Intuition

To check if two strings are anagrams, we need to verify that both strings contain the same characters with the same frequencies.

Approach

• Use a Counter to count the frequency of each character in t.
• For each character in s, decrement its count in the Counter.
• After processing, check if all counts in the Counter are zero.
• If so, the strings are anagrams; otherwise, they are not.

Complexity

• Time complexity: \(O(n)\), where $n$ is the length of the strings (each character is processed once).

• Space complexity: \(O(1)\) (since the alphabet size is fixed at 26 for lowercase English letters).

Code

from collections import Counter

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

        t_count = Counter(t)
        for c in s:
            if c in t_count:
                t_count[c] -= 1

        for k, v in t_count.items():
            if v != 0:
                return False

        return True

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