Valid anagram

We can split both string to characters, sort and then compare each characters at every position. If we don't find any match, we return False. After comparing every character, we will return True.

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

        s = list(s)
        t = list(t)
        s.sort()
        t.sort()        

        for i in range(len(s)):
            if s[i] != t[i]: 
                return False

        return True

For sorting we have complexity O(nlog(n)), and we will go through the whole list one by one, that is O(n). So, overall time complexity O(n). We don't use any extra memory, so space complexity is O(1).