0242.Valid Anagram

• Link: 242. Valid Anagram - Easy
• Track: NeetCode150

Question

Restate the problem

Anagram: same character, same frequency, different order

• Given two strings s and t
• Return true if t is an anagram of s, return false if not anagram.

Edge Case

• s = [aaa], t = [aa] → False, different frequency
• s = [], t = [] → True

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Method	Approach	Time Complexity	Space Complexity
Method 1 ⭐	Hashmap	O(n)	O(k)
Method 2	Sort	O($nlogn$)	O(n)

Method 1 - Hashmap

Approach

• Compare the s length and t length, if not, return False
• Counter s, count = {char: frequency}
• For each characterc in t
  • If c not in count or count is 0, return False
  • Decrement count
• return True

Complexity

• Time Complexity: O(n)
• Space Complexity: O(k)
  • k = number of distinct characters in s

Code

from collections import Counter
 
class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
 
        counter = Counter(s)
        
        for char in t:
            if counter[char] < 1:
                return False
            counter[char] -= 1
        
        return True

Method 2 - Sort

Complexity

• Time Complexity: O($nlogn$)
• Space Complexity: O(n)

Code

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        
        if len(s) != len(t):
            return False
        
        return sorted(s) == sorted(t)

Mistake:

• list.sort() works in-place, strings are immutable, cannot use .sort().

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History

• Feb-27-2026 Solved