Easy Python Solution Using HashMap || Step By Step Explanation  ✅ ✅
Explanation\nThis Python code defines a class Solution with a method isAnagram that checks if two input strings s and t are anagrams of each other. Anagrams are
# Explanation\nThis Python code defines a class `Solution` with a method `isAnagram` that checks if two input strings `s` and `t` are anagrams of each other. Anagrams are words or phrases formed by rearranging the letters of another, such that both strings contain the same characters with the same frequency. Let\'s break down the code step by step:\n\n1. **Method Definition**:\n   ```python\n   def isAnagram(self, s: str, t: str) -> bool:\n   ```\n   - This method `isAnagram` takes three arguments: `self` (which is a reference to the instance of the class, but it\'s not used in this method), `s` (the first input string), and `t` (the second input string).\n   - It also specifies that the method returns a boolean value (`True` or `False`).\n\n2. **Length Comparison**:\n   ```python\n   if len(s) != len(t):\n       return False\n   ```\n   - This code block checks whether the lengths of the two input strings `s` and `t` are equal. If they are not, it immediately returns `False`. This is because if the lengths are different, the strings cannot be anagrams.\n\n3. **Character Counting**:\n   ```python\n   countS, countT = {}, {}\n   ```\n   - Here, two empty dictionaries, `countS` and `countT`, are created to store the counts of each character in the respective strings.\n\n4. **Counting Characters in `s` and `t`**:\n   ```python\n   for i in range(len(s)):\n       countS[s[i]] = 1 + countS.get(s[i], 0)\n       countT[t[i]] = 1 + countT.get(t[i], 0)\n   ```\n   - This loop iterates through the characters of both strings, `s` and `t`.\n   - For each character in `s`, it updates the count in the `countS` dictionary. If the character is not yet in the dictionary, it initializes it with a count of 1. If the character is already in the dictionary, it increments the count by 1.\n   - The same process is performed for `t` and the `countT` dictionary.\n\n5. **Comparison**:\n   ```python\n   return countS == countT\n   ```\n   - Finally, the code compares the two dictionaries, `countS` and `countT`. If they are equal, it means that both input strings have the same characters with the same frequency, and the function returns `True`, indicating that `s` and `t` are anagrams. Otherwise, it returns `False`.\n\nIn summary, this code first checks if the lengths of the input strings are the same. If they are, it counts the occurrences of each character in both strings and compares these counts using dictionaries. If the dictionaries are equal, the strings are anagrams, and the function returns `True`, otherwise, it returns `False`.\n\n# Code\n```\nclass Solution:\n    def isAnagram(self, s: str, t: str) -> bool:\n        if len(s) != len(t):\n            return False\n\n        countS, countT = {}, {}\n\n        for i in range(len(s)):\n            countS[s[i]] = 1 + countS.get(s[i], 0)\n            countT[t[i]] = 1 + countT.get(t[i], 0)\n            \n        return countS == countT\n\n```\n\n**Please upvote if you like the solution.\nHappy Coding! \uD83D\uDE0A**\n
['Hash Table', 'String', 'Python', 'Python3']