# 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

```cpp
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;
    }
};
```


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