883. Projection Area of 3D Shapes

Description

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

Solutions

Solution 1: Mathematics

We can calculate the area of the three projections separately.

Projection area on the xy plane: Each non-zero value will be projected onto the xy plane, so the projection area on the xy plane is the count of non-zero values.
Projection area on the yz plane: The maximum value in each row.
Projection area on the zx plane: The maximum value in each column.

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 $O(1)$.

Python3JavaC++GoTypeScriptRust

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

class Solution {
    public int projectionArea(int[][] grid) {
        int xy = 0, yz = 0, zx = 0;
        for (int i = 0, n = grid.length; i < n; ++i) {
            int maxYz = 0;
            int maxZx = 0;
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] > 0) {
                    ++xy;
                }
                maxYz = Math.max(maxYz, grid[i][j]);
                maxZx = Math.max(maxZx, grid[j][i]);
            }
            yz += maxYz;
            zx += maxZx;
        }
        return xy + yz + zx;
    }
}

class Solution {
public:
    int projectionArea(vector<vector<int>>& grid) {
        int xy = 0, yz = 0, zx = 0;
        for (int i = 0, n = grid.size(); i < n; ++i) {
            int maxYz = 0, maxZx = 0;
            for (int j = 0; j < n; ++j) {
                xy += grid[i][j] > 0;
                maxYz = max(maxYz, grid[i][j]);
                maxZx = max(maxZx, grid[j][i]);
            }
            yz += maxYz;
            zx += maxZx;
        }
        return xy + yz + zx;
    }
};

func projectionArea(grid [][]int) int {
    xy, yz, zx := 0, 0, 0
    for i, row := range grid {
        maxYz, maxZx := 0, 0
        for j, v := range row {
            if v > 0 {
                xy++
            }
            maxYz = max(maxYz, v)
            maxZx = max(maxZx, grid[j][i])
        }
        yz += maxYz
        zx += maxZx
    }
    return xy + yz + zx
}

function projectionArea(grid: number[][]): number {
    const xy: number = grid.flat().filter(v => v > 0).length;
    const yz: number = grid.reduce((acc, row) => acc + Math.max(...row), 0);
    const zx: number = grid[0]
        .map((_, i) => Math.max(...grid.map(row => row[i])))
        .reduce((acc, val) => acc + val, 0);
    return xy + yz + zx;
}

impl Solution {
    pub fn projection_area(grid: Vec<Vec<i32>>) -> i32 {
        let xy: i32 = grid
            .iter()
            .map(|row| row.iter().filter(|&&v| v > 0).count() as i32)
            .sum();
        let yz: i32 = grid.iter().map(|row| *row.iter().max().unwrap_or(&0)).sum();
        let zx: i32 = (0..grid[0].len())
            .map(|i| grid.iter().map(|row| row[i]).max().unwrap_or(0))
            .sum();
        xy + yz + zx
    }
}

883. Projection Area of 3D Shapes

Description

Solutions

Solution 1: Mathematics

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