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

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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

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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}

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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).