Add Two Numbers Represented as Linked List in DSA Javascript - 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.

function addTwoNumbers(l1, l2) {
  let dummy = new ListNode(0);
  let current = dummy;
  let carry = 0;
  while (l1 || l2 || carry) {
    let sum = (l1 ? l1.val : 0) + (l2 ? l2.val : 0) + carry;
    carry = Math.floor(sum / 10);
    current.next = new ListNode(sum % 10);
    current = current.next;
    if (l1) l1 = l1.next;
    if (l2) l2 = l2.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 while loop that goes through each node of the linked lists.
• 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 loop cycles
100	About 100 loop cycles
1000	About 1000 loop cycles

Pattern observation: The number of steps grows directly with the number of digits.

Final Time Complexity

Time Complexity: O(n)

This means the time to add the numbers grows linearly with the length of the longer linked list.

Common Mistake

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

[OK] Correct: The code only walks 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 to efficiently process linked lists digit by digit, a common pattern in coding problems.

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

B. Multiply two numbers stored in arrays and return the product as an array.

C. Sort two linked lists and merge them into one sorted list.

D. Find the difference between two numbers represented as linked lists.

Solution

1. Step 1: Understand the problem statement

   The problem requires adding two numbers where each number is represented by a linked list. Each node contains a single digit, and digits are stored in reverse order.

2. Step 2: Identify the expected output

   The output should be a linked list representing the sum of the two numbers, also in reverse order.

3. Final Answer:

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

4. Quick Check:

   Adding digits node by node = Add linked lists [OK]

Hint: Remember digits are reversed in linked lists [OK]

Common Mistakes:

• Thinking digits are stored in normal order
• Confusing addition with multiplication
• Returning a number instead of a linked list

2. Which of the following is the correct way to create a node for a singly linked list in JavaScript for this problem?

easy

A. function ListNode(val) { this.val = val; this.next = null; }

B. class ListNode { constructor(val) { this.value = val; this.nextNode = null; } }

C. let node = { value: val, next: null };

D. function Node(val) { this.val = val; this.nextNode = null; }

Solution

1. Step 1: Check the standard node structure

   The node should have a property named 'val' for the digit and 'next' for the next node pointer.

2. Step 2: Verify the syntax correctness

   function ListNode(val) { this.val = val; this.next = null; } correctly defines a function constructor with 'val' and 'next' properties initialized properly.

3. Final Answer:

   function ListNode(val) { this.val = val; this.next = null; } -> Option A

4. Quick Check:

   Node with val and next properties = function ListNode(val) { this.val = val; this.next = null; } [OK]

Hint: Use 'val' and 'next' for node properties [OK]

Common Mistakes:

• Using 'value' instead of 'val'
• Using 'nextNode' instead of 'next'
• Incorrect constructor syntax

3. What is 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)

function addTwoNumbers(l1, l2) {
  let dummy = new ListNode(0);
  let current = dummy;
  let carry = 0;
  while (l1 || l2 || carry) {
    let sum = carry;
    if (l1) { sum += l1.val; l1 = l1.next; }
    if (l2) { sum += l2.val; l2 = l2.next; }
    carry = Math.floor(sum / 10);
    current.next = new ListNode(sum % 10);
    current = current.next;
  }
  return dummy.next;
}

medium

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

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

C. 7 -> 1 -> 7 -> null

D. 8 -> 0 -> 7 -> null

Solution

1. Step 1: Add digits node by node with carry

   2+5=7 (carry 0), 4+6=10 (carry 1), 3+4+1=8 (carry 0)

2. Step 2: Form the resulting linked list

   The digits are 7 -> 0 -> 8 representing 807 in reverse order.

3. Final Answer:

   7 -> 0 -> 8 -> null -> Option B

4. Quick Check:

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

Hint: Add digits with carry, build new nodes [OK]

Common Mistakes:

• Ignoring carry after adding digits
• Mixing digit order in output
• Returning sum as a number, not linked list

4. Identify the bug in the following code snippet for adding two numbers represented as linked lists:

function addTwoNumbers(l1, l2) {
  let dummy = new ListNode(0);
  let current = dummy;
  let carry = 0;
  while (l1 !== null && l2 !== null) {
    let sum = carry + l1.val + l2.val;
    carry = Math.floor(sum / 10);
    current.next = new ListNode(sum % 10);
    current = current.next;
    l1 = l1.next;
    l2 = l2.next;
  }
  return dummy.next;
}

medium

A. The function does not return the dummy node itself.

B. The carry is not calculated correctly using Math.floor.

C. The dummy node should be initialized with value 1 instead of 0.

D. The loop condition should use '||' instead of '&&' to handle lists of different lengths.

Solution

1. Step 1: Analyze the loop condition

   The loop stops when either l1 or l2 becomes null, missing remaining nodes if lists differ in length.

2. Step 2: Correct the loop condition

   Using '||' ensures the loop continues until both lists and carry are fully processed.

3. Final Answer:

   The loop condition should use '||' instead of '&&' to handle lists of different lengths. -> Option D

4. Quick Check:

   Loop condition must cover all nodes = The loop condition should use '||' instead of '&&' to handle lists of different lengths. [OK]

Hint: Use '||' to continue while any list or carry remains [OK]

Common Mistakes:

• Stopping loop too early with '&&'
• Ignoring leftover carry after loop
• Not advancing pointers correctly

5. Given two linked lists representing numbers in reverse order, how would you modify the addition function to handle the case where the input lists have different lengths and the final carry creates a new node? Choose the best approach to ensure correct output for inputs like: List 1: 9 -> 9 -> 9 -> null (represents 999) List 2: 1 -> null (represents 1)

function addTwoNumbers(l1, l2) {
  let dummy = new ListNode(0);
  let current = dummy;
  let carry = 0;
  while (l1 || l2 || carry) {
    let sum = carry;
    if (l1) { sum += l1.val; l1 = l1.next; }
    if (l2) { sum += l2.val; l2 = l2.next; }
    carry = Math.floor(sum / 10);
    current.next = new ListNode(sum % 10);
    current = current.next;
  }
  return dummy.next;
}

hard

A. Add zeros to the shorter list before starting the addition to equalize lengths.

B. Stop the loop when both lists end and ignore carry since it won't affect the sum.

C. Use a loop condition that continues while l1, l2, or carry exists, adding carry as a new node if needed.

D. Only add nodes while both lists have nodes, then append the remaining nodes without addition.

Solution

1. Step 1: Understand the need to handle different lengths and carry

   When lists differ in length, addition must continue until both lists and carry are fully processed.

2. Step 2: Use a loop condition that checks l1, l2, or carry

   This ensures the loop adds all digits and creates a new node if carry remains after lists end.

3. Final Answer:

   Use a loop condition that continues while l1, l2, or carry exists, adding carry as a new node if needed. -> Option C

4. Quick Check:

   Loop until all digits and carry processed = Use a loop condition that continues while l1, l2, or carry exists, adding carry as a new node if needed. [OK]

Hint: Loop while any list or carry remains to add all digits [OK]

Common Mistakes:

• Ignoring carry after lists end
• Trying to pad shorter list manually
• Appending leftover nodes without adding carry