<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>

l2, r2, u2, d2, c2, l5, r5, u5, d5, c5 = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        grid_factors = []
        for _ in range(m):
            row = [[0]*10 for _ in range(n)]
            grid_factors.append(row)
        for r in range(m):
            for c in range(n):
                num = grid[r][c]
                fac_2 = 0
                num_2, rem = num, 0
                while num_2 > 1:
                    num_2, rem = divmod(num_2, 2)
                    if rem != 0:
                        break
                    fac_2 += 1
                grid_factors[r][c][c2] = fac_2
                fac_5 = 0
                num_5, rem = num, 0
                while num_5 > 1:
                    num_5, rem = divmod(num_5, 5)
                    if rem != 0:
                        break
                    fac_5 += 1
                grid_factors[r][c][c5] = fac_5
        for r in range(m):
            row = grid_factors[r]
            cum_2, cum_5 = 0, 0
            for c in range(n):
                cell = row[c]
                cell[l2] = cum_2
                cell[l5] = cum_5
                cum_2 += cell[c2]
                cum_5 += cell[c5]
            cum_2, cum_5 = 0, 0
            for c in range(n-1,-1,-1):
                cell = row[c]
                cell[r2] = cum_2
                cell[r5] = cum_5
                cum_2 += cell[c2]
                cum_5 += cell[c5]
        for c in range(n):
            cum_2, cum_5 = 0, 0
            for r in range(m):
                cell = grid_factors[r][c]
                cell[u2] = cum_2
                cell[u5] = cum_5
                cum_2 += cell[c2]
                cum_5 += cell[c5]
            cum_2, cum_5 = 0, 0
            for r in range(m-1,-1,-1):
                cell = grid_factors[r][c]
                cell[d2] = cum_2
                cell[d5] = cum_5
                cum_2 += cell[c2]
                cum_5 += cell[c5]
        max_trailing_zeros = 0
        for r in range(m):
            for c in range(n):
                cell = grid_factors[r][c]
                try_1 = min(cell[c2] + cell[l2] + cell[d2], cell[c5] + cell[l5] + cell[d5])
                try_2 = min(cell[c2] + cell[l2] + cell[u2], cell[c5] + cell[l5] + cell[u5])
                try_3 = min(cell[c2] + cell[r2] + cell[d2], cell[c5] + cell[r5] + cell[d5])
                try_4 = min(cell[c2] + cell[r2] + cell[u2], cell[c5] + cell[r5] + cell[u5])
                best_for_cell = max(try_1, try_2, try_3, try_4)
                max_trailing_zeros = max(best_for_cell, max_trailing_zeros)
        return max_trailing_zeros

        
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