Group Anagrams Using Hash Map in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Grouping Anagrams by Sorted Key

What is the output of the following Python code that groups anagrams using a hash map with sorted strings as keys?

DSA Python

from collections import defaultdict
words = ["eat", "tea", "tan", "ate", "nat", "bat"]
groups = defaultdict(list)
for word in words:
    key = ''.join(sorted(word))
    groups[key].append(word)
result = list(groups.values())
print(result)

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

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

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

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

💡 Hint

Think about how sorting each word groups anagrams together.

The code sorts each word to create a key. Anagrams have the same sorted key, so they are grouped together in the dictionary. The final list contains groups of anagrams.

🧠 Conceptual

intermediate

Key Choice for Grouping Anagrams

Which of the following is the best key to use in a hash map to group anagrams efficiently?

A The first character of the word

B The length of the word

C The sorted string of the word

D The sum of ASCII values of characters in the word

💡 Hint

Anagrams have the same letters in different orders.

Sorting the string puts all letters in order, so anagrams become identical keys. Length or first character alone cannot uniquely identify anagrams. Sum of ASCII values can cause collisions.

🔧 Debug

advanced

Identify the Error in Anagram Grouping Code

What error does the following code produce when grouping anagrams using a hash map?

DSA Python

words = ["bat", "tab", "eat"]
groups = {}
for word in words:
    key = sorted(word)
    if key not in groups:
        groups[key] = []
    groups[key].append(word)
print(groups)

A KeyError: key not found

B TypeError: unhashable type: 'list'

C SyntaxError: invalid syntax

D No error, prints grouped anagrams

💡 Hint

Check the type of the key used in the dictionary.

The sorted(word) returns a list, which is unhashable and cannot be used as a dictionary key, causing a TypeError.

❓ Predict Output

advanced

Output of Grouping Anagrams Using Character Count as Key

What is the output of this Python code that groups anagrams using character count as the hash map key?

DSA Python

from collections import defaultdict
words = ["abc", "bca", "cab", "xyz", "zyx", "yxz"]
groups = defaultdict(list)
for word in words:
    count = [0]*26
    for ch in word:
        count[ord(ch) - ord('a')] += 1
    groups[tuple(count)].append(word)
result = list(groups.values())
print(result)

A [['abc', 'bca', 'cab'], ['xyz', 'zyx', 'yxz']]

B [['abc', 'xyz'], ['bca', 'zyx'], ['cab', 'yxz']]

C [['abc'], ['bca'], ['cab'], ['xyz'], ['zyx'], ['yxz']]

D [['abc', 'bca'], ['cab', 'xyz'], ['zyx', 'yxz']]

💡 Hint

Character count arrays uniquely identify anagrams.

The code counts characters for each word and uses the count tuple as key. Anagrams have identical counts, so they group together.

🧠 Conceptual

expert

Why Use Tuple Instead of List as Hash Map Key in Anagram Grouping?

Why is a tuple preferred over a list as a key in a hash map when grouping anagrams by character count?

A Tuples are immutable and hashable, lists are mutable and unhashable

B Tuples use less memory than lists

C Lists are faster to access than tuples

D Lists cannot store integers

💡 Hint

Dictionary keys must be hashable and immutable.

Lists can change after creation, so they cannot be used as dictionary keys. Tuples cannot change and are hashable, making them valid keys.