<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, A: List[List[int]]) -> int:
        m, n = len(A), len(A[0])
        left = [[[0, 0] for _ in range(n)] for _ in range(m)]
        top = [[[0, 0] for _ in range(n)] for _ in range(m)]
        
        def helper(num):
            a, b = 0, 0
            while num % 2 == 0:
                num //= 2
                a += 1
            while num % 5 == 0:
                num //= 5
                b += 1
            return [a, b]
        
        for i in range(m):
            for j in range(n):
                if j == 0:
                    left[i][j] = helper(A[i][j])
                else:
                    a, b = helper(A[i][j])
                    left[i][j][0] = left[i][j - 1][0] + a
                    left[i][j][1] = left[i][j - 1][1] + b
        for j in range(n):
            for i in range(m):
                if i == 0:
                    top[i][j] = helper(A[i][j])
                else:
                    a, b, = helper(A[i][j])
                    top[i][j][0] = top[i - 1][j][0] + a               
                    top[i][j][1] = top[i - 1][j][1] + b


        ans = 0
        for i in range(m):
            for j in range(n):
                a, b = top[m - 1][j]
                d, e= left[i][n - 1]
                x, y = helper(A[i][j])
                a1, b1 = top[i][j]
                a2, b2= left[i][j]
                tmp = [a1 + a2 - x, b1 + b2 - y]
                ans = max(ans, min(tmp))
                tmp = [d - a2 + a1, e - b2 + b1]
                ans = max(ans, min(tmp))             
                tmp = [a - a1 + a2, b - b1 + b2]
                ans = max(ans, min(tmp))
                tmp = [a + d - a1 - a2 + x, b + e - b1 - b2 + y]
                ans = max(ans, min(tmp))
                
        return ans
3319