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

Python Code
1
2
3
4
5
6

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

Java Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19

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;
    }
}

C++ Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17

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;
    }
};

Go Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

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
}

TypeScript Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

function projectionArea(grid: number[][]): number {
    const n = grid.length;
    let res = grid.reduce((r, v) => r + v.reduce((r, v) => r + (v === 0 ? 0 : 1), 0), 0);
    for (let i = 0; i < n; i++) {
        let xMax = 0;
        let yMax = 0;
        for (let j = 0; j < n; j++) {
            xMax = Math.max(xMax, grid[i][j]);
            yMax = Math.max(yMax, grid[j][i]);
        }
        res += xMax + yMax;
    }
    return res;
}

Rust Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

impl Solution {
    pub fn projection_area(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut res = 0;
        let mut x_max = vec![0; n];
        let mut y_max = vec![0; n];
        for i in 0..n {
            for j in 0..n {
                let val = grid[i][j];
                if val == 0 {
                    continue;
                }
                res += 1;
                x_max[i] = x_max[i].max(val);
                y_max[j] = y_max[j].max(val);
            }
        }
        res + y_max.iter().sum::<i32>() + x_max.iter().sum::<i32>()
    }
}