Group Anagrams Using Hash Map in DSA C# - Time & Space Complexity

Time Complexity: Group Anagrams Using Hash Map

O(n x k log k)

Understanding Time Complexity

We want to understand how the time needed to group anagrams grows as the list of words gets bigger.

Specifically, how does the work increase when we have more words or longer words?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public IList> GroupAnagrams(string[] strs) {
    var map = new Dictionary>();
    foreach (var s in strs) {
        var chars = s.ToCharArray();
        Array.Sort(chars);
        var key = new string(chars);
        if (!map.ContainsKey(key)) {
            map[key] = new List();
        }
        map[key].Add(s);
    }
    return new List>(map.Values);
}
    

This code groups words that are anagrams by sorting each word and using the sorted word as a key in a hash map.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each word and sorting its characters.
• How many times: Once for each word in the input array.

How Execution Grows With Input

For each word, sorting its characters takes time proportional to the word's length times log of that length.

Input Size (n = number of words)	Approx. Operations
10 (words of length k)	About 10 x k log k operations
100	About 100 x k log k operations
1000	About 1000 x k log k operations

Pattern observation: The total work grows linearly with the number of words, and sorting each word adds a small extra cost depending on word length.

Final Time Complexity

Time Complexity: O(n x k log k)

This means the time grows roughly in proportion to the number of words times the cost to sort each word.

Common Mistake

[X] Wrong: "Sorting each word is constant time, so overall time is just O(n)."

[OK] Correct: Sorting depends on word length, so it adds a factor of k log k, which can be significant if words are long.

Interview Connect

Understanding this time complexity helps you explain how your solution scales and why sorting each word is necessary to group anagrams efficiently.

Self-Check

"What if we used character count arrays instead of sorting each word? How would the time complexity change?"

Practice

1. What is the main idea behind grouping anagrams using a hash map?

easy

A. Use sorted strings as keys to group words with the same letters

B. Use the length of the word as the key to group anagrams

C. Use the first letter of the word as the key to group anagrams

D. Use the last letter of the word as the key to group anagrams

Solution

1. Step 1: Understand anagrams

   Anagrams are words with the same letters in different orders.

2. Step 2: Use sorted string as key

   Sorting letters of each word gives a unique key for all anagrams.

3. Final Answer:

   Use sorted strings as keys to group words with the same letters -> Option A

4. Quick Check:

   Sorted string key groups anagrams [OK]

Hint: Sort each word and use it as key in hash map [OK]

Common Mistakes:

• Using word length as key causes wrong grouping
• Using first or last letter misses many anagrams
• Not sorting letters before using as key

2. Which of the following C# code snippets correctly initializes a hash map to group anagrams?

easy

A. var map = new Dictionary();

B. var map = new List>();

C. var map = new Dictionary();

D. var map = new Dictionary();

Solution

1. Step 1: Identify correct dictionary type

   We need keys as strings (sorted words) and values as lists of strings (anagrams).

2. Step 2: Check each option

   var map = new Dictionary(); uses Dictionary which fits the requirement.

3. Final Answer:

   var map = new Dictionary(); -> Option A

4. Quick Check:

   Dictionary with string keys and list values [OK]

Hint: Use Dictionary for grouping anagrams [OK]

Common Mistakes:

• Using List of dictionaries instead of dictionary
• Using wrong key or value types
• Using Dictionary which can't group anagrams

3. What is the output of the following C# code?

var strs = new string[] {"eat", "tea", "tan", "ate", "nat", "bat"};
var map = new Dictionary();
foreach (var s in strs) {
  var key = String.Concat(s.OrderBy(c => c));
  if (!map.ContainsKey(key)) map[key] = new List<string>();
  map[key].Add(s);
}
foreach (var group in map.Values) {
  Console.WriteLine(string.Join(", ", group));
}

medium

A. "eat, tea\nate, tan\nbat, nat"

B. "eat, tan, bat\ntea, ate, nat"

C. "eat, tea, ate\ntan, nat\nbat"

D. "eat, tea, ate, tan, nat, bat"

Solution

1. Step 1: Sort each word to find keys

   "eat", "tea", "ate" all sort to "aet"; "tan", "nat" to "ant"; "bat" to "abt".

2. Step 2: Group words by keys and print

   Groups are ["eat", "tea", "ate"], ["tan", "nat"], ["bat"]. Printed line by line.

3. Final Answer:

   "eat, tea, ate\ntan, nat\nbat" -> Option C

4. Quick Check:

   Sorted keys group anagrams correctly [OK]

Hint: Sort words, group by sorted key, print groups [OK]

Common Mistakes:

• Printing all words in one line
• Grouping by unsorted words
• Mixing groups incorrectly

4. Identify the bug in this C# code snippet for grouping anagrams:

var map = new Dictionary();
foreach (var s in strs) {
  var key = s.OrderBy(c => c).ToString();
  if (!map.ContainsKey(key)) map[key] = new List<string>();
  map[key].Add(s);
}

medium

A. Not checking if key exists before adding

B. Missing initialization of map dictionary

C. Using List<string> instead of List<char> for values

D. Using ToString() on IEnumerable<char> does not produce sorted string key

Solution

1. Step 1: Understand s.OrderBy(c => c)

   This returns IEnumerable<char>, not a string.

2. Step 2: Using ToString() on IEnumerable<char>

   ToString() returns type name, not sorted string key. Need to join chars to string.

3. Final Answer:

   Using ToString() on IEnumerable<char> does not produce sorted string key -> Option D

4. Quick Check:

   Use String.Concat or new string() to convert chars to string [OK]

Hint: Convert sorted chars to string with String.Concat, not ToString() [OK]

Common Mistakes:

• Calling ToString() on IEnumerable<char> expecting string
• Not initializing dictionary before use
• Ignoring key existence check

5. Given a list of strings, how would you modify the grouping anagrams code to ignore case differences (e.g., "Eat" and "tea" should be grouped)?

hard

A. Group by string length only

B. Convert each string to lowercase before sorting and using as key

C. Sort strings but keep original case for keys

D. Use the original string as key without sorting

Solution

1. Step 1: Normalize case for grouping

   Convert each string to lowercase to treat "Eat" and "tea" the same.

2. Step 2: Sort the lowercase string and use as key

   Sorting lowercase letters groups anagrams ignoring case.

3. Final Answer:

   Convert each string to lowercase before sorting and using as key -> Option B

4. Quick Check:

   Lowercase + sort = case-insensitive grouping [OK]

Hint: Lowercase strings before sorting for case-insensitive grouping [OK]

Common Mistakes:

• Sorting without case normalization
• Using original case keys causing separate groups
• Grouping by length which is insufficient