Anagram Check Techniques in DSA Python - Build from Scratch

Anagram Check Techniques

📖 Scenario: You work in a word game company. You need to check if two words are anagrams. Anagrams are words made by rearranging the letters of another word, like 'listen' and 'silent'.

🎯 Goal: Build a simple program that checks if two given words are anagrams using a dictionary to count letters.

📋 What You'll Learn

Create two string variables with exact given words

Create a dictionary to count letters of the first word

Use a loop to update the dictionary counts

Compare the dictionary counts with the second word

Print True if anagrams, else False

💡 Why This Matters

🌍 Real World

Checking anagrams is useful in word games, puzzles, and text analysis to find related words.

💼 Career

Understanding how to count and compare elements in data structures is a key skill for software developers and data analysts.

1

Create the two words to check

Create two string variables called word1 and word2 with the exact values 'listen' and 'silent' respectively.

DSA Python

# Create variables word1 and word2 with given words
# Your code here

Hint

Use simple assignment like word1 = 'listen'.

2

Create a dictionary to count letters in word1

Create an empty dictionary called letter_count to store letter frequencies from word1.

DSA Python

word1 = 'listen'
word2 = 'silent'
# Create empty dictionary letter_count
# Your code here

Hint

Use {} to create an empty dictionary.

3

Count letters in word1 using a loop

Use a for loop with variable letter to go through each letter in word1. For each letter, update letter_count by adding 1 if the letter exists, or setting it to 1 if it does not.

DSA Python

word1 = 'listen'
word2 = 'silent'
letter_count = {}
# Loop through word1 and update letter_count
# Your code here

Hint

Use if letter in letter_count to check presence, then update count.

4

Check if word2 is an anagram and print result

Create a variable is_anagram and set it to True. Use a for loop with variable letter to check each letter in word2. If a letter is not in letter_count or its count is zero, set is_anagram to False and break the loop. Otherwise, decrease the count of that letter in letter_count by 1. Finally, print is_anagram.

DSA Python

word1 = 'listen'
word2 = 'silent'
letter_count = {}
for letter in word1:
    if letter in letter_count:
        letter_count[letter] += 1
    else:
        letter_count[letter] = 1
# Check if word2 is an anagram and print result
# Your code here

Hint

Use a loop to check letters and print is_anagram at the end.

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