Easy Python Solution 👨‍💻 || Briefly Explained ✅✅
Code Explanation\nThe provided Python code defines a class Solution with a method groupAnagrams. This method takes a list of strings strs as input and is suppos
# Code Explanation\nThe provided Python code defines a class `Solution` with a method `groupAnagrams`. This method takes a list of strings `strs` as input and is supposed to group the anagrams from the input list into lists of lists, where each inner list contains words that are anagrams of each other. Anagrams are words or phrases formed by rearranging the letters of another, such as "listen" and "silent."\n\nLet\'s break down the code step by step:\n\n1. **Initialize a Dictionary**: The code starts by initializing an empty dictionary called `ans`. This dictionary will be used to store groups of anagrams, with the keys being the character counts of the letters in the anagrams and the values being lists of strings that belong to each group.\n\n    ```python\n    ans = collections.defaultdict(list)\n    ```\n\n    Here, `collections.defaultdict(list)` creates a dictionary where the default value for each key is an empty list.\n\n2. **Iterate Through the Input List**: The code then enters a loop to iterate through each string `s` in the input list `strs`.\n\n    ```python\n    for s in strs:\n    ```\n\n3. **Initialize a Character Count List**: Inside the loop, a list called `count` is initialized with 26 zeros. This list will be used to count the occurrences of each letter in the current string `s`. Each index in the `count` list corresponds to a letter in the lowercase English alphabet.\n\n    ```python\n    count = [0] * 26\n    ```\n\n4. **Count the Occurrence of Letters in the String**: The code then enters another loop to iterate through each character `c` in the current string `s`.\n\n    ```python\n    for c in s:\n    ```\n\n5. **Increment the Count for Each Letter**: Inside the inner loop, the code calculates the position of the character `c` in the alphabet (by subtracting the ASCII value of \'a\' from the ASCII value of `c`) and increments the corresponding element in the `count` list. This effectively counts the occurrence of each letter in the current string `s`.\n\n    ```python\n    count[ord(c) - ord("a")] += 1\n    ```\n\n6. **Group Anagrams Using Character Counts as Keys**: After counting the occurrences of letters in the current string `s`, the code converts the `count` list into a tuple. This tuple is used as a key in the `ans` dictionary. The value associated with this key is a list, and the current string `s` is appended to this list. This step groups strings with the same character count together.\n\n    ```python\n    ans[tuple(count)].append(s)\n    ```\n\n7. **Return Grouped Anagrams**: After processing all strings in the input list `strs`, the code returns the values of the `ans` dictionary. These values are lists of grouped anagrams.\n\n    ```python\n    return ans.values()\n    ```\n\nSo, in summary, this code efficiently groups anagrams by counting the occurrences of each letter in each string and using these counts as keys in a dictionary. It ensures that anagrams with the same character counts are grouped together, and it returns these groups as a list of lists.\n\n# Python Code\n```\nclass Solution:\n    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:\n        ans = collections.defaultdict(list)\n\n        for s in strs:\n            count = [0] * 26\n            for c in s:\n                count[ord(c) - ord("a")] += 1                \n            ans[tuple(count)].append(s)\n\n        return ans.values()\n\n```\n\n**Please upvote if you like the solution.\nHappy Coding! \uD83D\uDE0A**\n
['Array', 'Hash Table', 'String', 'Python', 'Python3']