Efficient Solution for 'Rotating the Box' in Python
Intuition\nThe problem involves rotating a 2D grid (the "box") clockwise while ensuring gravity rules are applied to move the stones (#) down to their lowest po
# Intuition\nThe problem involves rotating a 2D grid (the "box") clockwise while ensuring gravity rules are applied to move the stones (#) down to their lowest possible positions. The idea is to first process each row to simulate gravity and then map the rows to their respective columns in the rotated grid.\n\n# Approach\n1. Initialization:\n    - Determine the dimensions of the box (m x n).\n    - Create an empty grid rotated of size n x m to store the rotated version of the box.\n2. Process Rows for Gravity:\n    - Iterate through each row of the box.\n    - Use a pointer (num) to track the lowest available position for stones (#) in the current row.\n    - Traverse the row in reverse to process obstacles (*) and stones (#).\n        - When encountering an obstacle (*), reset num to the position before the obstacle.\n        - When encountering a stone (#), move it to the num position and decrement num.\n3. Rotate the Box:\n    - Map the processed rows to columns of the rotated grid, ensuring the rotated grid represents a 90-degree clockwise rotation of the original box.\n4. Return the Result:\n    - Return the rotated grid containing the transformed and rotated box.\n\n# Complexity\n- Time complexity:\n$$O(m\xD7n)$$\n\n- Space complexity:\n$$O(m\xD7n)$$\n\n# Code\n```python3 []\nclass Solution:\n    def rotateTheBox(self, box: List[List[str]]) -> List[List[str]]:\n        m, n = len(box), len(box[0])\n        rotated = [[\'.\'] * m for _ in range(n)]\n        for row in range(m):\n            num = n - 1\n            for col in range(n - 1, -1, -1):\n                if box[row][col] == \'*\':\n                    rotated[col][m - 1 - row] = \'*\'\n                    num = col - 1\n                elif box[row][col] == \'#\':\n                    rotated[num][m - 1 - row] = \'#\'\n                    num -= 1\n        return rotated        \n```
['Array', 'Two Pointers', 'Matrix', 'Python3']