Surface Area of 3D Shapes

Description

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

Solutions

class Solution {
public:
    int surfaceArea(vector<vector<int>> &grid) {
        int N = grid.size();
        int area = 0;
        for (int i = 0; i < N; ++i)
            for (int j = 0; j < N; ++j) {
                if (grid[i][j] == 0)
                    continue;
                area += 2;
                area += i > 0 ? max(grid[i][j] - grid[i - 1][j], 0) : grid[i][j];
                area += j > 0 ? max(grid[i][j] - grid[i][j - 1], 0) : grid[i][j];
                area += i + 1 < N ? max(grid[i][j] - grid[i + 1][j], 0) : grid[i][j];
                area += j + 1 < N ? max(grid[i][j] - grid[i][j + 1], 0) : grid[i][j];
            }
        return area;
    }
};

class Solution {
public:
    int surfaceArea(vector<vector<int>> &grid) {
        int N = grid.size(), area = 0;
        for (int i = 0; i < N; ++i)
            for (int j = 0; j < N; ++j) {
                if (grid[i][j] == 0)
                    continue;
                area += 4 * grid[i][j] + 2;
                // subtract hidden areas
                if (i)
                    area -= min(grid[i][j], grid[i - 1][j]) * 2;
                if (j)
                    area -= min(grid[i][j], grid[i][j - 1]) * 2;
            }
        return area;
    }
};