892. Surface Area of 3D Shapes

892. Surface Area of 3D Shapes (Easy)

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

Solution 1.

// OJ: https://leetcode.com/problems/surface-area-of-3d-shapes/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    int surfaceArea(vector<vector<int>>& grid) {
        int N = grid.size(), area = 0;
        for (int i = 0; i < N; ++i) {
            int prev = 0;
            for (int j = 0; j < N; ++j) {
                if (grid[i][j]) area += 2;
                area += abs(grid[i][j] - prev);
                prev = grid[i][j];
            }
            area += prev;
        }
        for (int i = 0; i < N; ++i) {
            int prev = 0;
            for (int j = 0; j < N; ++j) {
                area += abs(grid[j][i] - prev);
                prev = grid[j][i];
            }
            area += prev;
        }
        return area;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/surface-area-of-3d-shapes/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
private:
    int dirs[4][2] = {{0,1}, {-1,0}, {0,-1}, {1,0}};
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]) continue;
                area += 2;
                for (auto &dir : dirs) {
                    int x = i + dir[0], y = j + dir[1];
                    int h = x < 0 || x >= N || y < 0 || y >= N ? 0 : grid[x][y];
                    area += max(grid[i][j] - h, 0);
                }
            }
        }
        return area;
    }
};