<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:

        REV = lambda x: tuple(reversed(tuple(x)))

        def factorCount(value: int, factor: int) -> int:
            answer = 0
            while value % factor == 0:
                answer += 1
                value //= factor
            return answer
        
        grid25 = [
            [
                (
                    factorCount(value, 2),
                    factorCount(value, 5),
                )
                for value in row
            ]
            for row in grid
        ]
        # print(f'{grid25=}')
        
        grid2 = [
            [
                v2
                for (v2, v5) in row
            ]
            for row in grid25
        ]
        # print(f'{grid2=}')
        grid5 = [
            [
                v5
                for (v2, v5) in row
            ]
            for row in grid25
        ]
        # print(f'{grid5=}')

        def accumulate_each_row_L(grid: List[List[int]]) -> List[List[int]]:
            return [
                list(accumulate(row))
                for row in grid
            ]

        def accumulate_each_row_R(grid: List[List[int]]) -> List[List[int]]:
            return [
                REV(accumulate(REV(row)))
                for row in grid
            ]
        
        def transpose_grid(grid: List[List[int]]) -> List[List[int]]:
            return tuple(zip(*grid))

        sumRows2L = accumulate_each_row_L(grid2)
        sumRows2R = accumulate_each_row_R(grid2)
        sumRows5L = accumulate_each_row_L(grid5)
        sumRows5R = accumulate_each_row_R(grid5)
        rowSums2 = [
            row[0]
            for row in sumRows2R
        ]
        rowSums5 = [
            row[0]
            for row in sumRows5R
        ]

        transpose2 = transpose_grid(grid2)
        transpose5 = transpose_grid(grid5)

        sumCols2L = accumulate_each_row_L(transpose2)
        sumCols2R = accumulate_each_row_R(transpose2)
        sumCols5L = accumulate_each_row_L(transpose5)
        sumCols5R = accumulate_each_row_R(transpose5)

        colSums2 = [
            row[0]
            for row in sumCols2R
        ]
        colSums5 = [
            row[0]
            for row in sumCols5R
        ]

        def sum3Pair(pair1: Tuple[int,int], pair2: Tuple[int,int], pair3: Tuple[int,int]) -> Tuple[int,int]:
            # print(f'sum3Pair({pair1},{pair2},{pair3}):')
            return tuple(
                map(
                    sum,
                    zip(pair1, pair2, pair3)
                )
            )

        answers = []
        for i in range(len(grid)):
            # print(f'{i=}')
            for j in range(len(grid[0])):
                # print(f'  {j=}')
                neg_IJ = (-grid2[i][j], -grid5[i][j])
                row_L = (sumRows2L[i][j], sumRows5L[i][j])
                row_R = (sumRows2R[i][j], sumRows5R[i][j])
                col_L = (sumCols2L[j][i], sumCols5L[j][i])  # these 2 "column" grids
                col_R = (sumCols2R[j][i], sumCols5R[j][i])  # are referenced backward: J,I
                for row in [row_L, row_R]:
                    for col in [col_L, col_R]:
                        RowCol = sum3Pair(row, col, neg_IJ)
                        Min = min(RowCol)
                        if i == 3 and j == 4:
                            print(f'{i=} {j=}')
                            print(f'{row=}')
                            print(f'{col=}')
                            print(f'{neg_IJ=}')
                            print(f'  {RowCol=}')
                            print(f'    {Min=}')
                        answers.append(Min)
        print(f'{answers=}')
        return max(answers)

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