[코테 풀이] Projection Area of 3D Shapes

Q_883) Projection Area of 3D Shapes

출처 : https://leetcode.com/problems/projection-area-of-3d-shapes/

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.

class Solution {
   public int projectionArea(int[][] grid) {
       int x = 0, y = 0, z = 0;
       for (int i = 0; i < grid.length; i++) {
           int maxY = 0;
           for (int j = 0; j < grid[i].length; j++) {
               if (grid[i][j] != 0) x++;
               if (maxY < grid[i][j]) maxY = grid[i][j];
           }
           y += maxY;
       }
       for (int q = 0; q < grid.length; q++) {
           int maxZ = 0;
           for (int p = 0; p < grid.length; p++) {
               if (grid[p][q] > maxZ) maxZ = grid[p][q];
           }
           z += maxZ;
       }
       return x + y + z;
   }
}