Surface Area of 3D Shapes

You are given an n x n grid where you have placed some 1 x 1 x 1 cubes. Each value v = grid[i][j] represents a tower of v cubes placed on top of cell (i, j). After placing these cubes, you have decided to glue any directly adjacent cubes to each other, forming several irregular 3D shapes. Return the total surface area of the resulting shapes. Note: The bottom face of each shape counts toward its surface area.

For example:

grid = [[1,2],[3,4]] should return 34 grid = [[1,1,1],[1,0,1],[1,1,1]] should return 32 grid = [[2,2,2],[2,1,2],[2,2,2]] should return 46

Can you write a function to solve this problem efficiently?

def surfaceArea(grid):
    n = len(grid)
    area = 0

    for i in range(n):
        for j in range(n):
            if grid[i][j] > 0:
                area += 2  # Top and bottom surfaces
                area += grid[i][j] * 4  # Initial side surfaces

                # Subtract overlapping surfaces
                if i > 0:
                    area -= min(grid[i][j], grid[i - 1][j])
                if i < n - 1:
                    area -= min(grid[i][j], grid[i + 1][j])
                if j > 0:
                    area -= min(grid[i][j], grid[i][j - 1])
                if j < n - 1:
                    area -= min(grid[i][j], grid[i][j + 1])

    return area

# Example usage
grid1 = [[1, 2], [3, 4]]
print(surfaceArea(grid1))  # Output: 34

grid2 = [[1, 1, 1], [1, 0, 1], [1, 1, 1]]
print(surfaceArea(grid2))  # Output: 32

grid3 = [[2, 2, 2], [2, 1, 2], [2, 2, 2]]
print(surfaceArea(grid3))  # Output: 46

Explanation:

1. Initialization: Initialize the surface area to 0.
2. Iterate through the grid: Loop through each cell in the grid.
3. Base Area: For each cell (i, j) with a tower of height v = grid[i][j], add 2 to the area (top and bottom surfaces).
4. Side Area: Add 4 * v to the area for the initial side surfaces.
5. Subtract Overlapping Areas: Check adjacent cells (up, down, left, right) and subtract the minimum of the heights between the current cell and the adjacent cell from the area. This accounts for the shared faces between adjacent towers.
6. Return: After iterating through all cells, return the total surface area.

Big(O) Run-time Analysis:

The code iterates through each cell of the n x n grid once. For each cell, it performs a constant number of operations (additions, subtractions, and comparisons). Therefore, the time complexity is O(n^2), where n is the size of the grid.

Big(O) Space Usage Analysis:

The code uses a constant amount of extra space, regardless of the size of the input grid. It only uses a few variables to keep track of the area and loop indices. Therefore, the space complexity is O(1).

Edge Cases and How to Handle Them:

1. Empty Grid: If the grid is empty (i.e., n = 0), the code will still work correctly, returning a surface area of 0, because the loops will not execute.
2. Zero Heights: If a cell has a value of 0 (i.e., grid[i][j] = 0), it does not contribute to the surface area, and the code correctly skips adding any area for that cell.
3. Grid Boundaries: The code checks for boundaries before accessing adjacent cells to avoid index out-of-bounds errors.
4. Large Heights: If the heights in the grid are very large, the area can become large as well. However, since the problem constraints limit the heights to be at most 50, integer overflow is not a concern in most programming languages.