Introduction

In this blog post, we will explore how to solve the LeetCode problem "1594" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a m x n matrix grid. Initially, you are located at the top-left corner (0, 0), and in each step, you can only move right or down in the matrix. Among all possible paths starting from the top-left corner (0, 0) and ending in the bottom-right corner (m - 1, n - 1), find the path with the maximum non-negative product. The product of a path is the product of all integers in the grid cells visited along the path. Return the maximum non-negative product modulo 109 + 7. If the maximum product is negative, return -1. Notice that the modulo is performed after getting the maximum product. Example 1: Input: grid = [[-1,-2,-3],[-2,-3,-3],[-3,-3,-2]] Output: -1 Explanation: It is not possible to get non-negative product in the path from (0, 0) to (2, 2), so return -1. Example 2: Input: grid = [[1,-2,1],[1,-2,1],[3,-4,1]] Output: 8 Explanation: Maximum non-negative product is shown (1 1 -2 -4 1 = 8). Example 3: Input: grid = [[1,3],[0,-4]] Output: 0 Explanation: Maximum non-negative product is shown (1 0 -4 = 0). Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 15 -4 <= grid[i][j] <= 4

Explanation

Here's a solution to the problem, adhering to the requested format:

• Key Idea: Use Dynamic Programming to track both the maximum and minimum possible products at each cell. This is crucial to handle negative numbers, as the smallest (most negative) product can become the largest if multiplied by another negative.

• Zero Handling: If a zero is encountered, the maximum product is immediately zero for paths going through it. We must account for this at each step.

• Modulo Arithmetic: Apply the modulo operator only at the very end, after finding the final maximum product, to avoid precision issues or incorrect results.

• Complexity: O(mn) runtime, O(mn) storage

Code

    def maxProductPath(grid):
    m, n = len(grid), len(grid[0])
    max_dp = [[0] * n for _ in range(m)]
    min_dp = [[0] * n for _ in range(m)]

    max_dp[0][0] = grid[0][0]
    min_dp[0][0] = grid[0][0]

    for i in range(1, m):
        max_dp[i][0] = max_dp[i - 1][0] * grid[i][0]
        min_dp[i][0] = min_dp[i - 1][0] * grid[i][0]

    for j in range(1, n):
        max_dp[0][j] = max_dp[0][j - 1] * grid[0][j]
        min_dp[0][j] = min_dp[0][j - 1] * grid[0][j]

    for i in range(1, m):
        for j in range(1, n):
            max_dp[i][j] = max(max_dp[i - 1][j] * grid[i][j], max_dp[i][j - 1] * grid[i][j],
                             min_dp[i - 1][j] * grid[i][j], min_dp[i][j - 1] * grid[i][j])
            min_dp[i][j] = min(max_dp[i - 1][j] * grid[i][j], max_dp[i][j - 1] * grid[i][j],
                             min_dp[i - 1][j] * grid[i][j], min_dp[i][j - 1] * grid[i][j])

    max_product = max_dp[m - 1][n - 1]
    if max_product < 0:
        return -1
    else:
        return max_product % (10**9 + 7)