#2245

Medium

LeetCode Problem Workspace

Maximum Trailing Zeros in a Cornered Path

Find the maximum trailing zeros in the product of a cornered path within a 2D grid.

Problem Statement

You are given a 2D integer array grid of size m x n, where each cell contains a positive integer. A cornered path in the grid is defined as a set of adjacent cells with at most one turn, with no backtracking to previously visited cells. The path should move horizontally or vertically up to the turn, and after the turn, the direction must switch to the alternate direction without revisiting any cell.

The goal is to find the maximum number of trailing zeros in the product of any valid cornered path. The product of a path is the product of the values in all the cells along the path. A trailing zero is produced by a factor of 10 in the product, which arises from multiplying a factor of 2 and a factor of 5. Your task is to calculate the maximum number of trailing zeros for all possible cornered paths in the grid.

Examples

Example 1

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

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

Input: grid = [[4,3,2],[7,6,1],[8,8,8]]

Output: 0

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 Approach

Factorizing the Product

To solve the problem, begin by calculating the number of factors of 2 and 5 in each grid cell. The product of any path will generate trailing zeros if it contains these factors. For each cornered path, calculate how many times the factors of 2 and 5 appear, and determine the number of trailing zeros based on the minimum of these counts.

Tracking Paths

Use a dynamic approach to track possible cornered paths from each cell. As you move along the grid, calculate the potential product along the path, keeping track of the factors of 2 and 5 for each direction. Switch directions when encountering the turn and ensure no cell is revisited.

Optimizing with Prefix Sum

Use a prefix sum technique to optimize the calculation of trailing zeros across multiple potential paths. By precomputing the factors of 2 and 5 for rows and columns, you can efficiently calculate the number of trailing zeros in any given path, minimizing the number of redundant calculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used for calculating the product and tracking paths. A brute-force method may involve iterating through all potential paths, resulting in O(m * n) complexity. Optimizations using prefix sums can reduce this complexity significantly, allowing for faster computation of trailing zeros for each path.

What Interviewers Usually Probe

Can the candidate efficiently handle large grid sizes and optimize the computation of trailing zeros?
Does the candidate demonstrate an understanding of how factors of 2 and 5 affect trailing zeros?
Is the candidate able to think critically about how to manage path direction and avoid redundant calculations?

Common Pitfalls or Variants

Common pitfalls

Forgetting to consider the factorization of both 2 and 5 when calculating trailing zeros.
Failing to optimize the solution with dynamic programming or prefix sums, leading to redundant calculations.
Not properly tracking the path direction and revisiting cells during the corner turn.

Follow-up variants

Find the maximum number of trailing zeros for paths with multiple turns.
Modify the grid to allow diagonal paths and calculate trailing zeros for these new paths.
Add constraints to limit the grid size, challenging the optimization of the solution.

FAQ

What is the key concept for solving the Maximum Trailing Zeros in a Cornered Path problem?

The key concept is understanding how trailing zeros are created by the factors of 2 and 5 in the product of path values. Efficiently calculating the product of these factors in a cornered path is critical.

How can I optimize the calculation of trailing zeros in a grid?

Optimizing the calculation can be achieved by using prefix sums to precompute the factors of 2 and 5 in rows and columns, reducing redundant calculations for each path.

What patterns are involved in solving the Maximum Trailing Zeros in a Cornered Path problem?

This problem involves Array and Matrix patterns, particularly using dynamic programming or prefix sums to efficiently handle large grids.

What is the time complexity of solving this problem?

The time complexity can vary depending on the approach, with a brute-force solution being O(m * n), but optimizations with prefix sums can significantly reduce this.

What are common pitfalls to avoid when solving this problem?

Common pitfalls include failing to account for both 2 and 5 factors in the product, not optimizing the path calculations, and revisiting cells during the corner turn.

terminal

Solution

Solution 1: Prefix Sum + Enumerate Turning Point

Firstly, we need to understand that for a product, the number of trailing zeros depends on the smaller count of $2$ and $5$ in its factors. Also, each corner path should cover as many numbers as possible, so it must start from a boundary, reach a turning point, and then reach another boundary.

class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        r2 = [[0] * (n + 1) for _ in range(m + 1)]
        c2 = [[0] * (n + 1) for _ in range(m + 1)]
        r5 = [[0] * (n + 1) for _ in range(m + 1)]
        c5 = [[0] * (n + 1) for _ in range(m + 1)]
        for i, row in enumerate(grid, 1):
            for j, x in enumerate(row, 1):
                s2 = s5 = 0
                while x % 2 == 0:
                    x //= 2
                    s2 += 1
                while x % 5 == 0:
                    x //= 5
                    s5 += 1
                r2[i][j] = r2[i][j - 1] + s2
                c2[i][j] = c2[i - 1][j] + s2
                r5[i][j] = r5[i][j - 1] + s5
                c5[i][j] = c5[i - 1][j] + s5
        ans = 0
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                a = min(r2[i][j] + c2[i - 1][j], r5[i][j] + c5[i - 1][j])
                b = min(r2[i][j] + c2[m][j] - c2[i][j], r5[i][j] + c5[m][j] - c5[i][j])
                c = min(r2[i][n] - r2[i][j] + c2[i][j], r5[i][n] - r5[i][j] + c5[i][j])
                d = min(
                    r2[i][n] - r2[i][j - 1] + c2[m][j] - c2[i][j],
                    r5[i][n] - r5[i][j - 1] + c5[m][j] - c5[i][j],
                )
                ans = max(ans, a, b, c, d)
        return ans

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