<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 copy import deepcopy

class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        """
        Number of zeros, we can easily find using number of 2, 5
        So in prefix sum way, for all 4 directions we need to find the number of 2,5
        up, down, left, right
        
        Now as we find the result, we can find 4 types. 
        up-right, up-left, down-right, down-left. 
        We need to subtract the curr-idx we iterating, as thats added 2 times. 
        Finally number of zeros is only min count of 2,5. That also needed to be accounted. 
        """ 
        @lru_cache(None)
        def factors(x):
            res = [0,0]
            while x%2==0:
                res[0]+=1
                x//=2
            while x%5==0:
                res[1]+=1
                x//=5
            return res
        
        def sum_factors(x, y):
            return [x[0]+y[0], x[1]+y[1]]
        
        def diff_and_min(x, y):
            return min(x[0]-y[0], x[1]-y[1])
        
        m, n = len(grid), len(grid[0])
        zeros = [[None for _ in range(n)] for _ in range(m)]
        up, down, left, right =  deepcopy(zeros), deepcopy(zeros), deepcopy(zeros),deepcopy(zeros)
        
        for i in range(m):
            for j in range(n):
                up[i][j] = factors(grid[i][j]) if i == 0 else sum_factors(up[i-1][j], factors(grid[i][j]))
                left[i][j] = factors(grid[i][j]) if j == 0 else sum_factors(left[i][j-1], factors(grid[i][j]))
                
        for i in range(m-1, -1, -1):
            for j in range(n-1, -1, -1):
                down[i][j] = factors(grid[i][j]) if i==m-1 else sum_factors(down[i+1][j], factors(grid[i][j]))
                right[i][j] = factors(grid[i][j]) if j==n-1 else sum_factors(right[i][j+1], factors(grid[i][j]))
                
        
        res = 0
        for i in range(m):
            for j in range(n):
                res = max(
                    res,
                    diff_and_min(sum_factors(up[i][j], right[i][j]), factors(grid[i][j])),
                    diff_and_min(sum_factors(up[i][j], left[i][j]), factors(grid[i][j])),
                    diff_and_min(sum_factors(down[i][j], right[i][j]), factors(grid[i][j])),
                    diff_and_min(sum_factors(down[i][j], left[i][j]), factors(grid[i][j]))
                )
                
        return res
4550