Group Anagrams: Finding Order in Chaos

The Group Anagrams problem is one of those classic interview questions that looks scary at first glance but is actually pretty satisfying once you see the trick. The goal? Take a messy list of words and organize them so that words made of the same letters sit together.

It’s a fantastic problem for learning how to design keys for hash maps and how to normalize data into what we call a “canonical form.”

If you’ve ever played Scrabble or solved word jumbles, you’ve essentially run this algorithm in your head: you look at a pile of letters and realize that “eat”, “tea”, and “ate” are just different arrangements of the same building blocks.

Real-World Analogy

The Sorting Approach

The most intuitive way to handle this? Sort the letters. If you take any anagram and sort its characters alphabetically, they become identical.

• “eat” $\rightarrow$ “aet”
• “tea” $\rightarrow$ “aet”
• “ate” $\rightarrow$ “aet”

By using this sorted string as a key in a Hash Map (or Dictionary), we can group all the matching original strings together. It’s simple, clean, and easy to explain.

Interactive Visualization: Hash Map Grouping

Explore how words are processed, sorted into keys, and stored in the hash map.

Core Concepts

To crack this efficiently, we need to lean on two main ideas:

1. Canonical Representation : Think of this as a unique ID tag. For anagrams, we need a version of the word that looks exactly the same for every member of the group, but different for anyone else.

   Key Concept: A Canonical Representation transforms complex data into a standard format (like a sorted string) so we can easily compare and group things.

2. Hash Map (Dictionary): This is where we store our groups. It gives us lightning-fast lookups ($O(1)$ on average), which is exactly what we need for performance.

The Frequency Count Approach (Optimization)

Sorting is the most readable solution, but it isn’t always the fastest. It costs $O(K \log K)$ per string (where $K$ is the string length). If you’re dealing with massive strings, that math starts to hurt.

A faster (though slightly more complex) way uses Character Counts. Since anagrams have the exact same count of every character, we can use the frequency array (e.g., a:1, b:0, c:0, ..., e:1, ...) as the key.

Interactive Visualization: Character Counting

See how we can generate keys without sorting by counting character frequencies.

Implementation

Here is the implementation using the Sorting Approach. I’m sticking with this one because it’s the most universal and, honestly, the easiest one to write out on a whiteboard without bugging out. Note that in Python, the dictionary simplifies the logic significantly.

class Solution:
  def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
      # Dictionary to map the sorted string (key) to list of anagrams
      anagram_map = defaultdict(list)
      
      for s in strs:
          # Create the key by sorting the string
          # 'eat' -> ['a', 'e', 't'] -> 'aet'
          key = "".join(sorted(s))
          
          # Append the original string to the correct group
          anagram_map[key].append(s)
          
      # Return just the grouped lists (values of the map)
      return list(anagram_map.values())

class Solution {
  public List<List<String>> groupAnagrams(String[] strs) {
      if (strs == null || strs.length == 0) return new ArrayList<>();
      
      // Map to store sorted string -> list of anagrams
      Map<String, List<String>> map = new HashMap<>();
      
      for (String s : strs) {
          // Convert string to char array to sort it
          char[] ca = s.toCharArray();
          Arrays.sort(ca);
          
          // Create a string key from sorted chars
          String key = String.valueOf(ca);
          
          // If key doesn't exist, initialize new list
          if (!map.containsKey(key)) {
              map.put(key, new ArrayList<>());
          }
          
          // Add original string to the list
          map.get(key).add(s);
      }
      
      return new ArrayList<>(map.values());
  }
}

class Solution {
public:
  vector<vector<string>> groupAnagrams(vector<string>& strs) {
      // Unordered map for O(1) average lookups
      unordered_map<string, vector<string>> map;
      
      for (string s : strs) {
          // Create a temporary string to serve as key
          string key = s;
          
          // Sort the key to find canonical form
          sort(key.begin(), key.end());
          
          // Add original string to the vector at this key
          map[key].push_back(s);
      }
      
      // Collect all values into the result vector
      vector<vector<string>> result;
      for (auto& p : map) {
          // Use reference 'auto& p' to avoid copying the vector
          result.push_back(p.second);
      }
      
      return result;
  }
};

Complexity Analysis

You need to know the trade-offs here, especially if your interviewer asks, “Can we make this faster?”

Time Complexity

• $N$ is the number of strings.
• $K$ is the maximum length of a string.
• We iterate through $N$ strings, and for each one, we sort it. That sorting step takes $O(K \log K)$.
• Note: The Character Count approach brings this down to $O(N \cdot K)$, which is better for very long strings.

Space Complexity

• We have to store every string in our Hash Map.
• In the worst case (where there are no anagrams at all), we end up storing $N$ keys and $N$ lists containing the original strings. The total space used matches the total size of the input.

Common Pitfalls & Edge Cases

I’ve seen a lot of candidates trip up on these specific edge cases. Watch out for them:

1. Empty Inputs: Make sure you handle strs = [] or strs = [""]. The logic usually handles [""] naturally (the key is just an empty string), but an empty list input [] might need an early return in some languages.
2. Array Hashing (Java): In Java, arrays int[] don’t hash by their content. If you try to use the count array as a key, it’ll use the memory address instead. You have to wrap it in Arrays.toString() or a List to get it to behave.
3. Case Sensitivity: LeetCode 49 specifies lowercase English letters, but the real world is messy (“Tea” vs “eat”). In an interview, always ask if you need to normalize case (e.g., s.toLowerCase()) before generating the key.

Practical Applications

Believe it or not, this isn’t just a toy problem. This pattern—grouping by a “normalized” key—shows up all over the place:

1. Data Deduplication: Identifying duplicate records that might have fields in different orders (like JSON objects where key order doesn’t matter).
2. Fuzzy Search: Grouping search terms that are just misspellings or variations of each other.
3. Log Aggregation: Grouping error logs that are identical except for the specific timestamp or ID.

Conclusion

The “Group Anagrams” problem is really just a lesson in data transformation. By converting messy, varied data into a strict canonical form, we can use Hash Maps to organize chaos instantly. Whether you choose sorting (for readability) or frequency counting (for raw speed), the core principle is the same: find the hidden structure that similar items share.