Solving the Group Anagrams Problem in C#
1. The Problem: Group Anagrams
The “Group Anagrams” challenge is a fundamental exercise in string manipulation and data structuring. The requirement is:
Given an array of strings
strs, 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, typically using all the original letters exactly once.
2. The Intuition: The “Sorted Key” Approach
The core idea is to find a canonical form for each string. Since anagrams consist of the same letters, they will all look identical if we sort their characters alphabetically.
- Take the words
"eat","tea", and"ate". - Sort each one:
"eat"→"aet""tea"→"aet""ate"→"aet"
- Now that they all share the same key (
"aet"), we can easily group them in a Dictionary.
3. The Implementation (C#)
Here is a clean implementation using System.Linq for character sorting and a Dictionary for grouping.
public class Solution {
public List<List<string>> GroupAnagrams(string[] strs) {
// 1. Initialize our dictionary to store grouped anagrams
// Key: The sorted version of the string (canonical key)
// Value: A list of original strings that match that key
var dict = new Dictionary<string, List<string>>();
foreach(var currString in strs) {
// 2. Create the canonical key by sorting characters
var key = new string(currString.OrderBy(c => c).ToArray());
// 3. Check if we've seen this anagram group before
if (!dict.TryGetValue(key, out var existing)) {
existing = new List<string>();
dict[key] = existing;
}
// 4. Add the original string to its respective group
existing.Add(currString);
}
// 5. Return all the groups as a List of Lists
return new List<List<string>>(dict.Values);
}
}
4. Step-by-Step Breakdown
Step 1: Initialize the Dictionary
We use a Dictionary<string, List<string>> where:
- Key: The sorted string (e.g.,
"aet"). - Value: A list containing all original words that sort into that key (e.g.,
["eat", "tea", "ate"]).
Step 2: Sorting as a Canonical Key
By using currString.OrderBy(c => c).ToArray(), we ensure that any string containing the same letters will produce the same character array. Converting it back to a new string(...) gives us our lookup key.
Step 3: Handle New Groups
TryGetValue is a performance-friendly way to check if a key exists and retrieve it in one go. If it’s a new key, we initialize a new List<string>.
Step 4: Storing Results
The original word (unmodified) is added to the list associated with its sorted key.
5. Complexity Analysis
| Metric | Complexity | Why? |
|---|---|---|
| Time Complexity | O(n * k log k) | n is the number of strings, and k is the max length of a string. We sort each string (k log k) for every string (n). |
| Space Complexity | O(n * k) | We store every string and its characters in the dictionary. |
6. Why use Sorting?
While you can also group anagrams by counting the frequency of each letter (which takes O(n * k) time), sorting is often:
- Simpler to implement: Standard LINQ methods make it concise.
- Fast enough: For most competitive programming constraints (where string length
kis relatively small),k log kis negligible compared to the overhead of building frequency maps.
7. Summary
The Group Anagrams problem is a perfect example of using a Hash Map (Dictionary) to categorize items based on a derived property. By transforming each input into a “canonical” form (sorting), we turn a complex comparison problem into a simple key lookup.
Leave a comment