<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) :
        matrix = grid 
        self.ans,prefix,suffix = 0,[i[:] for i in matrix ],[i[:] for i in matrix ]
      
        def f(n):
            x,y =0,0
            while n%2==0:
                n = n//2
                x+=1
            while n% 5==0:
                n = n // 5
                y+=1
            return x,y

        for i in prefix:
            for n,v in enumerate(i):
                if n==0 : i[0] =f(v)
                else:
                    x,y = f(v)
                    a,b = i[n-1]
                    i[n]=x+a,y+b

        for i in suffix :
            for n in range(len(i)-1,-1,-1):
                if n == len(i)-1:i[n]=f(i[n])
                else:
                    x,y = f(i[n])
                    a,b= i[n+1]
                    i[n]=x+a,y+b

        for j in range(len(matrix[0])):
            two,five =0,0
            for i in range(len(matrix)):
                self.ans = max(self.ans,min(two+prefix[i][j][0],five + prefix[i][j][1]))
                self.ans = max(self.ans , min(two + suffix[i][j][0] , five + suffix[i][j][1]))

                x,y = f(matrix[i][j])
                two +=x
                five +=y 
            
            two,five = 0,0
            for i in range(len(matrix)-1,-1,-1):
                self.ans = max(self.ans,min(two+prefix[i][j][0],five + prefix[i][j][1]))
                self.ans = max(self.ans , min(two + suffix[i][j][0] , five + suffix[i][j][1]))
                x,y = f(matrix[i][j])
                two +=x
                five +=y 
      
        return self.ans
3142