Anagram Check Techniques in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if two strings are anagrams by sorting.

DSA Python

def are_anagrams(s1, s2):
    return sorted(s1) [1] sorted(s2)

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == causes wrong result.

We compare sorted versions of both strings for equality to check anagrams.

2fill in blank

medium

Complete the code to count characters using a dictionary for anagram check.

DSA Python

def count_chars(s):
    counts = {}
    for ch in s:
        counts[ch] = counts.get(ch, 0) [1] 1
    return counts

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

A+=

B-=

C*=

D/=

Attempts:

3 left

💡 Hint

Common Mistakes

Using -= causes counts to decrease incorrectly.

We add 1 to the count for each character encountered.

3fill in blank

hard

Fix the error in the code to check anagrams by comparing character counts.

DSA Python

def are_anagrams(s1, s2):
    if len(s1) != len(s2):
        return False
    counts = {}
    for ch in s1:
        counts[ch] = counts.get(ch, 0) + 1
    for ch in s2:
        if ch not in counts or counts[ch] [1] 0:
            return False
        counts[ch] -= 1
    return True

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

A==

B!=

C<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using == 0 misses negative counts.

We check if count is less than 0 to detect extra characters in s2.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension counting characters only if they appear more than once.

DSA Python

def repeated_chars(s):
    return {ch: s.count(ch) [1] ch in s if s.count(ch) [2] 1}

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

Afor

Bif

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'if' instead of 'for' causes syntax error.

Use 'for' to iterate and '>' to filter characters appearing more than once.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps characters to their counts only if count is greater than zero.

DSA Python

def char_counts(s):
    return [1]: [2] for [3] in set(s) if s.count([3]) > 0}

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

Ach

Bs.count(ch)

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causes errors.

Use 'ch' as key and variable, and count characters with s.count(ch).

Practice

1. What does it mean if two strings are anagrams of each other?

easy

A. They contain the same characters in any order

B. They have the same length but different characters

C. They start with the same character

D. They end with the same character

Solution

1. Step 1: Understand the definition of anagrams

   Anagrams are words or phrases made by rearranging the letters of another, so they must have the same characters.

2. Step 2: Compare the character sets

   If two strings have exactly the same characters, just in different orders, they are anagrams.

3. Final Answer:

   They contain the same characters in any order -> Option A

4. Quick Check:

   Anagrams = same characters [OK]

Hint: Check if sorted strings are equal to confirm anagrams [OK]

Common Mistakes:

• Thinking same length means anagrams
• Only checking first or last character
• Ignoring character counts

2. Which of the following Python code snippets correctly checks if two strings s1 and s2 are anagrams using sorting?

easy

A. s1 == s2[::-1]

B. sorted(s1) == sorted(s2)

C. s1.sort() == s2.sort()

D. len(s1) == len(s2)

Solution

1. Step 1: Understand sorting strings in Python

   The sorted() function returns a new sorted list of characters from the string.

2. Step 2: Compare sorted lists for equality

   If sorted(s1) equals sorted(s2), both strings have the same characters in the same counts.

3. Final Answer:

   sorted(s1) == sorted(s2) -> Option B

4. Quick Check:

   sorted strings equal means anagrams [OK]

Hint: Use sorted() to compare strings for anagram check [OK]

Common Mistakes:

• Using .sort() on strings (strings have no sort method)
• Comparing reversed strings instead of sorted
• Only checking length equality

3. What will be the output of the following code?

from collections import Counter
s1 = 'listen'
s2 = 'silent'
print(Counter(s1) == Counter(s2))

medium

A. False

B. TypeError

C. True

D. None

Solution

1. Step 1: Understand Counter usage

   Counter counts the frequency of each character in the string.

2. Step 2: Compare Counters for equality

   Since 'listen' and 'silent' have the same characters with same counts, their Counters are equal.

3. Final Answer:

   True -> Option C

4. Quick Check:

   Counter equality means anagrams [OK]

Hint: Use Counter to count chars and compare for anagrams [OK]

Common Mistakes:

• Expecting False because strings differ in order
• Confusing Counter with sorting
• Not importing collections module

4. Find the error in this anagram check code:

def are_anagrams(s1, s2):
    return s1.sort() == s2.sort()

medium

A. Missing import statement

B. Function should return True always

C. Comparison should be with reversed strings

D. Strings do not have a sort() method

Solution

1. Step 1: Check string methods

   Strings in Python do not have a sort() method; only lists do.

2. Step 2: Identify correct approach

   Use sorted(s1) and sorted(s2) which return sorted lists for comparison.

3. Final Answer:

   Strings do not have a sort() method -> Option D

4. Quick Check:

   sort() not valid on strings [OK]

Hint: Use sorted() on strings, not .sort() method [OK]

Common Mistakes:

• Trying to call .sort() on strings
• Comparing reversed strings instead of sorted
• Ignoring return value of sorted()

5. You want to check if two strings are anagrams but ignore spaces and capitalization. Which code snippet correctly implements this?

hard

A. sorted(s1.lower().replace(' ', '')) == sorted(s2.lower().replace(' ', ''))

B. Counter(s1) == Counter(s2)

C. s1 == s2

D. len(s1) == len(s2)

Solution

1. Step 1: Normalize strings by lowering case and removing spaces

   Use s1.lower().replace(' ', '') and similarly for s2 to ignore case and spaces.

2. Step 2: Sort normalized strings and compare

   Sorting ensures characters are in order for easy comparison.

3. Final Answer:

   sorted(s1.lower().replace(' ', '')) == sorted(s2.lower().replace(' ', '')) -> Option A

4. Quick Check:

   Normalize then sort to check anagrams ignoring spaces/case [OK]

Hint: Lowercase and remove spaces before sorting for anagram check [OK]

Common Mistakes:

• Not removing spaces before comparison
• Not converting to lowercase
• Using Counter without normalization