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

Time Complexity: Group Anagrams Using Hash Map

O(n * m log m)

Understanding Time Complexity

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

Specifically, how does the code handle many words and their letters efficiently?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import java.util.*;

public List> groupAnagrams(String[] strs) {
    Map> map = new HashMap<>();
    for (String s : strs) {
        char[] chars = s.toCharArray();
        Arrays.sort(chars);
        String key = new String(chars);
        map.computeIfAbsent(key, k -> new ArrayList<>()).add(s);
    }
    return new ArrayList<>(map.values());
}
    

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

Identify Repeating Operations

• Primary operation: Loop over all words, and for each word, sorting its characters.
• How many times: The outer loop runs once per word (n times). Sorting runs once per word, each on word length m.

How Execution Grows With Input

As the number of words grows, the code sorts each word's letters and inserts into the map.

Input Size (n)	Approx. Operations
10	About 10 times sorting words of length m
100	About 100 times sorting words of length m
1000	About 1000 times sorting words of length m

Pattern observation: The total work grows roughly in direct proportion to the number of words times the cost to sort each word.

Final Time Complexity

Time Complexity: O(n * m log m)

This means the time grows with the number of words and the time to sort each word's letters.

Common Mistake

[X] Wrong: "Sorting all words is just O(n) because sorting is fast."

[OK] Correct: Sorting each word depends on its length m, so sorting is not constant time; it costs about m log m per word.

Interview Connect

Understanding this complexity helps you explain how your solution scales with input size, a key skill in interviews.

Self-Check

"What if instead of sorting, we used a frequency count of letters as the key? How would the time complexity change?"

Practice

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

easy

A. Count the length of each word and group by length

B. Group words by their first letter only

C. Use sorted strings as keys to group words that are anagrams

D. Use the original word as the key in the hash map

Solution

1. Step 1: Understand anagram grouping

   Anagrams are words with the same letters in different orders. Sorting letters gives a unique key for each group.

2. Step 2: Use sorted string as hash map key

   By sorting each word and using it as a key, all anagrams map to the same key and can be grouped together.

3. Final Answer:

   Use sorted strings as keys to group words that are anagrams -> Option C

4. Quick Check:

   Sorted string key = group anagrams [OK]

Hint: Sort each word and use as key to group anagrams [OK]

Common Mistakes:

• Grouping by word length only
• Using original word as key without sorting
• Grouping by first letter only

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

easy

A. HashMap map = new HashMap<>();

B. HashMap map = new HashMap<>();

C. Map map = new HashMap<>();

D. Map map = new HashMap<>();

Solution

1. Step 1: Identify correct map types for grouping

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

2. Step 2: Check each option's type correctness

   Map map = new HashMap<>(); uses Map, which is correct. Others have wrong key or value types.

3. Final Answer:

   Map map = new HashMap<>(); -> Option D

4. Quick Check:

   Key=String, Value=List<String> for grouping [OK]

Hint: Use Map for grouping anagrams [OK]

Common Mistakes:

• Using int as key type
• Using String as value instead of List<String>
• Using List<String> as key type

3. What is the output of the following Java code snippet?

String[] words = {"eat", "tea", "tan", "ate", "nat", "bat"};
Map map = new HashMap<>();
for (String word : words) {
  char[] chars = word.toCharArray();
  Arrays.sort(chars);
  String key = new String(chars);
  map.computeIfAbsent(key, k -> new ArrayList<>()).add(word);
}
System.out.println(map.values());

medium

A. [[eat, tea, ate], [tan, nat], [bat]]

B. [[eat, tea], [tan, nat], [bat, ate]]

C. [[eat, ate], [tea, tan], [nat, bat]]

D. [[eat], [tea], [tan], [ate], [nat], [bat]]

Solution

1. Step 1: Sort each word to get keys

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

2. Step 2: Group words by sorted keys

   Words with key "aet" grouped: [eat, tea, ate]; key "ant": [tan, nat]; key "abt": [bat].

3. Final Answer:

   [[eat, tea, ate], [tan, nat], [bat]] -> Option A

4. Quick Check:

   Sorted keys group anagrams correctly [OK]

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

Common Mistakes:

• Not sorting words before grouping
• Adding words to wrong groups
• Printing keys instead of grouped values

4. Identify the error in this Java code for grouping anagrams:

String[] words = {"bat", "tab", "eat"};
Map map = new HashMap<>();
for (String word : words) {
  char[] chars = word.toCharArray();
  String key = new String(chars);
  Arrays.sort(chars);
  map.putIfAbsent(key, new ArrayList<>());
  map.get(key).add(word);
}
System.out.println(map.values());

medium

A. Sorting is done after creating the key, so keys are incorrect

B. Using putIfAbsent instead of computeIfAbsent causes error

C. The map should use List<Integer> instead of List<String>

D. The loop should use a while instead of for

Solution

1. Step 1: Check order of sorting and key creation

   Key is created from chars before sorting, so key is unsorted word.

2. Step 2: Identify impact on grouping

   Since keys are unsorted, anagrams get different keys and won't group correctly.

3. Final Answer:

   Sorting is done after creating the key, so keys are incorrect -> Option A

4. Quick Check:

   Sort before key creation to group correctly [OK]

Hint: Sort chars before creating key string [OK]

Common Mistakes:

• Creating key before sorting chars
• Confusing putIfAbsent with computeIfAbsent
• Wrong map value type

5. Given a list of words, how would you modify the grouping anagrams approach to handle case-insensitive grouping (e.g., "Eat" and "tea" are anagrams)?

hard

A. Use the original word as key without any change

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

C. Group words by their length only

D. Sort words but keep original case for keys

Solution

1. Step 1: Understand case-insensitive grouping

   To treat "Eat" and "tea" as same group, ignore case by converting all words to lowercase.

2. Step 2: Apply lowercase conversion before sorting

   Convert word to lowercase, then sort chars to create key. This ensures case differences don't split groups.

3. Step 3: Group words using lowercase sorted keys

   Use these keys in hash map to group anagrams regardless of original case.

4. Final Answer:

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

5. Quick Check:

   Lowercase + sort = case-insensitive grouping [OK]

Hint: Lowercase words before sorting for case-insensitive groups [OK]

Common Mistakes:

• Not converting to lowercase before sorting
• Using original case keys causing split groups
• Grouping by length instead of letters