<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:
        def two(n):
            res = 0
            while n % 2 == 0:
                res += 1
                n /= 2
            return res
        
        def five(n):
            res = 0
            while n % 5 == 0:
                res += 1
                n /= 5
            return res

        m = len(grid)
        n = len(grid[0])
        twos_l = [[0] * n for i in range(m)]
        for i in range(m):
            twos_l[i][0] = two(grid[i][0])
        for i in range(m):
            for j in range(1, n):
                twos_l[i][j] = twos_l[i][j - 1] + two(grid[i][j])
        
        twos_r = [[0] * n for i in range(m)]
        for i in range(m):
            twos_r[i][n - 1] = two(grid[i][n - 1])
        for i in range(m):
            for j in range(n - 2, -1, -1):
                twos_r[i][j] = twos_r[i][j + 1] + two(grid[i][j])
        
        twos_u = [[0] * n for i in range(m)]
        for i in range(n):
            twos_u[0][i] = two(grid[0][i])
        for i in range(n):
            for j in range(1, m):
                twos_u[j][i] = twos_u[j - 1][i] + two(grid[j][i])
        
        twos_d = [[0] * n for i in range(m)]
        for i in range(n):
            twos_d[m - 1][i] = two(grid[m - 1][i])
        for i in range(n):
            for j in range(m - 2, -1, -1):
                twos_d[j][i] = twos_d[j + 1][i] + two(grid[j][i])

        fives_l = [[0] * n for i in range(m)]
        for i in range(m):
            fives_l[i][0] = five(grid[i][0])
        for i in range(m):
            for j in range(1, n):
                fives_l[i][j] = fives_l[i][j - 1] + five(grid[i][j])
        
        fives_r = [[0] * n for i in range(m)]
        for i in range(m):
            fives_r[i][n - 1] = five(grid[i][n - 1])
        for i in range(m):
            for j in range(n - 2, -1, -1):
                fives_r[i][j] = fives_r[i][j + 1] + five(grid[i][j])
        
        fives_u = [[0] * n for i in range(m)]
        for i in range(n):
            fives_u[0][i] = five(grid[0][i])
        for i in range(n):
            for j in range(1, m):
                fives_u[j][i] = fives_u[j - 1][i] + five(grid[j][i])
        
        fives_d = [[0] * n for i in range(m)]
        for i in range(n):
            fives_d[m - 1][i] = five(grid[m - 1][i])
        for i in range(n):
            for j in range(m - 2, -1, -1):
                fives_d[j][i] = fives_d[j + 1][i] + five(grid[j][i])

        ans = 0
        for i in range(m):
            for j in range(n):
                ans = max(ans, min(twos_l[i][j] + twos_u[i][j] - two(grid[i][j]), fives_l[i][j] + fives_u[i][j] - five(grid[i][j])))
                ans = max(ans, min(twos_l[i][j] + twos_d[i][j] - two(grid[i][j]), fives_l[i][j] + fives_d[i][j] - five(grid[i][j])))
                ans = max(ans, min(twos_r[i][j] + twos_u[i][j] - two(grid[i][j]), fives_r[i][j] + fives_u[i][j] - five(grid[i][j])))
                ans = max(ans, min(twos_r[i][j] + twos_d[i][j] - two(grid[i][j]), fives_r[i][j] + fives_d[i][j] - five(grid[i][j])))

        return ans
115200