Problem Description

You have a m x n matrix filled with integers. Your task is to find a path from the top-left corner (0, 0) to the bottom-right corner (m-1, n-1) that gives the maximum non-negative product.

Movement Rules:

• Start at position (0, 0) (top-left corner)
• You can only move right or down at each step
• End at position (m-1, n-1) (bottom-right corner)

Product Calculation:

• The product of a path is calculated by multiplying all the numbers in the cells you visit along that path
• You want to find the path that gives the maximum product that is non-negative (≥ 0)

Return Value:

• If the maximum product is non-negative, return it modulo 10^9 + 7
• If the maximum product is negative (all possible paths result in negative products), return -1

Important Note: The modulo operation is applied after finding the maximum product, not during the calculation.

Example: If you have a grid like:

[2, 1, 3]
[0, 1, -2]

Some possible paths from (0,0) to (1,2) are:

• Path 1: 2 → 1 → 3 → -2 with product = 2 × 1 × 3 × (-2) = -12
• Path 2: 2 → 0 → 1 → -2 with product = 2 × 0 × 1 × (-2) = 0

The maximum non-negative product would be 0 in this case.

Intuition

When dealing with products, we need to consider that negative numbers can become positive when multiplied by another negative number. This means a very small (most negative) value could potentially become the largest positive value if we multiply it by a negative number later in the path.

Consider this scenario: if we have a path with product -100 and we encounter a cell with value -2, the product becomes 200. Meanwhile, a path with product 50 encountering the same -2 would only result in -100.

This leads us to a key insight: we need to track both the minimum and maximum products at each cell, because:

• The minimum product (most negative) could become the maximum when multiplied by a negative number
• The maximum product could become the minimum when multiplied by a negative number

For each cell (i, j), we need to know:

1. The minimum product possible to reach this cell
2. The maximum product possible to reach this cell

When we move to a new cell with value v:

• If v ≥ 0:
  • The new minimum = (smaller of the two previous minimums) × v
  • The new maximum = (larger of the two previous maximums) × v
• If v < 0:
  • The new minimum = (larger of the two previous maximums) × v (positive becomes negative)
  • The new maximum = (smaller of the two previous minimums) × v (negative becomes positive)

By maintaining both extremes throughout our path, we ensure we don't miss any opportunity where a negative intermediate result could lead to the optimal positive result later.

The dynamic programming approach naturally fits here because:

• Each cell's result only depends on its immediate predecessors (cell above or cell to the left)
• We can build up the solution incrementally from top-left to bottom-right
• The subproblems (finding min/max products to reach each cell) overlap and can be reused

Learn more about Dynamic Programming patterns.

Solution Approach

We implement a dynamic programming solution using a 3D array dp[i][j][k] where:

• dp[i][j][0] stores the minimum product to reach cell (i, j)
• dp[i][j][1] stores the maximum product to reach cell (i, j)

Step 1: Initialize the DP array

dp = [[[grid[0][0]] * 2 for _ in range(n)] for _ in range(m)]

The starting cell (0, 0) has both minimum and maximum values equal to grid[0][0].

Step 2: Fill the first column

for i in range(1, m):
    dp[i][0] = [dp[i - 1][0][0] * grid[i][0]] * 2

For cells in the first column, we can only arrive from above. Since there's only one path, both min and max are the same: the product of all cells from (0, 0) to (i, 0).

Step 3: Fill the first row

for j in range(1, n):
    dp[0][j] = [dp[0][j - 1][0] * grid[0][j]] * 2

Similarly, for the first row, we can only arrive from the left, so both min and max are the cumulative product.

Step 4: Fill the rest of the grid For each cell (i, j) where i ≥ 1 and j ≥ 1:

We can arrive from either (i-1, j) (above) or (i, j-1) (left). The current cell value is v = grid[i][j].

If v ≥ 0 (non-negative):

dp[i][j][0] = min(dp[i - 1][j][0], dp[i][j - 1][0]) * v
dp[i][j][1] = max(dp[i - 1][j][1], dp[i][j - 1][1]) * v

• Minimum stays minimum when multiplied by positive
• Maximum stays maximum when multiplied by positive

If v < 0 (negative):

dp[i][j][0] = max(dp[i - 1][j][1], dp[i][j - 1][1]) * v
dp[i][j][1] = min(dp[i - 1][j][0], dp[i][j - 1][0]) * v

• The largest positive becomes the smallest (most negative) when multiplied by negative
• The smallest (most negative) becomes the largest positive when multiplied by negative

Step 5: Return the result

ans = dp[-1][-1][1]
return -1 if ans < 0 else ans % mod

The maximum product to reach the bottom-right corner is stored in dp[m-1][n-1][1]. If it's negative, return -1; otherwise, return the value modulo 10^9 + 7.

Time Complexity: O(m × n) - we visit each cell once Space Complexity: O(m × n) - for the DP array storing min/max for each cell

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Consider this 3×3 grid:

[-1,  2,  3]
[ 4, -5,  6]
[ 7,  8, -9]

We'll track both minimum and maximum products at each cell using our DP approach.

Step 1: Initialize

• dp[0][0]: min = -1, max = -1 (starting cell value)

Step 2: Fill first row

• dp[0][1]: Only reachable from left
  • min = max = -1 × 2 = -2
• dp[0][2]: Only reachable from left
  • min = max = -2 × 3 = -6

Step 3: Fill first column

• dp[1][0]: Only reachable from above
  • min = max = -1 × 4 = -4
• dp[2][0]: Only reachable from above
  • min = max = -4 × 7 = -28

Step 4: Fill remaining cells

For dp[1][1] (value = -5, negative):

• From above: min = -2, max = -2
• From left: min = -4, max = -4
• Since -5 is negative, we swap min/max logic:
  • New min = max(-2, -4) × (-5) = -2 × (-5) = 10
  • New max = min(-2, -4) × (-5) = -4 × (-5) = 20

For dp[1][2] (value = 6, positive):

• From above: min = -6, max = -6
• From left: min = 10, max = 20
• Since 6 is positive:
  • New min = min(-6, 10) × 6 = -6 × 6 = -36
  • New max = max(-6, 20) × 6 = 20 × 6 = 120

For dp[2][1] (value = 8, positive):

• From above: min = 10, max = 20
• From left: min = -28, max = -28
• Since 8 is positive:
  • New min = min(10, -28) × 8 = -28 × 8 = -224
  • New max = max(20, -28) × 8 = 20 × 8 = 160

For dp[2][2] (value = -9, negative):

• From above: min = -36, max = 120
• From left: min = -224, max = 160
• Since -9 is negative, we swap:
  • New min = max(120, 160) × (-9) = 160 × (-9) = -1440
  • New max = min(-36, -224) × (-9) = -224 × (-9) = 2016

Step 5: Return result The maximum product at dp[2][2] is 2016, which is non-negative. Return: 2016 % (10^9 + 7) = 2016

The optimal path that gives us 2016 is: -1 → 4 → -5 → 6 → 8 → -9 Product: (-1) × 4 × (-5) × 8 × (-9) = 20 × 8 × (-9) = -1440 (Wait, this doesn't match!)

Let me recalculate the path correctly: Actually, the path is: -1 → 4 → 7 → 8 → -9 Product: (-1) × 4 × 7 × 8 × (-9) = -28 × 8 × (-9) = 2016 ✓

This example demonstrates how tracking both minimum and maximum values is crucial - the negative minimum (-224) at cell (2,1) became our maximum (2016) when multiplied by the negative value (-9) at the destination.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m * n) where m is the number of rows and n is the number of columns in the grid.

The algorithm iterates through each cell of the grid exactly once after initialization:

• Initializing the first row takes O(m) time
• Initializing the first column takes O(n) time
• The nested loops iterate through the remaining (m-1) * (n-1) cells, performing constant time operations for each cell
• Total time: O(m) + O(n) + O(m * n) = O(m * n)

Space Complexity: O(m * n)

The space is dominated by the 3D DP array dp which has dimensions m × n × 2:

• Each cell in the grid corresponds to an entry in dp that stores two values (minimum and maximum products)
• Total space used: m * n * 2 = O(m * n)
• Additional variables (m, n, mod, v, ans) use O(1) space
• Overall space complexity: O(m * n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Handling of Zero Values

Pitfall: When a cell contains 0, the product becomes 0, which might seem straightforward. However, a common mistake is not recognizing that 0 can actually be the optimal path when all other paths lead to negative products. Some implementations might incorrectly skip or special-case zeros.

Solution: Treat 0 like any other non-negative number. When multiplying by 0, both min and max become 0, which correctly propagates through the DP table.

2. Applying Modulo During Computation

Pitfall: A critical error is applying the modulo operation during the DP computation:

# WRONG - Don't do this!
dp[i][j][1] = (max(dp[i-1][j][1], dp[i][j-1][1]) * v) % MOD

This breaks the algorithm because modulo changes the actual values, making comparisons invalid. For example, if one path gives 10^9 + 8 and another gives 3, after modulo they become 1 and 3 respectively, incorrectly suggesting the second path is better.

Solution: Only apply modulo at the very end, after finding the maximum product:

# CORRECT
max_product = dp[-1][-1][1]
return -1 if max_product < 0 else max_product % MOD

3. Integer Overflow in Languages with Fixed Integer Size

Pitfall: In languages like Java or C++, multiplying large numbers can cause integer overflow. The product can grow exponentially with grid size, easily exceeding the bounds of standard integer types.

Solution: Use appropriate data types:

• In Python: No issue as integers have arbitrary precision
• In Java: Use long or BigInteger
• In C++: Use long long or implement custom handling

4. Incorrect Sign Flip Logic for Negative Numbers

Pitfall: When encountering a negative number, forgetting to swap min/max logic:

# WRONG - Using same logic as positive numbers
if v < 0:
    dp[i][j][0] = min(dp[i-1][j][0], dp[i][j-1][0]) * v  # Wrong!
    dp[i][j][1] = max(dp[i-1][j][1], dp[i][j-1][1]) * v  # Wrong!

Solution: Remember that multiplying by a negative number flips the ordering:

# CORRECT
if v < 0:
    dp[i][j][0] = max(dp[i-1][j][1], dp[i][j-1][1]) * v  # Max becomes min
    dp[i][j][1] = min(dp[i-1][j][0], dp[i][j-1][0]) * v  # Min becomes max

5. Not Tracking Both Minimum and Maximum

Pitfall: Only tracking the maximum product, thinking it's sufficient:

# WRONG - Only tracking maximum
dp[i][j] = max(dp[i-1][j], dp[i][j-1]) * grid[i][j]

This fails when negative numbers are involved because a very negative number (minimum) can become very positive (maximum) when multiplied by another negative number.

Solution: Always track both minimum and maximum at each cell to handle all possible sign combinations correctly.