Group Anagrams Using Hash Map in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dictionary for grouping anagrams.

DSA Python

anagrams = [1]

Drag options to blanks, or click blank then click option'

A{}

B[]

C()

Dset()

Attempts:

3 left

💡 Hint

Common Mistakes

Using a list [] instead of a dictionary.

Using a set() which does not store key-value pairs.

We use an empty dictionary {} to store groups of anagrams with keys as sorted strings.

2fill in blank

medium

Complete the code to get the sorted string key for each word.

DSA Python

key = ''.join(sorted([1]))

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Aword

Bkey

Canagrams

Dgroup

Attempts:

3 left

💡 Hint

Common Mistakes

Using the dictionary name instead of the word.

Using an undefined variable.

We sort the characters of the current word to create a key for grouping anagrams.

3fill in blank

hard

Fix the error in adding the word to the anagram group.

DSA Python

anagrams.setdefault(key, []).[1](word)

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Aextend

Badd

Cinsert

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using add() which is for sets, not lists.

Using extend() which expects an iterable.

We use append() to add the word to the list of anagrams for the key.

4fill in blank

hard

Fill both blanks to create the final list of grouped anagrams.

DSA Python

result = [[1] for [2] in anagrams.values()]

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Agroup

Bkey

Cgroups

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using keys instead of values.

Using variable names that do not match the group concept.

We iterate over the values (groups) in the dictionary and collect each group into a list.

5fill in blank

hard

Fill all three blanks to complete the function that groups anagrams.

DSA Python

def group_anagrams([1]):
    anagrams = {}
    for [2] in [3]:
        key = ''.join(sorted(word))
        anagrams.setdefault(key, []).append(word)
    return list(anagrams.values())

Drag options to blanks, or click blank then click option'

Awords

Bword

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names.

Not matching the parameter and loop variable names.

The function takes a list 'words', loops over each 'word' in 'words', and groups anagrams.

Practice

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

easy

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

B. Use the length of the string as the key

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

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

Solution

1. Step 1: Understand anagrams

   Anagrams are words with the same letters in different orders.

2. Step 2: Use sorted strings as keys

   Sorting letters of each word gives a common key for all anagrams.

3. Final Answer:

   Use sorted strings as keys to group words with the same letters -> Option C

4. Quick Check:

   Sorted string key groups anagrams = C [OK]

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

Common Mistakes:

• Using string length as key groups wrong words
• Using first or last character misses many anagrams
• Not sorting letters before using as key

2. Which of the following Python code snippets correctly initializes an empty hash map (dictionary) to group anagrams?

easy

A. anagrams = {}

B. anagrams = ()

C. anagrams = set()

D. anagrams = []

Solution

1. Step 1: Identify dictionary initialization

   In Python, an empty dictionary is created with curly braces {}.

2. Step 2: Check other options

   List uses [], set uses set(), tuple uses (). Only {} is dictionary.

3. Final Answer:

   anagrams = {} -> Option A

4. Quick Check:

   Empty dictionary = {} [OK]

Hint: Use {} to create an empty dictionary in Python [OK]

Common Mistakes:

• Using [] which creates a list, not a dictionary
• Using set() which creates a set, not a dictionary
• Using () which creates a tuple, not a dictionary

3. What is the output of the following code?

words = ["eat", "tea", "tan", "ate", "nat", "bat"]
anagrams = {}
for word in words:
    key = ''.join(sorted(word))
    if key not in anagrams:
        anagrams[key] = []
    anagrams[key].append(word)
print(list(anagrams.values()))

medium

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

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

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

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

Solution

1. Step 1: Sort each word and group

   Words "eat", "tea", "ate" sorted become "aet"; grouped together. "tan", "nat" sorted "ant"; grouped. "bat" sorted "abt"; alone.

2. Step 2: Collect groups in dictionary values

   Values are lists of grouped anagrams: [['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]

3. Final Answer:

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

4. Quick Check:

   Sorted keys group words correctly = A [OK]

Hint: Sort letters, use as key, group words in list [OK]

Common Mistakes:

• Appending words to wrong keys
• Not sorting words before grouping
• Mixing groups or missing words

4. Find the error in this code that tries to group anagrams:

words = ["abc", "bca", "cab"]
anagrams = {}
for word in words:
    key = sorted(word)
    if key not in anagrams:
        anagrams[key] = []
    anagrams[key].append(word)
print(anagrams)

medium

A. The dictionary is not initialized correctly

B. The print statement syntax is wrong

C. The append method is used incorrectly

D. sorted(word) returns a list, which cannot be used as a dictionary key

Solution

1. Step 1: Check key type used in dictionary

   sorted(word) returns a list, but lists are not hashable and cannot be dictionary keys.

2. Step 2: Fix by converting list to string or tuple

   Use key = ''.join(sorted(word)) or key = tuple(sorted(word)) to make hashable key.

3. Final Answer:

   sorted(word) returns a list, which cannot be used as a dictionary key -> Option D

4. Quick Check:

   Dictionary keys must be hashable, list is not [OK]

Hint: Convert sorted list to string or tuple before using as key [OK]

Common Mistakes:

• Using list as dictionary key causes TypeError
• Not converting sorted result to string or tuple
• Ignoring error messages about unhashable types

5. Given the list words = ["listen", "silent", "enlist", "google", "gogole", "abc"], which code snippet correctly groups the anagrams and returns the groups sorted alphabetically by their first word?

hard

A. anagrams = {}
for w in words:
key = w
anagrams[key] = w
print(list(anagrams.values()))

B. anagrams = {}
for w in words:
key = ''.join(sorted(w))
anagrams.setdefault(key, []).append(w)
groups = list(anagrams.values())
groups.sort(key=lambda x: x[0])
print(groups)

C. anagrams = {}
for w in words:
key = ''.join(sorted(w))
anagrams[key] = w
print(list(anagrams.keys()))

D. anagrams = []
for w in words:
anagrams.append(sorted(w))
print(anagrams)

Solution

1. Step 1: Group words by sorted key using dictionary

   anagrams = {} for w in words: key = ''.join(sorted(w)) anagrams.setdefault(key, []).append(w) groups = list(anagrams.values()) groups.sort(key=lambda x: x[0]) print(groups) uses sorted word as key and appends words to list correctly.

2. Step 2: Sort groups by first word

   anagrams = {} for w in words: key = ''.join(sorted(w)) anagrams.setdefault(key, []).append(w) groups = list(anagrams.values()) groups.sort(key=lambda x: x[0]) print(groups) sorts the list of groups by the first word in each group, meeting requirement.

3. Final Answer:

   Correctly groups anagrams using sorted keys, setdefault, append, and sorts groups by first word -> Option B

4. Quick Check:

   Group by sorted key + sort groups by first word = A [OK]

Hint: Use setdefault for grouping and sort groups by first element [OK]

Common Mistakes:

• Using word itself as key loses grouping
• Overwriting dictionary values instead of appending
• Using list instead of dictionary for grouping