How do you solve Surface Area of 3D Shapes on LeetCode?

Surface Area of 3D Shapes (LeetCode #892) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Each cube contributes to its own surface area, but adjacent cubes share surfaces that need to be subtracted.

What is the brute force solution for Surface Area of 3D Shapes?

The brute force approach for Surface Area of 3D Shapes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach calculates the surface area of each cube independently and then accounts for overlaps with adjacent cubes. This is straightforward but can be inefficient as it doesn't optimize for shared surfaces. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Surface Area of 3D Shapes?

Surface Area of 3D Shapes uses the following concepts: Array, Math, Geometry, Matrix. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Surface Area of 3D Shapes?

Always explain your thought process while coding; it helps the interviewer understand your logic. Consider edge cases, such as grids with all zeros or maximum heights. Practice writing clean and efficient code, as readability is often evaluated in interviews.

What are common mistakes when solving Surface Area of 3D Shapes?

Forgetting to account for the bottom face of the cubes. Not properly handling the edges of the grid, leading to index errors.

#892

Surface Area of 3D Shapes

Easy

ArrayMathGeometryMatrixArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(1)

💡

Intuition

Time O(n²)Space O(1)

The optimal solution leverages the same principles as the brute-force approach but minimizes redundant calculations by directly calculating the contributions from each cube's height and its neighbors, reducing unnecessary checks.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable 'surface_area' to 0.
2Step 2: Loop through each cell in the grid. For each cell, calculate the total surface area contribution.
3Step 3: For each cube, add 4 times its height plus 2 for the top and bottom.
4Step 4: Subtract the contributions from adjacent cells only if they exist.
5Step 5: Return the total surface area.

solution.py16 lines

1def surfaceArea(grid):
2    n = len(grid)
3    surface_area = 0
4    for i in range(n):
5        for j in range(n):
6            if grid[i][j] > 0:
7                surface_area += 4 * grid[i][j] + 2
8                if i > 0:
9                    surface_area -= min(grid[i][j], grid[i-1][j])
10                if j > 0:
11                    surface_area -= min(grid[i][j], grid[i][j-1])
12                if i < n - 1:
13                    surface_area -= min(grid[i][j], grid[i+1][j])
14                if j < n - 1:
15                    surface_area -= min(grid[i][j], grid[i][j+1])
16    return surface_area

ℹ

Complexity note: The time complexity remains O(n²) since we still iterate through each cell in the grid. The space complexity is O(1) as we only use a few variables to store the surface area.

1Each cube contributes to its own surface area, but adjacent cubes share surfaces that need to be subtracted.
2The bottom face of each cube is always counted, regardless of its position.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.