<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>

def get_twos_fives(n):
    twos = 0
    while n % 2 == 0:
        n = n // 2
        twos += 1
    fives = 0
    while n % 5 == 0:
        n = n // 5
        fives += 1
    return (twos, fives)

def add_pairs(p1, p2):
    u, v = p1
    x, y = p2
    return (u+x, v+y)

def tens(pair):
    return min(pair)



class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        grid = [[get_twos_fives(grid[r][c]) for c in range(cols)] for r in range(rows)]
        
        accum_left = [[(0, 0) for c in range(cols)] for r in range(rows)]
        accum_right = [[(0, 0) for c in range(cols)] for r in range(rows)]
        accum_above = [[(0, 0) for c in range(cols)] for r in range(rows)]
        accum_below = [[(0, 0) for c in range(cols)] for r in range(rows)]

        for r in range(rows):  # sweep left
            for c in range(1, cols):
                accum_left[r][c] = add_pairs(accum_left[r][c-1], grid[r][c-1])
        for r in range(rows):  # sweep right
            for c in range(cols-2, -1, -1):
                accum_right[r][c] = add_pairs(accum_right[r][c+1], grid[r][c+1])
        
        for c in range(cols):  # sweep down
            for r in range(1, rows):
                accum_above[r][c] = add_pairs(accum_above[r-1][c], grid[r-1][c])
        for c in range(cols):  # sweep up
            for r in range(rows-2, -1, -1):
                accum_below[r][c] = add_pairs(accum_below[r+1][c], grid[r+1][c])

        most_tens_at_corner = [[0 for c in range(cols)] for r in range(rows)]
        most_tens_overall = 0
        for r in range(rows):
            for c in range(cols):
                horiz = [accum_left[r][c], accum_right[r][c]]
                vert = [accum_above[r][c], accum_below[r][c]]
                four_directions_vals = [add_pairs(h, v) for h in horiz for v in vert]
                four_directions_vals_with_self = [add_pairs(v, grid[r][c]) for v in four_directions_vals]
                most_tens = max(tens(v) for v in four_directions_vals_with_self)
                most_tens_at_corner[r][c] = most_tens
                most_tens_overall = max(most_tens, most_tens_overall)
        
        return most_tens_overall
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