LeetCode 242: Valid Anagram — Python Solution

Solve LeetCode 242: Valid Anagram in Python with a frequency equality approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.

This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.

What the problem is testing

Count characters in one string and subtract characters from the other; all counts must finish at zero.

Algorithm

1. Count characters in one string and subtract characters from the other; all counts must finish at zero.
2. Maintain this invariant: The counter equals the frequency difference for the processed characters.
3. Continue until every input item or reachable state has been resolved, then return the accumulated result.

Python solution

from collections import Counter, defaultdict, deque, OrderedDict
import random

class Solution:
    def isAnagram(self, s, t):
        return Counter(s) == Counter(t)

Why this is correct

The proof follows the maintained state: The counter equals the frequency difference for the processed characters. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.

Complexity

O(n) time and O(k) space. The stated auxiliary space excludes the returned output unless the output is the data structure being built.

Edge cases

Different lengths cannot be anagrams; Unicode characters work as dictionary keys.

Tested reference code

This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.

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