I Got This DP Problem Wrong — Here’s Why

The DP Mistake Most Developers Make

Most developers approach Dynamic Programming by tracking only the maximum value.

It feels natural.

But in some problems… that approach fails.

🧩 The Problem

I was solving:

👉 Maximum Non-Negative Product in a Matrix

You start from (0,0) and can move only right or down.
The goal is to find the path with the maximum non-negative product.

At first glance, it seems simple:

Just track the maximum product path.

That’s exactly what I did.

And it didn’t work.

❌ Why Tracking Only Max Fails

The issue comes from negative numbers.

Because:

Negative × Negative = Positive

This changes everything.

A path that has a very small (negative) value now
can become the maximum later.

So if you only track the maximum…
you lose important information.

💡 The Key Insight

To solve this problem, you must track both:

• ✅ Maximum product
• ✅ Minimum product

Why?

Because:

• min × negative → max
• max × negative → min

This is the core idea.

⚙️ Approach

Use Dynamic Programming.

At each cell (i, j):

• max[i][j] → maximum product till here
• min[i][j] → minimum product till here

Transition:

max = max(top_max * val, top_min * val, left_max * val, left_min * val)

min = min(top_max * val, top_min * val, left_max * val, left_min * val)

💻 Java Solution

class Solution {
    public int maxProductPath(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        long[][] max = new long[m][n];
        long[][] min = new long[m][n];

        max[0][0] = min[0][0] = grid[0][0];

        for (int i = 1; i < m; i++) {
            max[i][0] = min[i][0] = max[i-1][0] * grid[i][0];
        }

        for (int j = 1; j < n; j++) {
            max[0][j] = min[0][j] = max[0][j-1] * grid[0][j];
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                long val = grid[i][j];

                long a = max[i-1][j] * val;
                long b = min[i-1][j] * val;
                long c = max[i][j-1] * val;
                long d = min[i][j-1] * val;

                max[i][j] = Math.max(Math.max(a, b), Math.max(c, d));
                min[i][j] = Math.min(Math.min(a, b), Math.min(c, d));
            }
        }

        long result = max[m-1][n-1];
        int MOD = 1_000_000_007;

        return result < 0 ? -1 : (int)(result % MOD);
    }
}

🚨 Edge Case

• If final result is negative → return -1
• Else return result % (1e9 + 7)

⏱ Complexity

• Time: O(m × n)
• Space: O(m × n)

🧠 Final Thought

This problem highlights an important idea:

Sometimes your worst path can become your best.

If you're working with multiplication and negative numbers in DP,
always think beyond just the maximum.

✍️ Author

Ishwar Chandra Tiwari
🚀 CodeWithIshwar