Group Anagrams Using Hash Map in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an anagram?

An anagram is a word or phrase formed by rearranging the letters of another word or phrase, using all the original letters exactly once.

beginner

Why do we use a hash map to group anagrams?

A hash map helps group anagrams by using a key that represents the sorted letters of a word. Words with the same sorted key are anagrams and get grouped together.

beginner

How do you create a key for each word to group anagrams?

Sort the letters of the word alphabetically. The sorted string is used as the key in the hash map.

intermediate

What is the time complexity of grouping anagrams using a hash map?

The time complexity is O(N * K log K), where N is the number of words and K is the maximum length of a word (due to sorting each word).

beginner

What data structure stores the grouped anagrams in the hash map?

Each key in the hash map points to a list (or array) of words that are anagrams of each other.

What is the key used in the hash map to group anagrams?

A The length of the word

B The first letter of the word

C The sorted version of the word

D The original word itself

The sorted version of the word is used as the key because all anagrams share the same sorted letters.

Which data structure is best to store groups of anagrams?

A Binary tree

B Stack

C Queue

D Hash map with list values

A hash map with lists as values allows grouping words by their sorted key efficiently.

What is the main operation done on each word before inserting into the hash map?

A Sorting the letters

B Reversing the word

C Counting vowels

D Converting to uppercase

Sorting the letters creates a uniform key for all anagrams.

If two words have the same sorted key, what does it mean?

A They are anagrams

B They have the same length

C They start with the same letter

D They are identical

Same sorted key means the words contain the same letters in the same frequency, so they are anagrams.

What is the main benefit of using a hash map for grouping anagrams?

A Sorting all words at once

B Fast lookup and grouping by key

C Reducing word length

D Avoiding sorting

Hash maps provide fast insertion and lookup, making grouping efficient.

Explain how to group a list of words into anagrams using a hash map.

Think about how sorting helps identify anagrams.

Describe the role of sorting in the 'Group Anagrams Using Hash Map' algorithm.

Focus on why sorting is done before grouping.

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