Add Two Numbers Represented as Linked List in DSA Go - Time & Space Complexity

Time Complexity: Add Two Numbers Represented as Linked List

O(n)

Understanding Time Complexity

We want to understand how the time needed to add two numbers stored as linked lists grows as the numbers get longer.

How does the work change when the linked lists have more digits?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    dummy := &ListNode{}
    current := dummy
    carry := 0
    for l1 != nil || l2 != nil || carry != 0 {
        sum := carry
        if l1 != nil {
            sum += l1.Val
            l1 = l1.Next
        }
        if l2 != nil {
            sum += l2.Val
            l2 = l2.Next
        }
        carry = sum / 10
        current.Next = &ListNode{Val: sum % 10}
        current = current.Next
    }
    return dummy.Next
}
    

This code adds two numbers digit by digit, stored in linked lists, and returns the sum as a new linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The single loop that moves through both linked lists simultaneously.
• How many times: It runs once for each digit in the longer linked list, plus one extra if there is a carry at the end.

How Execution Grows With Input

As the number of digits (nodes) increases, the loop runs more times, roughly once per digit.

Input Size (n)	Approx. Operations
10	About 10 steps through the loop
100	About 100 steps through the loop
1000	About 1000 steps through the loop

Pattern observation: The work grows directly with the number of digits; doubling digits roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to add the numbers grows linearly with the number of digits in the longer list.

Common Mistake

[X] Wrong: "The time depends on both lists multiplied together, like O(n * m)."

[OK] Correct: The code only moves through each list once, so the time depends on the longer list length, not the product of lengths.

Interview Connect

Understanding this linear time complexity helps you explain how your code handles numbers of any size efficiently, a skill valued in many coding challenges.

Self-Check

"What if the digits were stored in forward order instead of reverse? How would the time complexity change?"

Practice

1. What does the problem "Add Two Numbers Represented as Linked List" ask you to do?

easy

A. Reverse the digits of a number stored in a linked list.

B. Multiply two numbers stored as linked lists and return the product as a linked list.

C. Add two numbers where each digit is stored in a node of a linked list, and return the sum as a linked list.

D. Find the maximum digit in a linked list representing a number.

Solution

1. Step 1: Understand the problem statement and identify the expected output

   The problem requires adding two numbers where each number is represented by a linked list, with each node holding a single digit. The sum should also be returned as a linked list, where each node contains a digit of the result.

2. Final Answer:

   Add two numbers represented as linked lists and return their sum as a linked list -> Option C

3. Quick Check:

   Adding digits node by node = Add two numbers where each digit is stored in a node of a linked list, and return the sum as a linked list. [OK]

Hint: Focus on adding digit by digit with carry [OK]

Common Mistakes:

• Confusing addition with multiplication
• Ignoring carry over between digits
• Returning a number instead of a linked list

2. Which of the following Go code snippets correctly defines a node for a singly linked list used to represent a digit in the "Add Two Numbers Represented as Linked List" problem?

easy

A. type ListNode struct { Val int; Next *ListNode }

B. type ListNode struct { Value string; Next ListNode }

C. type Node struct { Val int; Next int }

D. type ListNode struct { Val int; Next ListNode }

Solution

1. Step 1: Check struct field types and verify pointer usage

   The node should hold an integer value and a pointer to the next node, so Val is int and Next is *ListNode. Next must be a pointer to ListNode to allow linking nodes dynamically.

2. Final Answer:

   type ListNode struct { Val int; Next *ListNode } -> Option A

3. Quick Check:

   Pointer to next node = type ListNode struct { Val int; Next *ListNode } [OK]

Hint: Use *ListNode for next pointer in Go structs [OK]

Common Mistakes:

• Using value type instead of pointer for Next
• Using wrong field types like string for digits
• Missing semicolons or commas in struct

3. What will be the output linked list after adding the two numbers represented by these linked lists? List 1: 2 -> 4 -> 3 -> null (represents 342) List 2: 5 -> 6 -> 4 -> null (represents 465) Use the standard addition approach digit by digit with carry.

medium

A. 7 -> 0 -> 8 -> null

B. 8 -> 0 -> 7 -> null

C. 7 -> 10 -> 7 -> null

D. 7 -> 1 -> 8 -> null

Solution

1. Step 1: Add digits from head nodes with carry and form the result linked list

   2 + 5 = 7, carry 0; next digits 4 + 6 = 10, store 0 carry 1; next digits 3 + 4 + 1 = 8. The result digits are 7 -> 0 -> 8 representing 807.

2. Final Answer:

   7 -> 0 -> 8 -> null -> Option A

3. Quick Check:

   342 + 465 = 807 = 7->0->8 [OK]

Hint: Add digits with carry, store remainder as node value [OK]

Common Mistakes:

• Not carrying over when sum >= 10
• Adding digits in wrong order
• Confusing node values with actual number order

4. Consider this Go function snippet to add two numbers represented as linked lists. What is the bug? func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode { carry := 0 dummy := &ListNode{} current := dummy for l1 != nil && l2 != nil { sum := l1.Val + l2.Val + carry carry = sum / 10 current.Next = &ListNode{Val: sum % 10} current = current.Next l1 = l1.Next l2 = l2.Next } return dummy.Next }

medium

A. The dummy node is not initialized properly.

B. Carry is not calculated correctly after sum.

C. The function does not return the correct node.

D. The loop stops when either list ends, missing remaining nodes in longer list.

Solution

1. Step 1: Analyze loop condition and identify missing processing

   The loop runs only while both l1 and l2 are not nil, so it stops if one list is longer. Remaining nodes in the longer list and any carry after loop are not handled, causing incorrect results.

2. Final Answer:

   The loop stops when either list ends, missing remaining nodes in longer list -> Option D

3. Quick Check:

   Loop condition misses longer list nodes = The loop stops when either list ends, missing remaining nodes in longer list. [OK]

Hint: Use OR condition to continue until both lists and carry are done [OK]

Common Mistakes:

• Using AND instead of OR in loop condition
• Ignoring leftover carry after loop
• Not linking remaining nodes after loop

5. You have two linked lists representing numbers in reverse order: List 1: 9 -> 9 -> 9 -> 9 -> null (represents 9999) List 2: 1 -> null (represents 1) What will be the resulting linked list after adding these two numbers?

hard

A. 0 -> 0 -> 0 -> 0 -> null

B. 0 -> 0 -> 0 -> 0 -> 1 -> null

C. 1 -> 0 -> 0 -> 0 -> 0 -> null

D. 9 -> 9 -> 9 -> 10 -> null

Solution

1. Step 1: Add digits with carry step-by-step and add final carry as new node

   9+1=10, store 0 carry 1; next 9+0+1=10, store 0 carry 1; repeat for all 9s, each time store 0 carry 1. After last 9, carry is 1, so add a new node with value 1 at the end.

2. Final Answer:

   0 -> 0 -> 0 -> 0 -> 1 -> null -> Option B

3. Quick Check:

   9999 + 1 = 10000 = 0->0->0->0->1 [OK]

Hint: Remember to add carry node if leftover after last digit [OK]

Common Mistakes:

• Not adding extra node for leftover carry
• Stopping addition too early
• Misplacing the carry node in the list