You are given an n x n grid where we place some 1 x 1 x 1 cubes that are axis-aligned with the x, y, and z axes.
Each value v = grid[i][j] represents a tower of v cubes placed on top of the cell (i, j).
We view the projection of these cubes onto the xy, yz, and zx planes.
A projection is like a shadow, that maps our 3-dimensional figure to a 2-dimensional plane. We are viewing the "shadow" when looking at the cubes from the top, the front, and the side.
Return the total area of all three projections.
Example 1:
Input: grid = [[1,2],[3,4]]
Output: 17
Explanation: Here are the three projections ("shadows") of the shape made with each axis-aligned plane.
Example 2:
Input: grid = [[2]] Output: 5
Example 3:
Input: grid = [[1,0],[0,2]] Output: 8
Constraints:
n == grid.length == grid[i].length1 <= n <= 500 <= grid[i][j] <= 50Problem Overview: You are given an n x n grid where each cell represents a stack of cubes. The task is to compute the total projection area when the 3D structure is projected onto the XY (top), YZ (front), and ZX (side) planes. Each projection reveals different visible surfaces based on cube heights.
Approach 1: Simulation of Projections (O(m*n) time, O(1) space)
Directly simulate what each projection would look like. The top projection (XY plane) counts how many cells contain at least one cube. Iterate through the grid and increment the area whenever grid[i][j] > 0. The front projection (YZ plane) depends on the tallest stack in each row, so compute the maximum value in every row and add them together. The side projection (ZX plane) depends on the tallest stack in each column, which you can compute by scanning column-wise while iterating the grid.
This method works because each projection reduces the 3D structure to a simple 2D visibility rule: presence for the top view, row maximum for the front view, and column maximum for the side view. A single nested loop can compute the top area and track row/column maximums simultaneously. The approach relies only on simple iteration and comparisons over a matrix, making it straightforward and efficient.
Approach 2: Pre-compute Row and Column Maximums (O(m*n) time, O(m+n) space)
Instead of calculating projections during traversal, pre-compute auxiliary arrays. First iterate through the grid and maintain two arrays: rowMax[i] for the tallest stack in each row and colMax[j] for the tallest stack in each column. At the same time, count the number of non‑zero cells for the top projection. Once traversal is complete, sum all values in rowMax to get the front projection and sum colMax for the side projection.
This separation improves clarity because each projection becomes an independent calculation derived from precomputed information. The technique is common when working with arrays and matrix traversal problems where row and column aggregates are reused. Memory usage increases slightly due to the extra arrays, but the logic becomes easier to reason about.
Recommended for interviews: The simulation approach is typically expected. It demonstrates that you understand how each geometric projection works and how to derive them using simple iteration over the grid. The row/column pre-computation variant shows stronger organization and can make the code cleaner, but both approaches run in the same optimal O(m*n) time using basic math and geometry reasoning.
In this approach, we calculate each plane's projection separately, then sum them up for the total projection area. For the XY-plane projection, we count every cell with a positive value. For the YZ-plane, we determine the maximum value in each row (representing the view from the side). For the ZX-plane, we find the maximum value in each column (representing the view from the front).
This C implementation uses nested loops to traverse the grid and calculate projection areas separately for each plane, accumulating the resulting areas for the total projection.
Time Complexity: O(n^2) since we traverse each grid element once.
Space Complexity: O(1) since we use no additional space proportional to the input size.
This approach improves upon the earlier method by precomputing the maximum values in each row and column before calculating the projections. By storing these values in separate arrays, it helps avoid repeatedly calculating max values within nested loops for each axis projection.
This C solution uses arrays to store the maximums of rows and columns, improving efficiency slightly from not recalculating these values repeatedly.
Time Complexity: O(n^2) to fill the grid, row, and column arrays.
Space Complexity: O(n) to store column maximums.
We can calculate the area of the three projections separately.
Finally, add up the three areas.
The time complexity is O(n^2), where n is the side length of the grid grid. The space complexity is O(1).
| Approach | Complexity |
|---|---|
| Simulation of Projections | Time Complexity: O(n^2) since we traverse each grid element once. |
| Pre-compute Column and Row Maximums | Time Complexity: O(n^2) to fill the grid, row, and column arrays. |
| Mathematics | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Simulation of Projections | O(m*n) | O(1) | Best general solution. Minimal memory usage and easy to compute projections during traversal. |
| Pre-compute Row and Column Maximums | O(m*n) | O(m+n) | When cleaner code is preferred or row/column aggregates are reused later. |
LeetCode Algorithms Easy: Projection Area of 3D Shapes • Mike the Coder • 2,266 views views
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