The problem involves adding two numbers represented by linked lists, where each node contains a single digit. The digits are stored in reverse order, and the result must also be returned as a linked list.

For example:

Input:
l1 = [2,4,3]
l2 = [5,6,4]

Output:
[7,0,8]

This problem may look simple at first, but it requires careful handling of several important concepts:

🔹 Linked List Traversal
I learned how to traverse two linked lists simultaneously while processing their corresponding nodes.

🔹 Carry Management
Just like traditional addition, when the sum of two digits is greater than 9, a carry needs to be passed to the next position. Handling the final carry after both lists are exhausted is also important.

🔹 Different List Lengths
The two linked lists may not contain the same number of nodes. The solution needs to continue processing even when one list becomes null.

🔹 Creating the Result Dynamically
I used a dummy/head node approach to construct the resulting linked list efficiently and return the correct result.

📊 My LeetCode Result:

✅ Accepted
🧪 Test Cases: 1569 / 1569 passed
⚡ Runtime: 1 ms
🏆 Beats: 99.69% of JavaScript submissions
💾 Memory: 61.65 MB
📈 Memory: Beats 43.02%

One of the biggest takeaways from this problem is that understanding the underlying data structure is more important than simply writing code that passes the test cases.

Working with linked lists has helped me become more comfortable with pointers/references, node traversal, conditional logic, and edge-case handling.

I’m continuing to practice LeetCode problems to strengthen my problem-solving skills, JavaScript fundamentals, Data Structures & Algorithms knowledge, and coding efficiency.

Every problem solved is another opportunity to learn something new. The goal isn't just to get an “Accepted” result, but to understand why the solution works and how it can be improved.

I’m looking forward to solving more problems, exploring different approaches, and improving my coding skills one challenge at a time. 🚀

Consistency + Practice + Learning = Growth.

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