LeetCode in Kotlin

892. Surface Area of 3D Shapes

Easy

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

class Solution {
    fun surfaceArea(grid: Array<IntArray>): Int {
        var surfaceArea = 0
        for (i in grid.indices) {
            for (j in grid[i].indices) {
                if (grid[i][j] > 0) {
                    surfaceArea += 4 * grid[i][j] + 2
                    surfaceArea -= hiddenSides(i, j, grid)
                }
            }
        }
        return surfaceArea
    }

    private fun hiddenSides(i: Int, j: Int, grid: Array<IntArray>): Int {
        var hidden = 0
        val tower = grid[i][j]
        if (j + 1 < grid[i].size && grid[i][j + 1] > 0) {
            hidden += tower.coerceAtMost(grid[i][j + 1])
        }
        if (j - 1 >= 0 && grid[i][j - 1] > 0) {
            hidden += tower.coerceAtMost(grid[i][j - 1])
        }
        if (i + 1 < grid.size && grid[i + 1][j] > 0) {
            hidden += tower.coerceAtMost(grid[i + 1][j])
        }
        if (i - 1 >= 0 && grid[i - 1][j] > 0) {
            hidden += tower.coerceAtMost(grid[i - 1][j])
        }
        return hidden
    }
}

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