Python Anagram Program: Check if Two Strings Are Anagrams

Q: How do I check if two strings are anagrams in Python?

Lowercase both strings, remove spaces and compare `sorted(a) == sorted(b)`. You can also compare `Counter(a) == Counter(b)`.

Q: Is sorting or Counter better for anagrams?

`Counter` is faster for long strings because it is O(n), while sorting is O(n log n). For short words, `sorted()` is shorter and just as good.

By Yogesh Sharma (LinkedIn) · Every code example is run and checked on Python 3.13 (how we test) · Last updated 21 Oct 2026

Short answer

Two strings are anagrams if they contain the same letters the same number of times. In Python, compare sorted(a) == sorted(b) after lowercasing and removing spaces.

"Listen" and "silent" use exactly the same letters in a different order, so they are anagrams. Checking this is a popular exercise because it forces you to think about what "same letters" means: order does not matter, but the number of each letter does.

The simplest solution sorts both strings and compares them. A second solution counts letters with a dictionary or Counter, which is faster for long inputs. Both are shown below.

Quick facts
Input	Two strings
Output	`True` if they are anagrams, otherwise `False`
Approach	Clean both strings, then compare sorted letters or letter counts
Time complexity	O(n log n) with sorting, O(n) with counting

Approach to check for anagrams step by step

Clean each string: lowercase it and remove spaces so "Dormitory" and "dirty room" can match.
Compare lengths quickly. Different lengths can never be anagrams.
Sort the letters of each string with sorted(), which returns a list of characters.
Compare the two sorted lists with ==.

If both lists are equal, every letter appears the same number of times in both strings.

How the anagram code works

sorted("listen") returns ["e", "i", "l", "n", "s", "t"], and sorted("silent") returns the very same list, so == gives True. We do not need to join the lists back into strings because lists compare element by element.

The cleaning step uses a small helper so the same rule is applied to both inputs. Keeping it in one function avoids the bug where one string is lowercased and the other is not.

Python

def clean(text):
    return text.replace(" ", "").lower()

print(sorted(clean("Dormitory")))
print(sorted(clean("dirty room")))

Counting letters instead of sorting

collections.Counter turns a string into a mapping of letter to count, and two Counter objects are equal when all their counts match. This takes one pass over each string, so it is O(n) instead of O(n log n). For short words the difference is invisible; choose whichever you find more readable. Learn more about it in Python collections.

Method	Time	Notes
`sorted(a) == sorted(b)`	O(n log n)	Shortest and easiest to remember
`Counter(a) == Counter(b)`	O(n)	Best for long strings
Manual dictionary of counts	O(n)	Shows how counting works

Examples you can run

Press Run, then edit the code and run it again.

Anagram check with sorted()

Clean both strings, then compare their sorted letters.

Anagram check with sorted()

def is_anagram(a, b):
    a = a.replace(" ", "").lower()
    b = b.replace(" ", "").lower()
    return sorted(a) == sorted(b)

print(is_anagram("Listen", "Silent"))
print(is_anagram("Dormitory", "dirty room"))
print(is_anagram("apple", "papel"))
print(is_anagram("apple", "pale"))

Expected outputTrue
True
True
False

Anagram check with Counter

Counter compares letter counts and never sorts.

Anagram check with Counter

from collections import Counter

def is_anagram(a, b):
    a = a.replace(" ", "").lower()
    b = b.replace(" ", "").lower()
    return Counter(a) == Counter(b)

print(is_anagram("night", "thing"))
print(is_anagram("aab", "abb"))

Expected outputTrue
False

Edge cases: empty strings and repeated letters

Two empty strings are anagrams of each other; the same letters with different counts are not.

Edge cases: empty strings and repeated letters

def is_anagram(a, b):
    return sorted(a.lower()) == sorted(b.lower())

print(is_anagram("", ""))
print(is_anagram("a", "A"))
print(is_anagram("aabb", "abbb"))

Expected outputTrue
True
False

Interactive anagram checker

Ask for two words and print a message.

Interactive anagram checker

first = input("First word: ")
second = input("Second word: ")
if sorted(first.lower()) == sorted(second.lower()):
    print("Anagrams!")
else:
    print("Not anagrams.")

Input (stdin) — one line per input() call

Expected outputFirst word: Evil
Second word: Vile
Anagrams!

Common mistakes

Comparing sets instead of sorted letters

✗ Wrong

a, b = "aab", "abb"
print(set(a) == set(b))

True, although "aab" and "abb" are not anagrams

✓ Right

a, b = "aab", "abb"
print(sorted(a) == sorted(b))

A set drops duplicates, so it loses the letter counts. Use sorted() or Counter to keep them.

Forgetting to normalise case

✗ Wrong

print(sorted("Listen") == sorted("Silent"))

False (uppercase letters sort before lowercase ones)

✓ Right

print(sorted("Listen".lower()) == sorted("Silent".lower()))

Lowercase both strings before sorting.

Python Anagram Program: Check if Two Strings Are Anagrams: questions people ask

How do I check if two strings are anagrams in Python?

Lowercase both strings, remove spaces and compare sorted(a) == sorted(b). You can also compare Counter(a) == Counter(b).

Is sorting or Counter better for anagrams?

Counter is faster for long strings because it is O(n), while sorting is O(n log n). For short words, sorted() is shorter and just as good.

Do anagrams have to be real words?

No. The program only compares letters, so any two strings with the same letter counts are anagrams, such as "abc" and "cab".

Keep learning

Step-by-step lessons:
Python Strings Python Sorting Python collections Module

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Python Program to Check Palindrome (String and Number)Python Program to Count Vowels in a String Python Word Count Program: Count Words in a String Python Program to Count Character Frequency in a String See all →