Projection Area of 3D Shapes

Problem

You are given an n x n grid where we place some 1 x 1 x 1 cubes that are axis-aligned with the x, y, and z axes.

Each value v = grid[i][j] represents a tower of v cubes placed on top of the cell (i, j).

We view the projection of these cubes onto the xy, yz, and zx planes.

A projection is like a shadow, that maps our 3-dimensional figure to a 2-dimensional plane. We are viewing the “shadow” when looking at the cubes from the top, the front, and the side.

Return the total area of all three projections.

Constraints

• n == grid.length == grid[i].length
• 1 <= n <= 50
• 0 <= grid[i][j] <= 50

Solution

The problem Projection Area of 3D Shapes can be solved by counting the number of cubes that can be seen from the top, the front, and the side.

Implementation

static const int fast_io = []()
{
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    return 0;
}();

class Solution
{
  public:
    int projectionArea(vector<vector<int>> &grid)
    {
        int ret = 0;
        int len = grid.size();

        for (int i = 0; i < len; i++)
        {
            int xy = 0, yz = 0, zx = 0;
            for (int j = 0; j < len; j++)
            {
                if (grid[i][j])
                    xy += 1;
                yz = max(yz, grid[j][i]);
                zx = max(zx, grid[i][j]);
            }
            ret += xy + yz + zx;
        }

        return ret;
    }
};