The DP Mistake Most Developers Make

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.

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🧩 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.

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❌ 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.

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💡 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.

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⚙️ 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)

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💻 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);
    }
}

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🚨 Edge Case

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

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⏱ Complexity

• Time: O(m × n)
• Space: O(m × n)

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🧠 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.

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✍️ Author

Ishwar Chandra Tiwari
🚀 CodeWithIshwar