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: [[2]] 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[0].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)
