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### by Sumit Chourasia | Oct 01, 2020 | Category :coding | Tags : algorithmडेटा-संरचनाआसानleetcodeगणित #### Projection Area of 3D Shapes - Math - Easy - LeetCode - मिनी टीवी

On a N * N grid, we place some 1 * 1 * 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 grid cell (i, j).

Now 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.

Here, 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.

Example 1:

Input: []
Output: 5
Example 2:

Input: [[1,2],[3,4]]
Output: 17
Explanation:
Here are the three projections ("shadows") of the shape made with each axis-aligned plane.

Example 3:

Input: [[1,0],[0,2]]
Output: 8
Example 4:

Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 14
Example 5:

Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 21

Note:

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

``````public class Solution {
public int ProjectionArea(int[][] grid) {
int res = 0, n = grid.Length;
for (int i = 0; i < n; ++i) {
int x = 0, y = 0;
for (int j = 0; j < n; ++j) {
x = Math.Max(x, grid[i][j]);
y = Math.Max(y, grid[j][i]);
if (grid[i][j] > 0) ++res;
}
res += x + y;
}
return res;
}
}``````

Time Complexity: O(n^2)

Space Complexity: O(1)

Explanation
front-back projection area on xz = sum(max value for every col)
right-left projection area on yz = sum(max value for every row)
top-down projection area on xy = sum(1 for every v > 0)
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