Rotating the Box

This approach cleverly reorders the operations. We observe that simulating gravity in the final rotated grid is equivalent to simulating gravity horizontally (to the right) in the original grid. By applying this horizontal gravity first, we can modify the input grid boxGrid in-place. After all stones in each row have 'settled' to the right, we then perform the 90-degree clockwise rotation to get the final configuration. This avoids the need for a full intermediate matrix for the initial rotation.

1. Realize that a 90-degree clockwise rotation turns the horizontal 'right' direction into the vertical 'down' direction.
2. Instead of rotating first, apply gravity horizontally on the original boxGrid. This means shifting all stones (#) in each row to the rightmost possible positions, respecting obstacles (*).
3. Iterate through each row i of boxGrid.
4. For each row, use a write_pos pointer, initialized to the last column index (n - 1), to track the rightmost available spot for a stone.
5. Iterate through the row from right to left (j from n - 1 down to 0).
6. If box[i][j] is an obstacle '*', it blocks movement. Reset write_pos to j - 1.
7. If box[i][j] is a stone '#, move it to box[i][write_pos], set box[i][j] to '.' (if j != write_pos), and decrement write_pos.
8. After this in-place modification, the boxGrid now represents the state after horizontal gravity.
9. Create the final n x m result matrix.
10. Populate the result matrix by rotating the modified boxGrid: result[j][m - 1 - i] = box[i][j].
11. Return the result matrix.

The core idea is to handle the gravity effect before rotation. For each row in the m x n boxGrid, we make the stones fall to the 'right'. This can be done efficiently in-place. We iterate through each row from right to left. A write_pos pointer keeps track of where the next stone from the left should be placed. When we see a stone, we place it at write_pos and move write_pos one step to the left. When we see an obstacle, it acts as a wall, and write_pos is reset to the position just to the left of the obstacle.

After applying this transformation to all rows of boxGrid, we proceed with the rotation. A new n x m matrix result is created. We then perform the standard 90-degree clockwise rotation, copying elements from the modified boxGrid to the result matrix. This method is more space-efficient because the gravity simulation is done in-place, eliminating the need for an extra n x m matrix used in the first approach.

class Solution {
    public char[][] rotateTheBox(char[][] box) {
        int m = box.length;
        int n = box[0].length;

        // Step 1: Apply gravity to each row (stones fall to the right)
        for (int i = 0; i < m; i++) {
            int writePos = n - 1;
            for (int j = n - 1; j >= 0; j--) {
                if (box[i][j] == '*') {
                    writePos = j - 1;
                } else if (box[i][j] == '#') {
                    box[i][j] = '.';
                    box[i][writePos] = '#';
                    writePos--;
                }
            }
        }

        // Step 2: Rotate the modified box
        char[][] result = new char[n][m];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                result[j][m - 1 - i] = box[i][j];
            }
        }

        return result;
    }
}