Rotating the Box

class Solution {
public:
    vector<vector<char>> transformBox(vector<vector<char>>& container) {
        int rows = container.size();
        int cols = container[0].size();
        vector<vector<char>> rotatedContainer(cols, vector<char>(rows, '.'));
    
        // Process each row for possible movement of stones
        for (int i = 0; i < rows; i++) {
            int availableSpace = cols - 1;
            // Process the row from end to start
            for (int j = cols - 1; j >= 0; j--) {
                // Stone found; move it to the available space
                if (container[i][j] == '#') {
                    rotatedContainer[availableSpace][rows - 1 - i] = '#';
                    availableSpace--;
                }
                // Obstacle found; next space becomes just above the obstacle
                if (container[i][j] == '*') {
                    rotatedContainer[j][rows - 1 - i] = '*';
                    availableSpace = j - 1;
                }
            }
        }
        return rotatedContainer;
    }
};

The provided solution in C++ effectively handles the task of rotating a matrix, which represents a container, 90 degrees clockwise.

• Start by defining the size of the container using rows and cols.
• Create a new matrix rotatedContainer with dimensions switched (i.e., cols x rows) and initially filled with '.' to indicate empty spaces.

The rotation involves gravity affecting the stones (represented by #) in the container such that they fall to the bottom:

1. Iterate through each row starting from the bottom-most row to simulate the effect of gravity.
2. Maintain a variable availableSpace that tracks where the next stone should fall within a column after rotation.
3. Traverse each row from right to left:
   • When encountering a stone (#), place it in the rotatedContainer at the current availableSpace index, then decrement availableSpace.
   • When finding an obstacle (*), set the obstacle in the correct position in rotatedContainer. Reset availableSpace to the position just above the newly placed obstacle.

After processing each row and column using the above logic, the function returns rotatedContainer, which will be the input container rotated 90 degrees clockwise with all stones settled at the bottom of their respective columns, obeying the obstacles' positions. This process efficiently simulates the physical behavior of the stones and obstacles during rotation.

class Solution {
    
    public char[][] rotateBox(char[][] box) {
        int rows = box.length;
        int columns = box[0].length;
        char[][] rotatedBox = new char[columns][rows];
    
        // Fill the rotatedBox with empty spaces initially
        for (int i = 0; i < columns; i++) {
            for (int j = 0; j < rows; j++) {
                rotatedBox[i][j] = '.';
            }
        }
    
        // Move stones down according to gravity
        for (int i = 0; i < rows; i++) {
            int emptySpace = columns - 1;
            // Iterate over columns of box from last to first
            for (int j = columns - 1; j >= 0; j--) {
                // Move stones down
                if (box[i][j] == '#') {
                    rotatedBox[emptySpace][rows - 1 - i] = '#';
                    emptySpace--;
                }
                // Place obstacles and reset the empty space above the obstacle
                if (box[i][j] == '*') {
                    rotatedBox[j][rows - 1 - i] = '*';
                    emptySpace = j - 1;
                }
            }
        }
        return rotatedBox;
    }
}

In this Java program, you'll transform a matrix representing a box by rotating it 90 degrees clockwise. The challenge primarily revolves around implementing gravity for stones within the box ('#'), leaving obstacles ('*') intact, and ensuring empty spaces ('.') are correctly positioned.

Here’s how the rotation and gravity are managed:

• Define a matrix rotatedBox with dimensions switched from the original (column count becomes row count and vice versa) and initialize all its cells with '.' to represent empty spaces.

• Traverse through each row of the original matrix from the first row to the last. Inside each row, iterate from the last column to the first to simulate the effect of gravity when the box is rotated.

• During each iteration, if a stone '#' is encountered, place it in the lowest possible position in the rotatedBox. This position is tracked using the emptySpace variable, which is decremented each time a stone is placed, effectively "dropping" the stone to the bottom of the rotated box.

• If an obstacle '*' is encountered, place it directly in the corresponding position in rotatedBox and reset the emptySpace to just above this obstacle, replicating the obstacle’s blocking effect on any stones above it in the original orientation.

• Continue the above steps for each element in every row, thus achieving a 90-degree rotation of the box with gravity properly simulated for stones.

Upon completing the process, rotatedBox contains the rotated version of the original box with all the rules regarding stones, obstacles, and empty spaces applied correctly. This matrix is what the function returns, showcasing a successful transformation respecting both rotation and internal box mechanics.

class Solution:
    def transformBox(self, box):
        rows = len(box)
        cols = len(box[0])
        transformed = [["." for _ in range(rows)] for _ in range(cols)]
    
        for row in range(rows):
            available_spot = cols - 1
            for col in range(cols - 1, -1, -1):
                if box[row][col] == "#":
                    transformed[available_spot][rows - row - 1] = "#"
                    available_spot -= 1
                elif box[row][col] == "*":
                    transformed[col][rows - row - 1] = "*"
                    available_spot = col - 1
    
        return transformed

The provided solution in Python tackles the problem of rotating a box which contains stones ('#'), empty spaces ('.'), and obstacles ('*'). The rotation performed is 90 degrees clockwise. Here's a breakdown of how the solution operates:

• Initializes the dimensions of the box with rows for the number of rows and cols for the number of columns.
• Creates a transformed box initialized with periods (representing empty space) where its dimensions are flipped compared to the original (rows become columns and vice versa).
• Iterates through each element of the original box starting from the bottom to manage gravity effect after rotation:
  • If a stone ('#') is found, it is placed in the lowest available spot within the column of the transformed box, considering the reversed order due to the rotation.
  • When an obstacle ('*') is encountered, it's placed in the corresponding rotated position, and the spot below the obstacle is set as the next available spot for placing a stone.
• Returns the transformed box where stones have "fallen" to the bottom of their respective new columns due to gravity, and obstacles remain in place, blocking any stones that appear above them.

Instance the public function transformBox in the class Solution to utilize this algorithm effectively. Ensure the input box is a 2D list representation where dimensions and characters are correctly defined for meaningful results.