How do you solve Maximum Non Negative Product in a Matrix on LeetCode?

Maximum Non Negative Product in a Matrix (LeetCode #1594) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Dynamic programming is essential for optimizing path problems in grids.

What is the brute force solution for Maximum Non Negative Product in a Matrix?

The brute force approach for Maximum Non Negative Product in a Matrix is Brute Force, with O(2^(m+n)) time complexity and O(m+n) space complexity. In the brute force approach, we explore all possible paths from the top-left corner to the bottom-right corner of the matrix. We calculate the product of the numbers along each path and keep track of the maximum non-negative product found. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Maximum Non Negative Product in a Matrix?

Maximum Non Negative Product in a Matrix uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Matrix Traversal.

What are the interview tips for Maximum Non Negative Product in a Matrix?

Explain your thought process clearly while coding. Discuss edge cases and how your solution handles them. Practice dynamic programming problems to build intuition.

What are common mistakes when solving Maximum Non Negative Product in a Matrix?

Not considering the case where the maximum product is negative. Failing to initialize the DP arrays correctly.

#1594

Maximum Non Negative Product in a Matrix

Medium

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^(m+n))	O(m * n)
Space	O(m+n)	O(m * n)

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Intuition

Time O(m * n)Space O(m * n)

In the optimal solution, we use dynamic programming to keep track of both the maximum and minimum products at each cell. This allows us to efficiently calculate the maximum non-negative product while considering the effects of negative numbers.

⚙️

Algorithm

4 steps

1Step 1: Create two matrices, maxProd and minProd, to store the maximum and minimum products up to each cell.
2Step 2: Initialize the starting point (0, 0) in both matrices with the value of grid[0][0].
3Step 3: Iterate through each cell in the grid, updating maxProd and minProd based on the values from the top and left cells.
4Step 4: At the end, check the value in maxProd at the bottom-right corner. If it's negative, return -1; otherwise, return maxProd[m-1][n-1] % (10^9 + 7).

solution.py25 lines

1# Full working Python code
2from typing import List
3
4def maxNonNegativeProduct(grid: List[List[int]]) -> int:
5    MOD = 10**9 + 7
6    m, n = len(grid), len(grid[0])
7    maxProd = [[0] * n for _ in range(m)]
8    minProd = [[0] * n for _ in range(m)]
9    maxProd[0][0] = minProd[0][0] = grid[0][0]
10
11    for i in range(m):
12        for j in range(n):
13            if i == 0 and j == 0: continue
14            max_val, min_val = float('-inf'), float('inf')
15            if i > 0:
16                max_val = max(max_val, maxProd[i-1][j] * grid[i][j])
17                min_val = min(min_val, minProd[i-1][j] * grid[i][j])
18            if j > 0:
19                max_val = max(max_val, maxProd[i][j-1] * grid[i][j])
20                min_val = min(min_val, minProd[i][j-1] * grid[i][j])
21            maxProd[i][j] = max_val
22            minProd[i][j] = min_val
23
24    result = maxProd[m-1][n-1]
25    return result % MOD if result >= 0 else -1

ℹ

Complexity note: The time complexity is linear with respect to the number of cells in the grid, as we visit each cell once. The space complexity is also linear due to the storage of two matrices.

1Dynamic programming is essential for optimizing path problems in grids.
2Tracking both maximum and minimum products helps handle negative numbers effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.