892. Surface Area of 3D Shapes #

Problem #

On a N * N grid, we place some 1 * 1 * 1 cubes.

Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j).

Return the total surface area of the resulting shapes.

Example 1:

Input: [[2]]
Output: 10

Example 2:

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

Example 3:

Input: [[1,0],[0,2]]
Output: 16

Example 4:

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

Example 5:

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

Note:

• 1 <= N <= 50
• 0 <= grid[i][j] <= 50

Problem Summary #

On an N * N grid, we place some 1 * 1 * 1  cubes. Each value v = grid[i][j] represents v cubes stacked on the corresponding cell (i, j). Please return the surface area of the final shape.

Solution Approach #

• Given a grid array, where the values represent cubes stacked on their cells, find the surface area of the final stacked cubes.
• Easy problem. According to the problem statement, find the overlapping faces when stacking, then subtract these overlapping areas from the total surface area to get the final answer.

Code #

package leetcode

func surfaceArea(grid [][]int) int {
	area := 0
	for i := 0; i < len(grid); i++ {
		for j := 0; j < len(grid[0]); j++ {
			if grid[i][j] == 0 {
				continue
			}
			area += grid[i][j]*4 + 2
			// up
			if i > 0 {
				m := min(grid[i][j], grid[i-1][j])
				area -= m
			}
			// down
			if i < len(grid)-1 {
				m := min(grid[i][j], grid[i+1][j])
				area -= m
			}
			// left
			if j > 0 {
				m := min(grid[i][j], grid[i][j-1])
				area -= m
			}
			// right
			if j < len(grid[i])-1 {
				m := min(grid[i][j], grid[i][j+1])
				area -= m
			}
		}
	}
	return area
}

func min(a, b int) int {
	if a > b {
		return b
	}
	return a
}