#https://leetcode.com/problems/valid-anagram/ class Solution(object): #1 #O(NlogN) def isAnagram(self, s, t): if len(s)!=len(t): return False return sorted(s)==sorted(t) #2 #O(N) #using counter def isAnagram(self, s, t): if len(s)!=len(t): return False counter1 = collections.Counter() counter2 = collections.Counter() for c in s: counter1[c]+=1 for c in t: counter2[c]+=1 return counter1==counter2 #3 #this is the same as the above solution #but even more clean #O(N) def isAnagram(self, s, t): return collections.Counter(s)==collections.Counter(t) #4 #this is the best solution so far #python handle string fast bc its written in c a bottom #O(26N)≈O(N) def isAnagram(self, s, t): if len(s)!=len(t): return False for i in string.ascii_lowercase: if s.count(i) != t.count(i): return False return True