human
Write me a the Python program you requested given the following:
I found an interesting problem on https://leetcode.com/problems/projection-area-of-3d-shapes/:
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

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python program you requested.

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