<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:
        numRows = len(grid)
        numCols = len(grid[0])
        total25CntsPerRow = [[0,0] for _ in range(numRows)]
        total25CntsPercol = [[0,0] for _ in range(numCols)]
        div25Counts = [[[0, 0] for _ in range(numCols + 1)]]
        for ri, r in enumerate(grid):
            div25Row = [[0, 0]]
            for ci, v in enumerate(r):
                cnts = []
                for d in (2, 5):
                    cnt = 0
                    while v % d == 0:
                        cnt += 1
                        v //= d
                    cnts.append(cnt)
                div25Row.append(cnts)
                for i, c in enumerate(cnts):
                    total25CntsPerRow[ri][i] += c
                    total25CntsPercol[ci][i] += c
            div25Counts.append(div25Row)
        # print(div25Counts)
        # print(total25CntsPerRow)
        # print(total25CntsPercol)
        result = max(max(map(min, total25CntsPerRow)), max(map(min, total25CntsPercol)))
        rowCumSum = [0, 0]
        for ri in range(1, numRows + 1):
            for ci in range(1, numCols + 1):
                curDivCount = div25Counts[ri][ci]
                rowRight = [total - cur - prev for total, cur, prev in zip(total25CntsPerRow[ri - 1], curDivCount, rowCumSum)]
                colCumSum = div25Counts[ri - 1][ci]
                colDown = [total - cur - prev for total, cur, prev in zip(total25CntsPercol[ci - 1], curDivCount, colCumSum)]
                result = max(result,
                           min(cur + rowPart + colPart for cur, rowPart, colPart in zip(curDivCount, rowCumSum, colCumSum)),
                           min(cur + rowPart + colPart for cur, rowPart, colPart in zip(curDivCount, rowRight, colCumSum)),
                           min(cur + rowPart + colPart for cur, rowPart, colPart in zip(curDivCount, rowCumSum, colDown)),
                           min(cur + rowPart + colPart for cur, rowPart, colPart in zip(curDivCount, rowRight, colDown))
                )
                # update row cum sums
                rowCumSum[0] += curDivCount[0]
                rowCumSum[1] += curDivCount[1]
                # update col cum sum
                curDivCount[0] += colCumSum[0]
                curDivCount[1] += colCumSum[1]
            rowCumSum[0] = 0
            rowCumSum[1] = 0
        return result
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