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. All words with the same sorted key are anagrams and stored 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 with sorted keys?

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 is used to store the grouped anagrams before returning the result?

A vector of vectors is used to store groups of anagrams collected from the hash map values.

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

A Sorted string of the word

B Length of the word

C First letter of the word

D Number of vowels in the word

The sorted string 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 Linked list

B Stack

C Queue

D Hash map with sorted string keys and vector of strings as values

A hash map with sorted string keys groups anagrams efficiently, storing each group in a vector.

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

A Converting to uppercase

B Reversing the word

C Sorting the characters

D Counting vowels

Sorting the characters creates a uniform key for all anagrams.

If the input list has 5 words each of length 3, what is the approximate time complexity to group anagrams?

A O(5 + 3)

B O(5 * 3 log 3)

C O(5 * 3)

D O(3 log 5)

Time complexity is O(N * K log K), here N=5 and K=3.

What will be the output grouping for input: ["eat", "tea", "tan", "ate", "nat", "bat"]?

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

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

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

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

Words with same sorted letters are grouped together.

Explain how to group anagrams using a hash map step-by-step.

Think about how sorting helps identify anagrams.

Describe the time complexity of grouping anagrams using sorting and hash map.

Focus on sorting cost per word and number of words.

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 characters

B. Use the length of the string as the key to group words

C. Use the first character of the string as the key

D. Use the last character of the string as the key

Solution

1. Step 1: Understand anagram grouping

   Anagrams have the same characters in different orders, so sorting the string gives a unique key for all anagrams.

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 with the same characters -> Option A

4. Quick Check:

   Sorted string key = group anagrams [OK]

Hint: Sort each word to find its anagram group key [OK]

Common Mistakes:

• Using string length as key misses different anagrams
• Using first or last character groups unrelated words
• Not sorting strings before using as keys

2. Which of the following C++ code snippets correctly declares a hash map to group anagrams where the key is a string and the value is a vector of strings?

easy

A. std::unordered_map anagramGroups;

B. std::map> anagramGroups;

C. std::unordered_map> anagramGroups;

D. std::unordered_map anagramGroups;

Solution

1. Step 1: Identify key and value types

   The key should be a string (sorted word), and the value should be a vector of strings (group of anagrams).

2. Step 2: Match correct declaration

   std::unordered_map> anagramGroups; uses unordered_map with key as string and value as vector<string>, which is correct for grouping anagrams.

3. Final Answer:

   std::unordered_map> anagramGroups; -> Option C

4. Quick Check:

   Key=string, Value=vector<string> [OK]

Hint: Use unordered_map for grouping [OK]

Common Mistakes:

• Using int as key instead of string
• Using vector<char> as key (not hashable)
• Using string as value instead of vector<string>

3. What is the output of the following C++ code snippet?

std::vector words = {"eat", "tea", "tan", "ate", "nat", "bat"};
std::unordered_map> groups;
for (auto &word : words) {
    std::string sortedWord = word;
    std::sort(sortedWord.begin(), sortedWord.end());
    groups[sortedWord].push_back(word);
}
for (auto &pair : groups) {
    std::cout << "Group: ";
    for (auto &w : pair.second) std::cout << w << " ";
    std::cout << std::endl;
}

medium

A. Group: eat tea
Group: tan bat
Group: ate nat

B. Group: eat tea ate
Group: tan nat
Group: bat

C. Group: ate tea
Group: bat tan
Group: eat nat

D. Group: eat tan bat
Group: tea nat ate

Solution

1. Step 1: Sort each word and group by sorted key

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

2. Step 2: Print groups

   Groups printed are: "eat tea ate", "tan nat", and "bat".

3. Final Answer:

   Group: eat tea ate Group: tan nat Group: bat -> Option B

4. Quick Check:

   Sorted keys group anagrams correctly [OK]

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

Common Mistakes:

• Not sorting words before grouping
• Mixing groups by original word order
• Printing keys instead of grouped words

4. Identify the error in the following code snippet that tries to group anagrams:

std::vector words = {"abc", "bca", "cab"};
std::unordered_map> groups;
for (auto word : words) {
    std::string key = word;
    std::sort(word.begin(), word.end());
    groups[key].push_back(word);
}

medium

A. Sorting is done on 'word' but 'key' is used as map key, causing wrong grouping

B. The vector 'words' is not initialized correctly

C. The map should use vector<char> as key instead of string

D. The loop variable should be a reference to avoid copying

Solution

1. Step 1: Analyze sorting and key usage

   The code sorts 'word' but uses 'key' (unsorted) as the map key, so grouping is incorrect.

2. Step 2: Correct usage

   Sort 'key' and use it as the map key to group anagrams properly.

3. Final Answer:

   Sorting is done on 'word' but 'key' is used as map key, causing wrong grouping -> Option A

4. Quick Check:

   Sort key before using in map [OK]

Hint: Sort the key string, not the original word [OK]

Common Mistakes:

• Sorting wrong variable before using as key
• Using unsorted string as map key
• Ignoring reference vs copy in loop

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

hard

A. Sort strings without changing case

B. Use the original string as key without any changes

C. Group by string length only

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

Solution

1. Step 1: Normalize case for consistent keys

   Convert each string to lowercase so that case differences do not affect grouping.

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

   Sorting the lowercase string ensures all case variants of anagrams map to the same key.

3. Step 3: Group anagrams using normalized keys

   Use the lowercase sorted string as the hash map key to group anagrams case-insensitively.

4. Final Answer:

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

5. Quick Check:

   Lowercase + sort = case-insensitive grouping [OK]

Hint: Lowercase strings before sorting to ignore case [OK]

Common Mistakes:

• Not normalizing case before grouping
• Using original case strings as keys
• Grouping by length or unsorted strings