892. Surface Area of 3D Shapes

Problem

You are given an n x n grid where you have placed some 1 x 1 x 1 cubes. Each value v = grid[i][j] represents a tower of v cubes placed on top of cell (i, j).

After placing these cubes, you have decided to glue any directly adjacent cubes to each other, forming several irregular 3D shapes.

Return the total surface area of the resulting shapes.

Note: The bottom face of each shape counts toward its surface area.

  Example 1:

Input: grid = [[1,2],[3,4]]
Output: 34

Example 2:

Input: grid = [[1,1,1],[1,0,1],[1,1,1]]
Output: 32

Example 3:

Input: grid = [[2,2,2],[2,1,2],[2,2,2]]
Output: 46

  Constraints:

• n == grid.length == grid[i].length

• 1 <= n <= 50

• 0 <= grid[i][j] <= 50

Solution (Java)

class Solution {
    public int surfaceArea(int[][] grid) {
        int surfaceArea = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] > 0) {
                    surfaceArea += 4 * grid[i][j] + 2;
                    surfaceArea -= hiddenSides(i, j, grid);
                }
            }
        }
        return surfaceArea;
    }

    private int hiddenSides(int i, int j, int[][] grid) {
        int hidden = 0;
        int tower = grid[i][j];
        if (j + 1 < grid[i].length && grid[i][j + 1] > 0) {
            hidden += Math.min(tower, grid[i][j + 1]);
        }
        if (j - 1 >= 0 && grid[i][j - 1] > 0) {
            hidden += Math.min(tower, grid[i][j - 1]);
        }
        if (i + 1 < grid.length && grid[i + 1][j] > 0) {
            hidden += Math.min(tower, grid[i + 1][j]);
        }
        if (i - 1 >= 0 && grid[i - 1][j] > 0) {
            hidden += Math.min(tower, grid[i - 1][j]);
        }
        return hidden;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).