How do you solve Add Two Numbers on LeetCode?

Add Two Numbers (LeetCode #2) can be solved using Brute Force or Optimal Solution (Iterative). The optimal approach is Optimal Solution (Iterative) with O(n) time complexity and O(1) space complexity. Key insight: The key observation is that the linked lists represent numbers in reverse order, allowing us to add them digit by digit directly.

What is the brute force solution for Add Two Numbers?

The brute force approach for Add Two Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves converting the linked lists into integers, adding them, and then converting the result back into a linked list. This is straightforward but not optimal because it requires extra space and time for conversions. The optimal Optimal Solution (Iterative) approach improves this to O(n).

What algorithm or data structure is used to solve Add Two Numbers?

Add Two Numbers uses the following concepts: Linked List, Math, Recursion. The recommended patterns to study are: Hash Map, Array, Two Pointers.

What are the interview tips for Add Two Numbers?

When you first see this problem, clarify the input format and confirm the linked list structure. Communicate your approach clearly, explaining how you'll handle the carry and the traversal of the lists. Mention edge cases like both lists being of different lengths or one of them being empty.

What are common mistakes when solving Add Two Numbers?

Students often forget to handle the carry when the sum exceeds 10. Another common mistake is not considering the case where one linked list is longer than the other.

Add Two Numbers

Medium

Linked ListMathRecursionHash MapArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Iterative)★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses an iterative method to add the two numbers digit by digit, taking care of the carry. This avoids the overhead of converting to and from integers, making it more efficient in both time and space.

⚙️

Algorithm

6 steps

1Step 1: Initialize a dummy head for the result linked list and a carry variable set to 0.
2Step 2: While both linked lists are not empty, extract the values and add them along with the carry.
3Step 3: Calculate the new digit (sum % 10) and update the carry (sum // 10).
4Step 4: Create a new node with the new digit and attach it to the result list.
5Step 5: Move to the next nodes in both linked lists.
6Step 6: If there's a carry left after processing both lists, add a new node with the carry.

solution.py26 lines

1# Full working Python code with comments
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7class Solution:
8    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
9        dummy_head = ListNode(0)
10        current = dummy_head
11        carry = 0
12
13        while l1 or l2 or carry:
14            val1 = l1.val if l1 else 0
15            val2 = l2.val if l2 else 0
16            total = val1 + val2 + carry
17            carry = total // 10
18            current.next = ListNode(total % 10)
19            current = current.next
20
21            if l1:
22                l1 = l1.next
23            if l2:
24                l2 = l2.next
25
26        return dummy_head.next

ℹ

Complexity note: The time complexity is O(n) because we traverse both linked lists once, and the space complexity is O(1) because we are using a constant amount of space for the carry and the result list nodes are created in place.

1The key observation is that the linked lists represent numbers in reverse order, allowing us to add them digit by digit directly.
2Another important insight is that we can handle carry values during addition, similar to how we add numbers on paper.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.