力扣242.有效的字母异位词

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        n_s = len(s)
        n_t = len(t)
        if n_s != n_t:
            return False
        
        char_freq = {}
        for char in s:
            if char in char_freq:
                char_freq[char] += 1
            else:
                char_freq[char] = 1
        
        for char in t:
            if char not in char_freq:
                return False
            char_freq[char] -= 1
        
        for freq in char_freq.values():
            if freq != 0:
                return False
        return True