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.