Anagram Check Techniques in DSA Python

Mental Model

Two words are anagrams if they have the exact same letters in the same amounts, just mixed up.

Analogy: Imagine two bags of colored beads. If both bags have the same colors and the same number of each color, they are anagrams, even if the beads are arranged differently.

word1: c -> a -> t -> null
word2: t -> a -> c -> null

We want to check if these two chains have the same beads (letters) in any order.

Dry Run Walkthrough

Input: word1: 'listen', word2: 'silent'

Goal: Check if 'listen' and 'silent' are anagrams by comparing their letter counts.

Step 1: Count letters in 'listen': l=1, i=1, s=1, t=1, e=1, n=1

counts1 = {'l':1, 'i':1, 's':1, 't':1, 'e':1, 'n':1}

Why: We need to know how many times each letter appears in the first word.

Step 2: Count letters in 'silent': s=1, i=1, l=1, e=1, n=1, t=1

counts2 = {'s':1, 'i':1, 'l':1, 'e':1, 'n':1, 't':1}

Why: We count letters in the second word to compare with the first.

Step 3: Compare counts1 and counts2: all letters and counts match

counts1 == counts2 -> True

Why: If counts match exactly, words are anagrams.

Result:

'listen' and 'silent' are anagrams

Annotated Code

DSA Python

from collections import Counter

class AnagramChecker:
    def are_anagrams(self, word1: str, word2: str) -> bool:
        # Count letters in first word
        counts1 = Counter(word1)
        # Count letters in second word
        counts2 = Counter(word2)
        # Compare both counts
        return counts1 == counts2

if __name__ == '__main__':
    checker = AnagramChecker()
    word1 = 'listen'
    word2 = 'silent'
    result = checker.are_anagrams(word1, word2)
    print(f"'{word1}' and '{word2}' are anagrams" if result else f"'{word1}' and '{word2}' are NOT anagrams")

counts1 = Counter(word1)

count letters in first word to know frequency

counts2 = Counter(word2)

count letters in second word to compare frequencies

return counts1 == counts2

check if both words have exactly same letter counts

Output

'listen' and 'silent' are anagrams

Complexity Analysis

Time: O(n) because we count letters in both words once, where n is the length of the words

Space: O(1) because the letter counts use fixed space limited by alphabet size

vs Alternative: Sorting both words takes O(n log n) time, so counting letters is faster and more efficient

Edge Cases

empty strings

Both empty strings are considered anagrams

DSA Python

return counts1 == counts2

different lengths

Letter counts differ, so not anagrams

DSA Python

return counts1 == counts2

same letters, same counts but different order (e.g. 'aabb' and 'abab')

Counts match, so they are anagrams

DSA Python

return counts1 == counts2

When to Use This Pattern

When you need to check if two words have the same letters in any order, use letter counting to quickly compare frequencies.

Common Mistakes

Mistake: Comparing sorted strings without considering case sensitivity
Fix: Convert both words to the same case (e.g., lower) before counting or sorting

Mistake: Not handling empty strings or different lengths explicitly
Fix: Counting letters naturally handles these cases, so rely on counts equality

Summary

Checks if two words have the same letters in the same amounts regardless of order.

Use when you want to verify if two strings are anagrams efficiently.

Counting letter frequencies and comparing them is the key insight for fast anagram checking.