<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 countTwoFactor(self,x):
        cnt = 0
        
        while (x%2==0):
            x=x//2
            cnt+=1
        return cnt

    def countFiveFactor(self,x):
        cnt = 0
        while x%5==0:
            x=x//5
            cnt+=1
        return cnt
    




    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        m = len(grid)
        n = len(grid[0])
        h_prefix = []
        v_prefix = []
        for i in range(m):
            temp = []
            for j in range(n):
                temp.append((0, 0))
            
            h_prefix.append(temp.copy())
            v_prefix.append(temp.copy())

        

        
        # fill the h_prefix array horizonattaly
        for i in range(m):
            for j in range(n):
                two_count = self.countTwoFactor(grid[i][j])
                five_count = self.countFiveFactor(grid[i][j])
                
                ## for h_prefix
                if j==0:
                    h_prefix[i][j] = (two_count, five_count)
                else:
                    h_prefix[i][j] = (two_count+h_prefix[i][j-1][0], five_count+h_prefix[i][j-1][1])
                
                ## for v_prefix
                if i==0:
                    v_prefix[i][j] = (two_count, five_count)
                else:
                    v_prefix[i][j] = (two_count+v_prefix[i-1][j][0], five_count+v_prefix[i-1][j][1])

        
        '''
        1, 2,  3,  4,           |
        5, 6,  7,  8            v1
        9, 10, 11, 12           |
                                |
        h1 .          <---h1--(i,j)---h2-->
                                |
                                |
                                v2
                                |
                                |

        horizonatal prefixes include i, j whereas vertical perfixes exclude, so that we dont count multiple times
        h1 = h_prefix[i][j]
        h2 = h_prefix[i][n-1]-h_prefix[i][j-1]
        v1 = v_prefix[i-1][j] ### we dont include element i, j for reason mentioned above
        v2 = v_prefix[m-1][j] - v_prefix[i][j]

        combinations possible = h1,v1; h1;v2; h2;v1; h2;v2

        '''
        ans = 0
        for i in range(m):
            for j in range(n):
                h1 = h_prefix[i][j]
                h2 = h_prefix[i][n-1]
                if j!=0:
                    h2 = (h2[0]-h_prefix[i][j-1][0], h2[1]-h_prefix[i][j-1][1])
                
                v1 = (0, 0)
                if i!=0:
                    v1=v_prefix[i-1][j]

                v2 = (v_prefix[m-1][j][0] - v_prefix[i][j][0], v_prefix[m-1][j][1] - v_prefix[i][j][1])

                option1 = min(h1[0]+v1[0], h1[1]+v1[1])
                option2 = min(h1[0]+v2[0], h1[1]+v2[1])
                option3 = min(h2[0]+v1[0], h2[1]+v1[1])
                option4 = min(h2[0]+v2[0], h2[1]+v2[1])


                #final ans = 
                curr_trailing_zeros = max(option1, max(option2, max(option3, option4)))
                ans = max(curr_trailing_zeros, ans)
        return ans
                
                 


         


        

                
        
2404