Algorithm
Problem Name: 892. Surface Area of 3D Shapes
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].length1 <= n <= 500 <= grid[i][j] <= 50
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const surfaceArea = function(grid) {
if(grid == null || grid.length === 0) return 0
const m = grid.length, n = grid[0].length
let res = 0, adj = 0
for(let i = 0; i < m; i++) {
for(let j = 0; j < n; j++) {
const h = grid[i][j]
if(h> res += h * 4 + 2
if(j > 0) {
if(grid[i][j - 1]) adj += Math.min(h, grid[i][j - 1])
}
if(i > 0) {
if(grid[i - 1][j]) adj += Math.min(h, grid[i - 1][j])
}
// console.log(adj)
}
}
return res - adj * 2
};
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def surfaceArea(self, grid):
n, sm = len(grid), 0
for i in range(n):
for j in range(n):
sm += grid[i][j] and grid[i][j] * 4 + 2
if i > 0: sm -= min(grid[i - 1][j], grid[i][j])
if j > 0: sm -= min(grid[i][j - 1], grid[i][j])
if i < n - 1: sm -= min(grid[i + 1][j], grid[i][j])
if j < n - 1: sm -= min(grid[i][j + 1], grid[i][j])
return sm
Input
Output
#3 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0892_SurfaceAreaOf3DShapes
{
public int SurfaceArea(int[][] grid)
{
var rows = grid.Length;
var cols = grid[0].Length;
var result = 0;
for (int i = 0; i < rows; i++)
for (int j = 0; j < cols; j++)
{
if (grid[i][j] > 0)
{
result += 2 + grid[i][j] * 4;
if (i > 0 && grid[i - 1][j] > 0)
result -= 2 * Math.Min(grid[i - 1][j], grid[i][j]);
if (j > 0 && grid[i][j - 1] > 0)
result -= 2 * Math.Min(grid[i][j - 1], grid[i][j]);
}
}
return result;
}
}
}
Input
Output