<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>

class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        dp1 = [[None] * n for _ in range(m)]
        dp2 = [[None] * n for _ in range(m)]

        @cache
        def cal_divisible(num: int, divisor: int) -> int:
            ans = 0
            while num % divisor == 0:
                ans += 1
                num //= divisor
            return ans

        def process(nums: List[int]) -> List:
            ans = [[] for _ in range(len(nums))]      
            twos, fives = 0, 0
            for i, num in enumerate(nums):
                twos += cal_divisible(num, 2)
                fives += cal_divisible(num, 5)
                ans[i].append([twos, fives])
            twos, fives = 0, 0
            for i in range(len(nums) - 1, -1, -1):
                twos += cal_divisible(nums[i], 2)
                fives += cal_divisible(nums[i], 5)
                ans[i].append([twos, fives])

            return ans        

        for i in range(m):
            rows = process(grid[i])
            for j in range(n):
                dp1[i][j] = rows[j]

        for j in range(n):
            cols = [grid[i][j] for i in range(m)]
            cols = process(cols)
            for i in range(m):
                dp2[i][j] = cols[i]

        ans = -math.inf
        for i in range(m):
            for j in range(n):
                rows, cols = dp1[i][j], dp2[i][j]
                for row in rows:
                    for col in cols:
                        twos = row[0] + col[0] - cal_divisible(grid[i][j], 2)
                        fives = row[1] + col[1] - cal_divisible(grid[i][j], 5)
                        ans = max(ans, min(twos, fives))

        return ans
5277