<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 collections import defaultdict

class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        # build 2s, 5s cnt lookup
        lkup2 = defaultdict(int)
        lkup5 = defaultdict(int)
        for i in range(1, 1001):
            if i % 2 == 0:
                lkup2[i] = lkup2[i//2] + 1
            if i % 5 == 0:
                lkup5[i] = lkup5[i//5] + 1
        # accumulative 2 and 5 counts
        m, n = len(grid), len(grid[0])
        cnt2 = [[(0, 0)] * (n+1) for i in range(m+1)]
        cnt5 = [[(0, 0)]* (n+1) for i in range(m+1)]
        for i in range(1, m+1):
            for j in range(1, n+1):
                cnt2[i][j] = (
                    cnt2[i][j-1][0] + lkup2[grid[i-1][j-1]],
                    cnt2[i-1][j][1] + lkup2[grid[i-1][j-1]],
                )
                cnt5[i][j] = (
                    cnt5[i][j-1][0] + lkup5[grid[i-1][j-1]],
                    cnt5[i-1][j][1] + lkup5[grid[i-1][j-1]],
                )
        # at each cell, pick the optimal directions (ex, up vs down)
        best = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                l2s, l5s = cnt2[i+1][j+1][0], cnt5[i+1][j+1][0]
                r2s, r5s = cnt2[i+1][n][0] - cnt2[i+1][j][0], cnt5[i+1][n][0] - cnt5[i+1][j][0]
                u2s, u5s = cnt2[i+1][j+1][1], cnt5[i+1][j+1][1]
                d2s, d5s = cnt2[m][j+1][1] - cnt2[i][j+1][1], cnt5[m][j+1][1] - cnt5[i][j+1][1]
                best[i][j] =max(
                    min(l2s+u2s-lkup2[grid[i][j]], l5s+u5s-lkup5[grid[i][j]]),
                    min(r2s+u2s-lkup2[grid[i][j]], r5s+u5s-lkup5[grid[i][j]]),
                    min(l2s+d2s-lkup2[grid[i][j]], l5s+d5s-lkup5[grid[i][j]]),
                    min(r2s+d2s-lkup2[grid[i][j]], r5s+d5s-lkup5[grid[i][j]]),
                )
        # calc optimal cell
        res = 0
        for i in range(m):
            for j in range(n):
                res = max(res, best[i][j])
        return res
2694