Maximum Trailing Zeros in a Cornered Path
Tags: leetcode, dsa, theabbie
Published: 2022-06-17 01:00:45

You are given a 2D integer array `grid` of size `m x n`, where each cell contains a positive integer.

A **cornered path** is defined as a set of adjacent cells with **at most** one turn. More specifically, the path should exclusively move either **horizontally** or **vertically** 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 **alternate** direction: move vertically if it moved horizontally, and vice versa, also without returning to a previously visited cell.

The **product** of a path is defined as the product of all the values in the path.

Return _the **maximum** number of **trailing zeros** in the product of a cornered path found in_ `grid`.

Note:

*   **Horizontal** movement means moving in either the left or right direction.
*   **Vertical** movement means moving in either the up or down direction.

**Example 1:**

![](https://assets.leetcode.com/uploads/2022/03/23/ex1new2.jpg)

**Input:** grid = \[\[23,17,15,3,20\],\[8,1,20,27,11\],\[9,4,6,2,21\],\[40,9,1,10,6\],\[22,7,4,5,3\]\]
**Output:** 3
**Explanation:** 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.

**Example 2:**

![](https://assets.leetcode.com/uploads/2022/03/25/ex2.jpg)

**Input:** grid = \[\[4,3,2\],\[7,6,1\],\[8,8,8\]\]
**Output:** 0
**Explanation:** 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.

**Constraints:**

*   `m == grid.length`
*   `n == grid[i].length`
*   `1 <= m, n <= 105`
*   `1 <= m * n <= 105`
*   `1 <= grid[i][j] <= 1000`

**SOLUTION:**

    class Solution:
        def num5divs(self, num):
            t = 0
            f = 0
            while num % 5 == 0:
                num = num // 5
                f += 1
            while num % 2 == 0:
                num = num // 2
                t += 1
            return (t, f)
        
        def maxTrailingZeros(self, grid: List[List[int]]) -> int:
            m = len(grid)
            n = len(grid[0])
            vertprefix = [[(0, 0)] for i in range(n)]
            horprefix = [[(0, 0)] for i in range(m)]
            for i in range(m):
                for j in range(n):
                    grid[i][j] = self.num5divs(grid[i][j])
                    a, b = grid[i][j]
                    ha, hb = horprefix[i][-1]
                    va, vb = vertprefix[j][-1]
                    horprefix[i].append((ha + a, hb + b))
                    vertprefix[j].append((va + a, vb + b))
            maxtrail = 0
            for i in range(m):
                for j in range(n):
                    hors = [vertprefix[j][i], (vertprefix[j][-1][0] - vertprefix[j][i + 1][0], vertprefix[j][-1][1] - vertprefix[j][i + 1][1])]
                    verts = [horprefix[i][j], (horprefix[i][-1][0] - horprefix[i][j + 1][0], horprefix[i][-1][1] - horprefix[i][j + 1][1])]
                    for ha, hb in hors:
                        for va, vb in verts:
                            a, b = grid[i][j]
                            maxtrail = max(maxtrail, min(a + ha + va, b + hb + vb))
            return maxtrail