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

implSolution{pubfnprojection_area(grid:Vec<Vec<i32>>) -> i32{letmut top = 0;letmut front = 0;letmut side = 0;for x in0..grid.len(){letmut front_max = 0;letmut side_max = 0;for y in0..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 }}