Projection Area of 3D Shapes

Projection Area of 3D Shapes

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

Input: grid = [[2]]
Output: 5
Example 3:

Input: grid = [[1,0],[0,2]]
Output: 8

class Solution {
    public int projectionArea(int[][] grid) {
     int n=grid.length;
        int ans=0;
        for(int i=0;i<n;++i)
        {
            int best_row=0;
            int best_col=0;
            for(int j=0;j<n;++j)
            {
                if(grid[i][j]>0)
                {
                    ans++;

                }
                best_row=Math.max(best_row,grid[i][j]);
                best_col=Math.max(best_col,grid[j][i]);
            }
            ans=ans+best_row+best_col;
        } 
         return ans;
    }   
}