<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>

from typing import List


class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        def count_factors(n, factor):
            count = 0
            while n % factor == 0:
                n //= factor
                count += 1
            return count

        m, n = len(grid), len(grid[0])
        
        # Precompute number of 2's and 5's in each grid element
        two_count = [[0]*n for _ in range(m)]
        five_count = [[0]*n for _ in range(m)]
        
        for i in range(m):
            for j in range(n):
                two_count[i][j] = count_factors(grid[i][j], 2)
                five_count[i][j] = count_factors(grid[i][j], 5)
        
        # Prefix sums to facilitate quick path factor calculation
        horizontal_two_prefix = [[0]*(n+1) for _ in range(m)]
        horizontal_five_prefix = [[0]*(n+1) for _ in range(m)]
        vertical_two_prefix = [[0]*(n+1) for _ in range(m+1)]
        vertical_five_prefix = [[0]*(n+1) for _ in range(m+1)]
        
        for i in range(m):
            for j in range(1, n+1):
                horizontal_two_prefix[i][j] = horizontal_two_prefix[i][j-1] + two_count[i][j-1]
                horizontal_five_prefix[i][j] = horizontal_five_prefix[i][j-1] + five_count[i][j-1]
        
        for j in range(n):
            for i in range(1, m+1):
                vertical_two_prefix[i][j] = vertical_two_prefix[i-1][j] + two_count[i-1][j]
                vertical_five_prefix[i][j] = vertical_five_prefix[i-1][j] + five_count[i-1][j]
        
        max_trailing_zeros = 0
        
        for i in range(m):
            for j in range(n):
                # Calculate trailing zeros for each possible cornered path
                right_and_down_two = horizontal_two_prefix[i][j+1] + vertical_two_prefix[m][j] - vertical_two_prefix[i+1][j]
                right_and_down_five = horizontal_five_prefix[i][j+1] + vertical_five_prefix[m][j] - vertical_five_prefix[i+1][j]
                
                right_and_up_two = horizontal_two_prefix[i][j+1] + vertical_two_prefix[i][j]
                right_and_up_five = horizontal_five_prefix[i][j+1] + vertical_five_prefix[i][j]
                
                left_and_down_two = horizontal_two_prefix[i][n] - horizontal_two_prefix[i][j] + vertical_two_prefix[m][j] - vertical_two_prefix[i+1][j]
                left_and_down_five = horizontal_five_prefix[i][n] - horizontal_five_prefix[i][j] + vertical_five_prefix[m][j] - vertical_five_prefix[i+1][j]
                
                left_and_up_two = horizontal_two_prefix[i][n] - horizontal_two_prefix[i][j] + vertical_two_prefix[i][j]
                left_and_up_five = horizontal_five_prefix[i][n] - horizontal_five_prefix[i][j] + vertical_five_prefix[i][j]
                
                max_trailing_zeros = max(max_trailing_zeros,
                                         min(right_and_down_two, right_and_down_five),
                                         min(right_and_up_two, right_and_up_five),
                                         min(left_and_down_two, left_and_down_five),
                                         min(left_and_up_two, left_and_up_five))
        
        return max_trailing_zeros
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