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👋 Hello, Harman here Welcome to this week’s Fullstack Insider newsletter.
If you’re preparing for coding interviews, this guide will walk you through solving the Add Two Numbers problem (linked lists addition) with detailed explanations, code, and dry runs.
Hey Builders 👋,
After Two Sum (#1), the very next LeetCode problem is Add Two Numbers (#2).
It’s slightly trickier but equally important because it tests your ability to:
Work with linked lists
Handle carryovers in addition
Write clean iterative logic
At first glance, it looks like simple math: add two numbers digit by digit.
But since the numbers are represented as linked lists in reverse order, you have to carefully simulate how addition works by hand.
🚀 Why This Matters
Interview Favorite → It’s a standard FAANG question. Interviewers love it because it checks linked list fundamentals.
Teaches Core Concepts → You’ll practice pointer manipulation, loops, and handling carry.
Real-World Use Cases →
Processing large numbers digit by digit (e.g., big integer math libraries).
Handling streams of data in chunks instead of whole numbers.
🛠 Problem Statement
Here’s the official LeetCode statement:
You are given two non-empty linked lists representing two non-negative integers.
The digits are stored in reverse order, and each node contains a single digit.
Add the two numbers and return the sum as a linked list.
Example
🔑 Step 1: Understanding the Problem
Imagine you’re adding numbers by hand:
342
+ 465
------
807But instead of writing them normally, they’re stored like this:
l1 = 2 → 4 → 3
l2 = 5 → 6 → 4 Reverse order makes it easier to start from the least significant digit (the ones place).
🔑 Step 2: Brute Force Thinking
A naive way would be:
Convert linked lists → integers.
Add them.
Convert back → linked list.
But this fails for very large numbers (bigger than integer limits).
That’s why LeetCode wants us to add digit by digit like we do manually.
🔑 Step 3: Optimized Iterative Solution
The idea:
Use a dummy head node to simplify result building.
Traverse both lists at once.
Keep a carry for sums greater than 9.
Create new nodes for each digit in the result.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def addTwoNumbers(l1, l2):
dummy = ListNode()
current = dummy
carry = 0
while l1 or l2 or carry:
val1 = l1.val if l1 else 0
val2 = l2.val if l2 else 0
total = val1 + val2 + carry
carry = total // 10
current.next = ListNode(total % 10)
current = current.next
l1 = l1.next if l1 else None
l2 = l2.next if l2 else None
return dummy.next
Walkthrough Example
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⏱ Complexity
Time Complexity: O(max(m,n)) → we traverse both lists once.
Space Complexity: O(max(m,n)) → result list size is proportional to longer input.
🔑 Step 4: Edge Cases
One list longer than the other → handled by
if l1 else 0.Carry leftover at the end → handled by
while l1 or l2 or carry.Very large numbers → safe because we never convert to int directly.
🔑 Step 5: Interview Tips
Don’t jump into code. First explain the manual addition process.
Use a dummy node — simplifies handling head node edge cases.
Always test with:
Equal lengths
Different lengths
Carry at the very end
✅ Pro Tip:
Interviewers don’t just want working code. They want to see that you think in steps.
Explain brute force first → then optimize with linked list addition logic.
💬 Question for You
When solving linked list problems, do you prefer to visualize with diagrams first or jump into code immediately?
Reply and let me know—I’ll feature the best answers in the next edition 🚀
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That’s a wrap! Catch you in next edition. 👋
—Harman