README.md

883. Projection Area of 3D Shapes

On a N * N grid, we place some 1 * 1 * 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 grid cell (i, j).

Now 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.

Here, 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: [[2]]
Output: 5

Example 2:

Input: [[1,2],[3,4]]
Output: 17
Explanation:
Here are the three projections ("shadows") of the shape made with each axis-aligned plane.

Example 3:

Input: [[1,0],[0,2]]
Output: 8

Example 4:

Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 14

Example 5:

Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 21

Note:

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

Solutions (Rust)

1. Mathematical

impl Solution {
    pub fn projection_area(grid: Vec<Vec<i32>>) -> i32 {
        let mut top = 0;
        let mut front = 0;
        let mut side = 0;

        for x in 0..grid.len() {
            let mut front_max = 0;
            let mut side_max = 0;

            for y in 0..grid[0].len() {
                if grid[x][y] > 0 {
                    top += 1;
                }
                front_max = front_max.max(grid[x][y]);
                side_max = side_max.max(grid[y][x]);
            }

            front += front_max;
            side += side_max;
        }

        top + front + side
    }
}