883. Projection Area of 3D Shapes

Problem Description

In this problem, we are given an n x n grid where each cell in the grid contains a non-negative integer value representing the height of a tower of cubes placed in that cell. The task is to calculate the total surface area of all three orthogonal projections (top-down, front, and side) of the arrangement of the cubes.

To understand this better, imagine looking at the cubes from three different viewpoints: from directly above (the xy-plane), from the front (the yz-plane), and from the side (the zx-plane). The area of the projection is simply the shadow of these cubes on the respective planes. The top-down view counts all the cubes that are visible from above, the front view counts the tallest tower in each row, and the side view counts the tallest tower in each column.

Intuition

For the solution, we can tackle each projection separately and sum up their areas:

• xy-plane (Top-Down View): To compute the area of the projection on the xy-plane, we count the number of cubes that are visible from the top. Each cube contributes to the area if its height is greater than zero. We simply iterate over the grid and count the number of cells with a value greater than zero.

• yz-plane (Front View): For the yz-plane, we are looking at the grid from the front, so we need to consider the tallest tower in each row since only the tallest tower is visible from this viewpoint. We iterate through each row and take the maximum value of the row, representing the height of the tallest tower.

• zx-plane (Side View): Similarly, for the zx-plane projection, which represents the view from the side, we need to find the tallest tower in each column. We can achieve this by zipping the columns together and finding the maximum height of each column.

The final area of all projections is the sum of the three calculated areas: the top-down, the front, and the side projection areas.

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Solution Approach

The implementation of the solution uses simple yet effective techniques to calculate the projection areas on all three planes. The choice of data structures is straightforward - since the input is a grid (a list of lists in Python), we use this same structure to operate on the data. Here's how the solution is structured:

• xy-plane (Top-Down View): The xy variable is calculated by using a nested list comprehension. The outer loop iterates over each row, and the inner loop iterates over each value in the row, checking if v > 0. This check effectively counts the number of cells with cubes visible from the top view, contributing to the xy area. The use of list comprehension makes this part compact and efficient.

  1xy = sum(v > 0 for row in grid for v in row)  

• yz-plane (Front View): The yz projection is calculated by iterating over each row in the grid and selecting the maximum value found in that row using the max function. This represents the highest tower when looking from the front, and the sum of these maximum values across all rows gives the yz area.

  1yz = sum(max(row) for row in grid)

• zx-plane (Side View): To find the zx projection, we need the tallest tower in each column. Python's zip function is used here, which effectively transposes the grid, grouping all elements at similar column indices together. By iterating over these transposed columns and selecting the maximum value found in each, we compute the zx area.

  1zx = sum(max(col) for col in zip(*grid))

After computing the areas individually, we sum them to get the total area of all three projections, which is returned as the final result. The algorithm's complexity is linear in the number of cells in the grid, making it O(n^2), where n is the dimension of the grid. The solution is not only intuitive but also showcases the power and readability of Python's list comprehensions and built-in functions.

No complex patterns or algorithms are necessary for this problem, as it's more about understanding what each projection looks like and selecting the appropriate elements from the grid to count. The solution, therefore, combines basic iteration and aggregation methods to achieve an efficient and understandable result.

1class Solution:
2    def projectionArea(self, grid: List[List[int]]) -> int:
3        xy = sum(v > 0 for row in grid for v in row)
4        yz = sum(max(row) for row in grid)
5        zx = sum(max(col) for col in zip(*grid))
6        return xy + yz + zx

Example Walkthrough

Let's go through an example to illustrate the solution approach using a 3 x 3 grid as an example:

Suppose we have the following grid representing the heights of the towers of cubes:

1grid = [
2    [1, 2, 3],
3    [4, 5, 6],
4    [7, 8, 9]
5]

We will walk through the solution by calculating the area of each orthogonal projection.

Top-Down View (xy-plane):

First, we look at how many cubes contribute to the top view. Since all cubes are visible from the top, we count all cells with a height greater than zero.

11 2 3
24 5 6
37 8 9

From this, we can see that all 9 cells contribute to the top view (as all heights are non-zero). Thus, the xy area is 9.

Front View (yz-plane):

Now, we look at each row to find the tallest tower, as this would be visible from the front.

11 2 3  <- Tallest in 1st row is 3
24 5 6  <- Tallest in 2nd row is 6
37 8 9  <- Tallest in 3rd row is 9

The heights of the tallest towers are 3, 6, and 9, and the yz area is the sum of these which is 3 + 6 + 9 = 18.

Side View (zx-plane):

For the side view, we need the tallest tower in each column. By looking at each column:

11
24
37 <- Tallest in 1st column is 7
4
52
65
78 <- Tallest in 2nd column is 8
8
93
106
119 <- Tallest in 3rd column is 9

The tallest towers from each column have heights 7, 8, and 9. Thus, the zx area is the sum of these which is 7 + 8 + 9 = 24.

Finally, add up all the areas to get the total surface area of all projections:

1xy + yz + zx = 9 + 18 + 24 = 51

So according to our solution, the final result for the given 3 x 3 grid is 51.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(N^2) where N is the number of rows (or columns, assuming the grid is square). This is due to the three separate loops that iterate over all N*N elements of the grid or its transposition. The first loop iterates over all elements for the xy shadow, the second loop iterates over each row to find the maximum for yz shadow, and the third loop iterates over each transposed column to find the maximum for the zx shadow.

The space complexity of the code is O(N) because of the space required for the transposition of the grid (zip(*grid)). This creates a list of tuples where each tuple is a column from the original grid. The space is linear with the number of columns N.

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