<p>You are given a 2D integer array <code>grid</code> of size <code>m x n</code>, where each cell contains a positive integer.</p>

<p>A <strong>cornered path</strong> is defined as a set of adjacent cells with <strong>at most</strong> one turn. More specifically, the path should exclusively move either <strong>horizontally</strong> or <strong>vertically</strong> up to the turn (if there is one), without returning to a previously visited cell. After the turn, the path will then move exclusively in the <strong>alternate</strong> direction: move vertically if it moved horizontally, and vice versa, also without returning to a previously visited cell.</p>

<p>The <strong>product</strong> of a path is defined as the product of all the values in the path.</p>

<p>Return <em>the <strong>maximum</strong> number of <strong>trailing zeros</strong> in the product of a cornered path found in </em><code>grid</code>.</p>

<p>Note:</p>

<ul>
	<li><strong>Horizontal</strong> movement means moving in either the left or right direction.</li>
	<li><strong>Vertical</strong> movement means moving in either the up or down direction.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/23/ex1new2.jpg" style="width: 577px; height: 190px;" />
<pre>
<strong>Input:</strong> grid = [[23,17,15,3,20],[8,1,20,27,11],[9,4,6,2,21],[40,9,1,10,6],[22,7,4,5,3]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The grid on the left shows a valid cornered path.
It has a product of 15 * 20 * 6 * 1 * 10 = 18000 which has 3 trailing zeros.
It can be shown that this is the maximum trailing zeros in the product of a cornered path.

The grid in the middle is not a cornered path as it has more than one turn.
The grid on the right is not a cornered path as it requires a return to a previously visited cell.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/25/ex2.jpg" style="width: 150px; height: 157px;" />
<pre>
<strong>Input:</strong> grid = [[4,3,2],[7,6,1],[8,8,8]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The grid is shown in the figure above.
There are no cornered paths in the grid that result in a product with a trailing zero.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 1000</code></li>
</ul>

import numpy as np

# directions
U = 0 # Up
D = 1 # Down
L = 2 # Left
R = 3 # Right
N_dir = 4

# factors
T = 0 # Two
F = 1 # Five
N_fac = 2

def f(x, p):
    '''returns the highest power of `p` that divides `x`'''
    c = 0
    while x % p == 0:
        c += 1
        x //= p
    return c

class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        grid = np.array(grid)

        # y[i, j, p] = 
        #   highest power of `p` that divides `grid[i, j]`
        y = np.zeros((m, n, N_fac), dtype=int)
        for i in range(m):
            for j in range(n):
                y[i, j, T] = f(grid[i, j], 2)
                y[i, j, F] = f(grid[i, j], 5)

        # x[i, j, d, p] = 
        #   count of times that `p` appears in leg `(i,j,d)`
        x = np.zeros((m, n, N_dir, N_fac), dtype=int)
        for i in range(m):
            for j in range(n):
                if i > 0:
                    x[i, j, U] = y[i-1, j] + x[i-1, j, U]
                    x[m-1-i, j, D] = y[m-i, j] + x[m-i, j, D]
                if j > 0:
                    x[i, j, L] = y[i, j-1] + x[i, j-1, L]
                    x[i, n-1-j, R] = y[i, n-j] + x[i, n-j, R]

        # the cornered path's legs can be oriented in one of four ways
        path_orientations = [(U,R), (U,L), (D,R), (D,L)]

        # for each orientation `(d1, d2)`, 
        #   for each corner `(i, j)` (vectorized):
        #       find the the number of trailing zeros in the product
        #       of the path `(i, j, d1, d2)`.
        #   max-reduce over corners (vectorized)
        # max-reduce over orientations.
        return max(
            (y + x[:, :, d1] + x[:, :, d2]).min(-1).max()
            for d1, d2 in path_orientations
        )
71700