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

Time Complexity: Add Two Numbers Represented as Linked List

O(n)

Understanding Time Complexity

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

How does the number of steps change when the linked lists have more digits?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode* dummy = new ListNode(0);
    ListNode* current = dummy;
    int carry = 0;
    while (l1 || l2 || carry) {
        int sum = carry;
        if (l1) { sum += l1->val; l1 = l1->next; }
        if (l2) { sum += l2->val; l2 = l2->next; }
        carry = sum / 10;
        current->next = new ListNode(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 while loop that goes through each node of both 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 number of steps grows roughly the same amount.

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

Pattern observation: The time grows linearly with the number of digits.

Final Time Complexity

Time Complexity: O(n)

This means the time to add the numbers grows directly in proportion to the number of digits.

Common Mistake

[X] Wrong: "The time complexity is O(n²) because we create new nodes inside the loop."

[OK] Correct: Creating a new node each time is a constant time operation, done once per digit, so it does not multiply the steps; the loop still runs only once per digit.

Interview Connect

Understanding this linear time complexity helps you explain how your solution scales with input size, a key skill in interviews.

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" mainly involve?

easy

A. Adding two numbers where each digit is stored in a node of a linked list

B. Sorting two linked lists in ascending order

C. Finding the middle node of a linked list

D. Reversing a linked list

Solution

1. Step 1: Understand the problem context

   The problem involves two linked lists where each node contains a single digit of a number, stored in reverse order.

2. Step 2: Identify the required operation

   We need to add these two numbers digit by digit, handling carry, and return the sum as a linked list.

3. Final Answer:

   Adding two numbers where each digit is stored in a node of a linked list -> Option A

4. Quick Check:

   Adding digits in linked lists = Adding two numbers where each digit is stored in a node of a linked list [OK]

Hint: Think of linked list nodes as digits of a number [OK]

Common Mistakes:

• Confusing addition with sorting
• Thinking the problem is about reversing lists
• Assuming the digits are stored in normal order

2. Which of the following C++ code snippets correctly declares a node structure for a singly linked list used in this problem (using nullptr)?

easy

A. struct Node { int val; Node* next; Node(int x) : val(x), next(NULL) {} };

B. struct Node { int val; Node next; Node(int x) : val(x), next(nullptr) {} };

C. struct Node { int val; Node* next; Node(int x) : val(x), next(nullptr) {} };

D. struct Node { int val; Node* next; Node(int x) { val = x; next = NULL; } };

Solution

1. Step 1: Check pointer usage in struct

   Node* next is correct to point to the next node; struct Node { int val; Node next; Node(int x) : val(x), next(nullptr) {} }; uses Node next which is incorrect (causes infinite recursion).

2. Step 2: Check constructor initialization

   struct Node { int val; Node* next; Node(int x) : val(x), next(nullptr) {} }; uses member initializer list correctly with nullptr; struct Node { int val; Node* next; Node(int x) { val = x; next = NULL; } }; uses assignment inside constructor body which is valid but less modern; struct Node { int val; Node* next; Node(int x) : val(x), next(NULL) {} }; uses NULL which is legacy.

3. Final Answer:

   struct Node { int val; Node* next; Node(int x) : val(x), next(nullptr) {} }; -> Option C

4. Quick Check:

   Use pointer and nullptr in constructor [OK]

Hint: Use Node* next and initialize with nullptr [OK]

Common Mistakes:

• Using Node next instead of Node* next
• Using NULL instead of nullptr
• Not initializing next pointer

3. What is the output linked list after adding the two numbers represented by these linked lists? List1: 2 -> 4 -> 3 -> null (represents 342) List2: 5 -> 6 -> 4 -> null (represents 465) Choose the correct resulting linked list:

medium

A. 7 -> 10 -> 7 -> null

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

C. 8 -> 0 -> 7 -> null

D. 7 -> 1 -> 8 -> 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

   Digits are 7, 0 (from 10 carry), 8 so list is 7 -> 0 -> 8 -> null

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Not handling carry correctly
• Adding digits in wrong order
• Confusing node values with actual numbers

4. Consider the following code snippet for adding two numbers represented as linked lists. What is the error?

Node* addTwoNumbers(Node* l1, Node* l2) {
  Node* dummy = new Node(0);
  Node* current = dummy;
  int carry = 0;
  while (l1 != nullptr && l2 != nullptr) {
    int sum = l1->val + l2->val + carry;
    carry = sum / 10;
    current->next = new Node(sum % 10);
    current = current->next;
    l1 = l1->next;
    l2 = l2->next;
  }
  return dummy->next;
}

medium

A. The loop condition should be while (l1 != nullptr || l2 != nullptr)

B. Carry is not updated correctly

C. Dummy node should be deleted before return

D. The sum calculation is incorrect

Solution

1. Step 1: Analyze loop condition

   The loop runs only while both l1 and l2 are not null, so it stops early if one list is longer.

2. Step 2: Correct loop condition

   It should continue while either l1 or l2 has nodes to process all digits, so use || instead of &&.

3. Final Answer:

   The loop condition should be while (l1 != nullptr || l2 != nullptr) -> Option A

4. Quick Check:

   Use || to handle different length lists [OK]

Hint: Use || in loop to cover all nodes [OK]

Common Mistakes:

• Using && causes missing nodes in longer list
• Ignoring carry after loop ends
• Not advancing pointers correctly

5. Given two linked lists representing numbers in reverse order, how would you modify the addition algorithm to handle the case when the final carry is non-zero after processing all nodes? Choose the best approach:

hard

A. Ignore the carry since it is not needed

B. Subtract the carry from the last node value

C. Restart the addition from the beginning

D. Add a new node with the carry value at the end of the result list

Solution

1. Step 1: Understand carry after full addition

   After adding all nodes, if carry is still 1, it means the sum has an extra digit.

2. Step 2: Add new node for carry

   To represent this extra digit, create a new node with carry value and link it at the end.

3. Final Answer:

   Add a new node with the carry value at the end of the result list -> Option D

4. Quick Check:

   Carry leftover means add new node [OK]

Hint: Add carry node if leftover after loop ends [OK]

Common Mistakes:

• Ignoring leftover carry causing wrong sum
• Trying to subtract carry instead of adding node
• Restarting addition unnecessarily