Leetcode - 242. Valid Anagram

Easy

Dec 22, 2025

#hashmap

#frequency

242. Valid Anagram

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

Example 1:

Input: s = "anagram", t = "nagaram"

Output: true

Example 2:

Input: s = "rat", t = "car"

Output: false

Constraints:

1 <= s.length, t.length <= 5 * 104
s and t consist of lowercase English letters.

Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

Notes

Intuition

Two strings are anagrams if they contain the same characters with the same frequencies. A fixed-size frequency array can track character counts efficiently.

Implementation

Create a 26-element array for lowercase letters. Iterate through string s, incrementing the count at each character's index (ord(char) - ord('a')). Then iterate through t, decrementing the counts. If the strings are anagrams, all counts will be zero at the end.

Edge-cases

Strings of different lengths cannot be anagrams. This is implicitly checked—the frequency array won't be all zeros if lengths differ.

Time: O(n) — two passes through the strings
Space: O(1) — frequency array is fixed at 26 elements

Complexity

Solution

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        freq = [0] * 26

        for char in s:
            freq[ord(char) - ord('a')] += 1

        for char in t:
            freq[ord(char) - ord('a')] -= 1

        return all(f == 0 for f in freq)
        



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()