Valid Anagram

• Difficulty: Easy
• Type: Sorting
• link

Problem

Solution

• Sort two words and check if they are equal
• Time complexity: O(n log n)

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

• Count every occurence of alphabets of two words and see if the counts are equal
• Time complexity: O(n)

import collections
class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if collections.Counter(s) == collections.Counter(t):
            return True
        else:
            return False