<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:
        # Time complexity: O(m*n*log M), space complexity: O(m*n),
        # where M is the maximum value in grid.
        m, n = len(grid), len(grid[0])
        # cum_r[i][j] contains the multiplicities of 2 and 5 in
        # grid[i][:j].
        # cum_c[i][j] contains the multiplicities of 2 and 5 in
        # grid[:i][j].
        cum_r = [[[0,0] for _ in range(n + 1)] for _ in range(m)]
        cum_c = [[[0,0] for _ in range(n)] for _ in range(m + 1)]
        for i in range(m):
            for j in range(n):
                v, f2, f5 = grid[i][j], 0, 0
                while v % 2 == 0:
                    v //= 2
                    f2 += 1
                while v % 5 == 0:
                    v //= 5
                    f5 += 1
                cum_r[i][j + 1][0] = cum_r[i][j][0] + f2
                cum_r[i][j + 1][1] = cum_r[i][j][1] + f5
                cum_c[i + 1][j][0] = cum_c[i][j][0] + f2
                cum_c[i + 1][j][1] = cum_c[i][j][1] + f5
        
        ans = 0
        for i in range(m):
            for j in range(n):
            # Consider all cornered paths that turn at (i,j).
            # ul contains the multiplicities of 2 and 5 in the path
            # that extends to the top and to the far left. The # of
            # trailing zeros for this path is min(ul).
            # Similar for the other three paths.
                ul, ur, dl, dr = [0,0], [0,0], [0,0], [0,0]
                for k in range(2):
                    ul[k] += cum_c[i + 1][j][k] + cum_r[i][j][k]
                    ur[k] += cum_c[i + 1][j][k] + cum_r[i][n][k] - \
                             cum_r[i][j + 1][k]
                    dl[k] += cum_c[m][j][k] - cum_c[i][j][k] + \
                             cum_r[i][j][k]
                    dr[k] += cum_c[m][j][k] - cum_c[i][j][k] + \
                             cum_r[i][n][k] - cum_r[i][j + 1][k]
                z = max(min(ul), min(ur), min(dl), min(dr))
                ans = max(ans, z)
        return ans
3179