#883

Easy

LeetCode Problem Workspace

Projection Area of 3D Shapes

Calculate the projection area of a 3D shape defined by a grid of towers with varying heights.

Problem Statement

You are given a grid representing a 3D shape where each value in the grid corresponds to the height of a tower of cubes. The grid is of size n x n, and each value grid[i][j] indicates the height of a tower at position (i, j).

Your goal is to compute the projection area of the shape when viewed from three orthogonal planes: the xy-plane, the yz-plane, and the zx-plane. The total projection area is the sum of the areas projected onto these planes.

Examples

Example 1

Input: grid = [[1,2],[3,4]]

Output: 17

Here are the three projections ("shadows") of the shape made with each axis-aligned plane.

Example 2

Input: grid = [[2]]

Output: 5

Example details omitted.

Example 3

Input: grid = [[1,0],[0,2]]

Output: 8

Example details omitted.

Constraints

n == grid.length == grid[i].length
1 <= n <= 50
0 <= grid[i][j] <= 50

Solution Approach

Sum Projections on the xy-Plane

To compute the projection onto the xy-plane, simply sum the number of cubes visible from each cell of the grid. This is done by adding the maximum height in each column.

Sum Projections on the yz-Plane

For the yz-plane, we need to consider the tallest tower in each row. The projection for this plane is the sum of the maximum heights from each row.

Sum Projections on the zx-Plane

The projection onto the zx-plane is calculated similarly to the yz-plane but column-wise. For each column, the maximum height value is added to the total projection area.

Complexity Analysis

Metric	Value
Time	O(N^2)
Space	O(1)

The time complexity of this solution is O(N^2) because we need to iterate over the entire grid to compute the projections on each plane. The space complexity is O(1) as the solution only uses a constant amount of extra space for intermediate calculations.

What Interviewers Usually Probe

Candidate can efficiently compute projections in O(N^2) time.
Candidate uses appropriate methods to calculate max values in rows and columns.
Candidate avoids unnecessary space usage, demonstrating space optimization.

Common Pitfalls or Variants

Common pitfalls

Confusing the dimensions of the projections and treating them incorrectly.
Misunderstanding the projection onto the yz- and zx-planes, treating them the same.
Failing to handle edge cases, such as grid size 1x1 or the presence of zero heights.

Follow-up variants

Handle grids of different sizes and edge cases, such as grids with minimum values.
Extend the problem to higher-dimensional projections, considering additional axes.
Optimize further for larger grids, testing performance for n up to 50.

FAQ

How can I calculate the projection area of a 3D shape?

To calculate the projection area, you sum the maximum heights in each row and column for the yz- and zx-planes, and the number of non-zero elements for the xy-plane.

What is the time complexity of the Projection Area of 3D Shapes problem?

The time complexity is O(N^2) because we must iterate through the grid to compute the projections on each plane.

How do I handle zero heights in the grid for the projection?

Zero heights in the grid should be ignored for the xy-plane projection, as they do not contribute to the area. They still affect the yz- and zx-planes by ensuring that the maximum height is correctly calculated.

What is the space complexity of this problem?

The space complexity is O(1), as the solution only requires constant space for intermediate calculations.

What are the variants of the Projection Area of 3D Shapes problem?

Variants include handling larger grids, optimizing performance for grid sizes up to 50, and extending the problem to higher-dimensional projections.

terminal

Solution

Solution 1: Mathematics

We can calculate the area of the three projections separately.

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class Solution:
    def projectionArea(self, grid: List[List[int]]) -> int:
        xy = sum(v > 0 for row in grid for v in row)
        yz = sum(max(row) for row in grid)
        zx = sum(max(col) for col in zip(*grid))
        return xy + yz + zx

Continue Practicing

#892 surface area of 3d shapes #1030 matrix cells in distance order #840 magic squares in grid