How can I get my desired output for this code? Time Limit Exceeded

Question

code-->

class Solution:
    def isAnagram(self, s1: str,t1: str) -> bool:
        flag=1
        s = list(s1)
        t = list(t1)
        for i in range(len(s)):
            for j in range(len(s)):
                if s[i]==t[j]:
                    s[i]=t[j]='0'
                    break
        for i in range(len(s)):
            if s[i] !='0' :
                flag=0
                break
        if flag:
          print(True)
        else:
          print(False)          

Constraints:

1 <= s.length, t.length <= 5 * 104

Please someone help to fix this code.

Answer 1

You could utilize the fact that two anagrams when sorted will be the same:

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

Or that they would have the same counts of characters:

from collections import Counter

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        return Counter(s) == Counter(t)

The latter is more efficient since sorting is O(n log n), while constructing the Counter/Dictionary/Map/HashTable is linear time i.e., O(n).

---

If you don't want to use collections.Counter since it feels like it's abstracting away the core part of the problem, you could replicate its functionality with a list (assumes both strings will only contain lowercase characters):

class Solution:
    def lowercase_counts(self, s: str) -> list[int]:
        counts = [0] * 26
        for ch in s:
            counts[ord(ch) - ord('a')] += 1
        return counts
    
    def isAnagram(self, s: str, t: str) -> bool:
        return self.lowercase_counts(s) == self.lowercase_counts(t)

Or more robustly supporting all Unicode characters using a dictionary:

class Solution:
    def character_counts(self, s: str) -> dict[str, int]:
        counts = {}
        for ch in s:
            counts[ch] = counts.get(ch, 0) + 1
        return counts
    
    def isAnagram(self, s: str, t: str) -> bool:
        return self.character_counts(s) == self.character_counts(t)

Answer 2

Simply use a Counter.

from collections import Counter

def is_anagram(s1:str, s2:str) -> bool:
    return Counter(s1) == Counter(s2)