<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])
        

        def decomp(v):
            nbTwo = 0
            nbFive = 0
            while v > 0 and v % 2 == 0:
                nbTwo +=1 
                v //= 2
            while v > 0 and v % 5 == 0:
                nbFive +=1 
                v //= 5
            return (nbTwo, nbFive)
        
        rows = []
        for r in range(m):
            d2, d5 = 0,0
            row = [(0,0)]
            for c in range(n):
                d = decomp(grid[r][c])
                d2 += d[0]
                d5 += d[1]
                row.append((d2, d5))
            rows.append(row)
        cols = []
        for c in range(n):
            d2, d5 = 0,0
            col = [(0,0)]
            for r in range(m):
                d = decomp(grid[r][c])
                d2 += d[0]
                d5 += d[1]
                col.append((d2, d5))
            cols.append(col)
        
        # print(rows)
        # print(cols)
        def add(t1, t2):
            return (t1[0] + t2[0], t1[1] + t2[1])
        def sub(t1, t2):
            return (t1[0] - t2[0], t1[1] - t2[1])

        res = 0
        for r in range(m):
            for c in range(n):
                tl = rows[r][c]
                tr = sub(rows[r][n],rows[r][c+1])
                tu = cols[c][r]
                td = sub(cols[c][m],cols[c][r+1])
                d = decomp(grid[r][c])

                p1 = add(add(tl,tu), d)
                p2 = add(add(tl,td), d)
                p3 = add(add(tr,tu), d)
                p4 = add(add(tr,td), d)

                v1 = min(p1[0], p1[1])
                v2 = min(p2[0], p2[1])
                v3 = min(p3[0], p3[1])
                v4 = min(p4[0], p4[1])

                res = max(res, v1, v2, v3, v4)

                
                
    
        return res
            


        
        
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