Add same Neetcode question to Leetcode folder

Author: Haleshot

Leetcode/Group_Anagrams/Group_Anagrams.py

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+# Neetcode
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+
+        # # Step 1
+        # anagrams = defaultdict(list)  # Initialize a dictionary to hold lists of anagrams
+
+        # # Iterate over each word in the list
+        # for word in strs:
+        #     # Sort the characters of the word to use as the key
+        #     sorted_word = ''.join(sorted(word))
+        #     # Group anagrams together using the sorted word as the key
+        #     anagrams[sorted_word].append(word)
+
+        # # Convert the dictionary values (which are lists of anagrams) to a list of lists
+        # return list(anagrams.values())
+
+        # # Step 2:
+        ans = collections.defaultdict(list)
+
+        for s in strs:
+            count = [0] * 26
+            for c in s:
+                count[ord(c) - ord("a")] += 1
+            ans[tuple(count)].append(s)
+        return ans.values()
+
+
+# Leetcode
+class Solution(object):
+    def groupAnagrams(self, strs):
+        """
+        :type strs: List[str]
+        :rtype: List[List[str]]
+        """
+        # # Step 1
+        # anagrams = defaultdict(list)  # Initialize a dictionary to hold lists of anagrams
+
+        # # Iterate over each word in the list
+        # for word in strs:
+        #     # Sort the characters of the word to use as the key
+        #     sorted_word = ''.join(sorted(word))
+        #     # Group anagrams together using the sorted word as the key
+        #     anagrams[sorted_word].append(word)
+
+        # # Convert the dictionary values (which are lists of anagrams) to a list of lists
+        # return list(anagrams.values())
+
+        # # Step 2:
+        ans = collections.defaultdict(list)
+
+        for s in strs:
+            count = [0] * 26
+            for c in s:
+                count[ord(c) - ord("a")] += 1
+            ans[tuple(count)].append(s)
+        return ans.values()

Leetcode/Group_Anagrams/README.md

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+# Group Anagrams (Medium)
+
+## Table of Contents
+
+- [Problem Statement](#problem-statement)
+- [Examples](#examples)
+- [Constraints](#constraints)
+- [Solutions](#solutions)
+  - [Approach 1: Using Sorted Strings as Keys](#approach-1-using-sorted-strings-as-keys)
+  - [Approach 2: Hash Map with Character Count as Key](#approach-2-hash-map-with-character-count-as-key)
+- [Complexity Analysis](#complexity-analysis)
+- [Code Explanation](#code-explanation)
+- [Related Resources](#related-resources)
+
+## Problem Statement
+
+Given an array of strings `strs`, group all anagrams together into sublists. You may return the output in any order.
+
+An anagram is a string that contains the exact same characters as another string, but the order of the characters can be different.
+
+- [View on Neetcode.io](https://neetcode.io/problems/anagram-groups)
+- [View on LeetCode](https://leetcode.com/problems/group-anagrams/description/)
+
+## Examples
+
+**Example 1:**
+```
+Input: strs = ["eat","tea","tan","ate","nat","bat"]
+Output: [["bat"],["nat","tan"],["ate","eat","tea"]]
+```
+
+**Example 2:**
+```
+Input: strs = [""]
+Output: [[""]]
+```
+
+**Example 3:**
+```
+Input: strs = ["a"]
+Output: [["a"]]
+```
+
+## Constraints
+
+- $1 \leq strs.length \leq 10^4$
+- $0 \leq strs[i].length \leq 100$
+- `strs[i]` consists of lowercase English letters.
+
+## Solutions
+
+### Approach 1: Using Sorted Strings as Keys
+
+```python
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        if len(nums) == 2:
+            return [0, 1]
+        for i in range(len(nums)):
+            for j in range(i + 1, len(nums)):
+                if nums[i] + nums[j] == target:
+                    if nums.index(nums[i]) == nums.index(nums[j]):
+                        return [nums.index(nums[i]), nums.index(nums[j], i + 1, len(nums))]
+                    return [nums.index(nums[i]), nums.index(nums[j])]
+```
+
+### Approach 2: Hash Map with Character Count as Key
+
+```python
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        seenMap = {}
+        for i in range(len(nums)):
+            remain = target - nums[i]
+            if remain in seenMap:
+                return [seenMap[remain], i]
+            seenMap[nums[i]] = i
+        return nums
+```
+
+## Complexity Analysis
+
+### Approach 1:
+- Time Complexity: $O(N * M * \log(M))$, where N is the number of strings and M is the maximum length of a string.
+- Space Complexity: $O(N * M)$
+
+### Approach 2:
+- Time Complexity: $O(N * M)$, where N is the number of strings and M is the maximum length of a string.
+- Space Complexity: $O(N * M)$
+
+## Code Explanation
+
+### Approach 1: Using Sorted Strings as Keys
+This approach involves sorting each string and using the sorted string as a key in a dictionary. 
+
+### Approach 2: Hash Map with Character Count as Key
+This solution uses a hash map to group anagrams:
+1. We iterate through each string in the input array.
+2. For each string, we create a key by counting the occurrences of each character.
+3. We use this count as a key in our hash map. The value is a list of all strings that have the same character count.
+4. Finally, we return the values of the hash map, which are the grouped anagrams.
+
+This approach is efficient and passes on both Neetcode.io and LeetCode.
+
+## Related Resources
+
+- [Neetcode Solution](https://github.com/neetcode-gh/leetcode/blob/main/python/0049-group-anagrams.py)
+- [LeetCode Solution](https://leetcode.com/problems/group-anagrams/solutions/5729062/solution)
+- [Personal Submission](https://leetcode.com/submissions/detail/1377240624/)
+- [Video Explanation (Neetcode)](https://youtu.be/vzdNOK2oB2E?si=aFbSKVZz4WfzrNg1)
+
+> [!NOTE]
+> This problem is part of a larger collection following the roadmap on [Neetcode Roadmap](https://neetcode.io/roadmap). For more details and related problems, please refer to the Neetcode.io website.