#1594

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Maximum Non Negative Product in a Matrix

Find the maximum non-negative product path in a matrix using dynamic programming and state transition.

Problem Statement

You are given an m x n matrix grid, starting from the top-left corner (0, 0), with moves restricted to right or down. You must find the path from (0, 0) to the bottom-right corner (m - 1, n - 1) that yields the maximum non-negative product.

Each move involves multiplying the current product by the value in the corresponding grid cell. If at any point the product becomes negative, it can impact future steps. The goal is to return the maximum non-negative product modulo 10^9 + 7. If no such path exists, return -1.

Examples

Example 1

Input: grid = [[-1,-2,-3],[-2,-3,-3],[-3,-3,-2]]

Output: -1

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

Maximum non-negative product is shown (1 * 1 * -2 * -4 * 1 = 8).

Example 3

Input: grid = [[1,3],[0,-4]]

Output: 0

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

Solution Approach

Dynamic Programming with State Transition

Use dynamic programming to track both the maximum and minimum products that can be obtained up to each cell. This is necessary because negative numbers can flip the sign of the product, so the minimum product at a point can become the maximum when multiplied by a negative value.

Path Transition Consideration

At each cell, consider the current value and the product results from the cell above or to the left. The transition requires taking both maximum and minimum values because of negative numbers potentially altering the product sign.

Modulo Operations for Large Results

Since the results can become very large, use modulo 10^9 + 7 to keep the product within the required bounds. Ensure that the final result respects the modulo constraint.

Complexity Analysis

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

The time complexity depends on the number of cells, which is O(m * n). For each cell, you perform constant-time calculations, considering two prior cells (above and left). The space complexity is O(m * n) because a 2D table is used to store the maximum and minimum products up to each point.

What Interviewers Usually Probe

Candidate's understanding of state transition dynamic programming should be evident in how they handle both maximum and minimum products.
They should be able to explain why maintaining both max and min products is crucial in the context of negative numbers.
Look for clarity in explaining how modulo operations are applied and how it affects large results.

Common Pitfalls or Variants

Common pitfalls

Failing to account for negative numbers and their effect on the product, which may lead to incorrect results.
Not handling the modulo operation correctly, leading to overflow or incorrect final results.
Incorrectly assuming that all paths will have non-negative products, leading to incorrect edge case handling.

Follow-up variants

Consider using this approach for similar grid-based dynamic programming problems with path constraints.
Variant could involve adding diagonal movement, requiring an additional transition step.
Extend this to 3D grids, where moves can be made in three dimensions instead of two.

FAQ

How do I manage negative numbers in the Maximum Non Negative Product in a Matrix problem?

By maintaining both the maximum and minimum products at each cell, you can properly handle negative numbers that may flip the sign of the product.

What is the time complexity of solving the Maximum Non Negative Product in a Matrix problem?

The time complexity is O(m * n), where m and n are the number of rows and columns in the matrix.

Why do I need to track both max and min products in this problem?

Tracking both allows you to properly handle negative numbers, as multiplying by a negative can turn the minimum product into a maximum one.

What happens if the product becomes negative at any point in the grid?

If the product is negative and no non-negative product path is possible, the function will return -1.

What is the purpose of the modulo operation in this problem?

The modulo operation ensures that large product values are reduced within the range of 10^9 + 7, as required by the problem's constraints.

terminal

Solution

Solution 1: Dynamic Programming

We define a 3D array $f$, where $f[i][j][0]$ and $f[i][j][1]$ represent the minimum and maximum product of all paths from the top-left corner $(0, 0)$ to position $(i, j)$, respectively. For each position $(i, j)$, we can transition from above $(i - 1, j)$ or from the left $(i, j - 1)$, so we need to consider the results of multiplying the minimum and maximum products of these two paths by the value of the current cell.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

class Solution:
    def maxProductPath(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        f = [[[0, 0] for _ in range(n)] for _ in range(m)]

        for i in range(m):
            for j in range(n):
                x = grid[i][j]
                if i == 0 and j == 0:
                    f[i][j][0] = x
                    f[i][j][1] = x
                    continue

                mn, mx = inf, -inf

                if i > 0:
                    a, b = f[i - 1][j]
                    mn = min(mn, a * x, b * x)
                    mx = max(mx, a * x, b * x)

                if j > 0:
                    a, b = f[i][j - 1]
                    mn = min(mn, a * x, b * x)
                    mx = max(mx, a * x, b * x)

                f[i][j][0], f[i][j][1] = mn, mx

        ans = f[m - 1][n - 1][1]
        mod = 10**9 + 7
        return -1 if ans < 0 else ans % mod

Continue Practicing

#1595 minimum cost to connect two groups of points #1504 count submatrices with all ones #1463 cherry pickup ii