Group Anagrams Using Hash Map in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the hash map variable for grouping anagrams.

DSA C

HashMap* [1] = createHashMap();

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AgroupMap

Bmap

CanagramGroups

DhashTable

Attempts:

3 left

💡 Hint

Common Mistakes

Using generic names like 'map' or 'hashTable' which are less descriptive.

The variable 'groupMap' is used to store groups of anagrams in the hash map.

2fill in blank

medium

Complete the code to sort the characters of the string key for grouping anagrams.

DSA C

qsort(key, strlen(key), sizeof(char), [1]);

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AcompareChars

BcharCompare

Ccmpfunc

DcompareStrings

Attempts:

3 left

💡 Hint

Common Mistakes

Using a function name that does not match the expected qsort comparator signature.

The 'cmpfunc' is the comparison function passed to qsort to sort characters.

3fill in blank

hard

Fix the error in the code to insert the original string into the hash map under the sorted key.

DSA C

insertHashMap([1], originalStr);

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Akey, originalStr

BgroupMap, key

CgroupMap, originalStr

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the order of arguments or missing the map variable.

The function insertHashMap takes the map and the key to insert the original string into the group.

4fill in blank

hard

Fill both blanks to check if the sorted key exists in the map and then add the original string to the group.

DSA C

if (hashMapContains([1], [2])) {
    addToGroup([1], [2], originalStr);
}

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AgroupMap

Bkey

CoriginalStr

DhashMap

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original string instead of the key for lookup.

We check if 'groupMap' contains 'key' and then add the original string to that group.

5fill in blank

hard

Fill all three blanks to create a new group list for the sorted key and insert the original string.

DSA C

List* newGroup = createList();
addToList(newGroup, [1]);
insertHashMap([2], [3], newGroup);

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AoriginalStr

BgroupMap

Ckey

DnewGroup

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up the order of arguments or inserting the wrong variables.

We add the original string to the new group list, then insert the list into the map with the sorted key.

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