Surface Area of 3D Shapes
<p>You are given an <code>n x n</code> <code>grid</code> where you have placed some <code>1 x 1 x 1</code> cubes. Each value <code>v = grid[i][j]</code> represents a tower of <code>v</code> cubes placed on top of cell <code>(i, j)</code>.</p>

<p>After placing these cubes, you have decided to glue any directly adjacent cubes to each other, forming several irregular 3D shapes.</p>

<p>Return <em>the total surface area of the resulting shapes</em>.</p>

<p><strong>Note:</strong> The bottom face of each shape counts toward its surface area.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0892.Surface%20Area%20of%203D%20Shapes/images/tmp-grid2.jpg" style="width: 162px; height: 162px;" />
<pre>
<strong>Input:</strong> grid = [[1,2],[3,4]]
<strong>Output:</strong> 34
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0892.Surface%20Area%20of%203D%20Shapes/images/tmp-grid4.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1],[1,0,1],[1,1,1]]
<strong>Output:</strong> 32
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0892.Surface%20Area%20of%203D%20Shapes/images/tmp-grid5.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[2,2,2],[2,1,2],[2,2,2]]
<strong>Output:</strong> 46
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 50</code></li>
</ul>
Geometry; Array; Math; Matrix
class Solution {
public:
    int surfaceArea(vector<vector<int>>& grid) {
        int n = grid.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j]) {
                    ans += 2 + grid[i][j] * 4;
                    if (i) ans -= min(grid[i][j], grid[i - 1][j]) * 2;
                    if (j) ans -= min(grid[i][j], grid[i][j - 1]) * 2;
                }
            }
        }
        return ans;
    }
};