be a single list containing all strings.

Examples of Edge Cases:

Input:   strings = []
Output:  []

Input:   strings = ["a"]
Output:  [["a"]]

Input:   strings = ["abc", "bca", "cab"]
Output:  [["abc", "bca", "cab"]]

Testing

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

Simple cases with a few strings.
Cases with strings of varying lengths.
Edge cases such as empty input or single-character strings.

Using a testing framework like JUnit can help automate and validate these test cases.

Thinking and Problem-Solving Tips

When approaching such problems:

Break down the problem into smaller parts.
Consider different approaches and their trade-offs.
Practice similar problems to improve 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 analyzing their complexities are crucial steps in developing an optimal solution.

Additional Resources

For further reading and practice:

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