<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:
    m = len(grid)
    n = len(grid[0])
    # leftPrefix2[i][j] := the number of 2 in grid[i][0..j]
    # leftPrefix5[i][j] := the number of 5 in grid[i][0..j]
    # topPrefix2[i][j] := the number of 2 in grid[0..i][j]
    # topPrefix5[i][j] := the number of 5 in grid[0..i][j]
    leftPrefix2 = [[0] * n for _ in range(m)]
    leftPrefix5 = [[0] * n for _ in range(m)]
    topPrefix2 = [[0] * n for _ in range(m)]
    topPrefix5 = [[0] * n for _ in range(m)]

    def getCount(num: int, factor: int) -> int:
      count = 0
      while num % factor == 0:
        num //= factor
        count += 1
      return count

    for i in range(m):
      for j in range(n):
        leftPrefix2[i][j] = getCount(grid[i][j], 2)
        leftPrefix5[i][j] = getCount(grid[i][j], 5)
        if j:
          leftPrefix2[i][j] += leftPrefix2[i][j - 1]
          leftPrefix5[i][j] += leftPrefix5[i][j - 1]

    for j in range(n):
      for i in range(m):
        topPrefix2[i][j] = getCount(grid[i][j], 2)
        topPrefix5[i][j] = getCount(grid[i][j], 5)
        if i:
          topPrefix2[i][j] += topPrefix2[i - 1][j]
          topPrefix5[i][j] += topPrefix5[i - 1][j]

    ans = 0
    for i in range(m):
      for j in range(n):
        curr2 = getCount(grid[i][j], 2)
        curr5 = getCount(grid[i][j], 5)
        l2 = leftPrefix2[i][j]
        l5 = leftPrefix5[i][j]
        r2 = leftPrefix2[i][n - 1] - (0 if j == 0 else leftPrefix2[i][j - 1])
        r5 = leftPrefix5[i][n - 1] - (0 if j == 0 else leftPrefix5[i][j - 1])
        t2 = topPrefix2[i][j]
        t5 = topPrefix5[i][j]
        d2 = topPrefix2[m - 1][j] - (0 if i == 0 else topPrefix2[i - 1][j])
        d5 = topPrefix5[m - 1][j] - (0 if i == 0 else topPrefix5[i - 1][j])
        ans = max(ans,
                  min(l2 + t2 - curr2, l5 + t5 - curr5),
                  min(r2 + t2 - curr2, r5 + t5 - curr5),
                  min(l2 + d2 - curr2, l5 + d5 - curr5),
                  min(r2 + d2 - curr2, r5 + d5 - curr5))

    return ans
        
2831