Time: O(n), Space: O(1)..using only ONE hashmap and no built in functions.
Using one hashmap only. Add keys and count values into hashmap for s, then delete count values for respective keys for t. Check if all values are 0.\n\nclass So
Using one hashmap only. Add keys and count values into hashmap for s, then delete count values for respective keys for t. Check if all values are 0.\n```\nclass Solution:\n    def isAnagram(self, s: str, t: str) -> bool:\n    \n        counter = {}\n            \n        if len(s) != len(t):\n            return False\n        \n        for char in s:\n            if char not in counter:\n                counter[char] = 1\n            else:\n                counter[char] += 1\n        \n        for char in t:\n            if char in counter:\n                counter[char] -= 1\n            else:\n                return False\n            \n        for val in counter.values():\n            if val != 0:\n                return False\n        \n        return True\n   ```
['Python', 'Python3']