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

Source: leetcode | Difficulty: EASY
Tags: Geometry

Matrices

Data structures

Mathematics

Starter code: class Solution:
    def projectionArea(self, grid: List[List[int]]) -> int:
        

Solutions:
class Solution:

	def projectionArea(self, grid: List[List[int]]) -> int:
		ans = 0
		for i in range(len(grid)):
			row = 0
			col = 0
			for j in range(len(grid[i])):
				if grid[i][j]:
					ans += 1
				row = max(row, grid[i][j])
				col = max(col, grid[j][i])
			ans += row + col
		return ans


class Solution:

	def projectionArea(self, grid: List[List[int]]) -> int:
		max_rows = [0 for _ in grid]
		max_cols = [0 for _ in grid]
		total_nonzero = 0
		for r in range(len(grid)):
			for c in range(len(grid[r])):
				if grid[r][c] > max_cols[c]:
					max_cols[c] = grid[r][c]
				if grid[r][c] > max_rows[r]:
					max_rows[r] = grid[r][c]
				if grid[r][c] > 0:
					total_nonzero += 1
		print(max_rows)
		print(max_cols)
		print(total_nonzero)
		return sum(max_rows) + sum(max_cols) + total_nonzero