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

Time Complexity: Group Anagrams Using Hash Map

O(n * 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 code handle checking and grouping words that are anagrams?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void groupAnagrams(char* strs[], int strsSize) {
    // Create a hash map to group anagrams
    for (int i = 0; i < strsSize; i++) {
        char key[100];
        strcpy(key, strs[i]);
        qsort(key, strlen(key), sizeof(char), cmpChar); // sort characters in the string
        hashmapInsert(key, strs[i]);
    }
    // Output groups from hashmap
}
    

This code groups words by sorting each word's letters 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: Sorting each word's characters inside the loop.
• How many times: Once for each word in the list (strsSize times).

How Execution Grows With Input

As the number of words grows, the code sorts each word, so the time grows with both the number of words and their length.

Input Size (n)	Approx. Operations
10 words	10 × sorting each word (depends on word length)
100 words	100 × sorting each word
1000 words	1000 × sorting each word

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

Final Time Complexity

Time Complexity: O(n * k log k)

This means the time grows with the number of words (n) times the cost to sort each word of length k.

Common Mistake

[X] Wrong: "Sorting all words is just O(n) because we do it once per word."

[OK] Correct: Sorting each word takes time depending on its length (k log k), so total time depends on both number of words and word length.

Interview Connect

Understanding how sorting inside loops affects time helps you explain your solution clearly and shows you know how to analyze code performance.

Self-Check

"What if we used counting sort for each word instead of general sorting? How would the time complexity change?"

Practice

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

easy

A. Use the length of words as keys to group anagrams

B. Store words in a list without any sorting or grouping

C. Group words by their first character only

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

Solution

1. Step 1: Understand anagrams and their properties

   Anagrams are words made of the same letters in any order, so sorting letters gives a unique key for each group.

2. Step 2: Use sorted strings as hash map keys

   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 with the same letters -> Option D

4. Quick Check:

   Sorted word keys group anagrams [OK]

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

Common Mistakes:

• Grouping by word length only
• Using first character as key
• Not sorting words before grouping

2. Which of the following is the correct way to declare a hash map (dictionary) in C for grouping anagrams?

easy

A. map anagramGroups;

B. unordered_map anagramGroups;

C. unordered_map anagramGroups;

D. vector<string> anagramGroups;

Solution

1. Step 1: Identify the data types needed for grouping anagrams

   We need a map from a string key (sorted word) to a list of strings (anagrams).

2. Step 2: Choose the correct C++ STL container

   unordered_map fits this need perfectly.

3. Final Answer:

   unordered_map anagramGroups; -> Option B

4. Quick Check:

   Map from string to vector<string> groups anagrams [OK]

Hint: Use unordered_map for grouping anagrams [OK]

Common Mistakes:

• Using map which mismatches key and value types
• Using vector<string> alone without mapping keys
• Using unordered_map which is unrelated

3. What is the output of the following C++ code snippet? vector<string> words = {"eat", "tea", "tan", "ate", "nat", "bat"}; unordered_map groups; for (auto &word : words) { string sortedWord = word; sort(sortedWord.begin(), sortedWord.end()); groups[sortedWord].push_back(word); } for (auto &pair : groups) { sort(pair.second.begin(), pair.second.end()); for (auto &w : pair.second) cout << w << " "; cout << endl; }

medium

A. bat
nat tan
eat tea ate

B. eat tea ate
bat
tan nat

C. tea ate eat
bat
tan nat

D. ate eat tea
bat
nat tan

Solution

1. Step 1: Group words by sorted keys

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

2. Step 2: Sort each group and print

   Each group is sorted alphabetically before printing: "ate eat tea", "bat", "nat tan".

3. Final Answer:

   bat nat tan eat tea ate -> Option A

4. Quick Check:

   Sorted keys group anagrams; groups sorted before print [OK]

Hint: Sort words and groups before printing for consistent output [OK]

Common Mistakes:

• Not sorting groups before printing
• Printing groups in wrong order
• Confusing sorted keys with original words

4. Identify the bug in this C++ code for grouping anagrams: vector<string> words = {"abc", "bca", "cab"}; unordered_map groups; for (auto word : words) { string key = word; sort(word.begin(), word.end()); groups[key].push_back(word); }

medium

A. Sorting is done on 'word' but key uses unsorted string

B. Using unordered_map instead of map causes error

C. Loop variable should be reference to avoid copies

D. Missing initialization of groups map

Solution

1. Step 1: Check how key and sorted string are assigned

   Key is assigned before sorting, so key is unsorted original word.

2. Step 2: Sorting is done on 'word' but key remains unsorted

   Groups use unsorted key, so anagrams won't group correctly.

3. Final Answer:

   Sorting is done on 'word' but key uses unsorted string -> Option A

4. Quick Check:

   Key must be sorted string for correct grouping [OK]

Hint: Sort string before assigning key for hash map [OK]

Common Mistakes:

• Assigning key before sorting
• Sorting wrong variable
• Not using reference in loop (less critical here)

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

hard

A. Group words by their first letter ignoring case

B. Use word length as key instead of sorted string

C. Convert all words to lowercase before sorting and grouping

D. Sort words without changing case and group normally

Solution

1. Step 1: Understand case-insensitive grouping

   "Eat" and "tea" differ in case but are anagrams ignoring case.

2. Step 2: Normalize words by converting to lowercase before sorting

   This ensures keys match regardless of original case.

3. Final Answer:

   Convert all words to lowercase before sorting and grouping -> Option C

4. Quick Check:

   Lowercase normalization groups case-insensitive anagrams [OK]

Hint: Lowercase words before sorting to group case-insensitive anagrams [OK]

Common Mistakes:

• Grouping by length only
• Ignoring case normalization
• Sorting without case conversion