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;
    struct ListNode *next;
};

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode* dummy = malloc(sizeof(struct ListNode));
    struct 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 = malloc(sizeof(struct ListNode));
        current = current->next;
        current->val = sum % 10;
        current->next = NULL;
    }
    return dummy->next;
}
    

This code adds two numbers digit by digit, moving through each linked list once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves 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

Each step processes one digit from the input lists, so the total steps grow directly with the number of digits.

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

Pattern observation: The number of steps grows in a straight line as the input size increases.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "The time depends on the sum of the numbers, not the length of the lists."

[OK] Correct: The code processes each digit once, so the time depends on how many digits there are, not the actual number values.

Interview Connect

Understanding this helps you explain how linked list traversal affects performance, a key skill in many coding problems.

Self-Check

"What if the input lists were doubly linked lists? How would the time complexity change?"

Practice

1. What does the problem "Add Two Numbers Represented as Linked List" mean?
Each node contains a single digit and digits are stored in reverse order. What is the goal?

easy

A. Reverse both linked lists and then add the digits.

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

C. Add the two numbers digit by digit and return the sum as a linked list in reverse order.

D. Sort the linked lists before adding.

Solution

1. Step 1: Understand the representation of numbers in linked lists

   Each linked list node holds one digit, and the digits are stored in reverse order, meaning the first node is the ones place.

2. Step 2: Define the goal of the problem

   The goal is to add these two numbers digit by digit, handling any carry, and return the result as a new linked list in the same reverse order format.

3. Final Answer:

   Add the two numbers digit by digit and return the sum as a linked list in reverse order. -> Option C

4. Quick Check:

   Adding reversed digit lists = sum list reversed [OK]

Hint: Add digits node-wise with carry, keep result reversed [OK]

Common Mistakes:

• Trying to add numbers as integers directly
• Ignoring carry between digits
• Not returning the result in reverse order

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

easy

A. struct Node { int val; struct Node* prev; };

B. struct Node { int val; Node* next; };

C. struct Node { int val; struct Node next; };

D. struct Node { int val; struct Node* next; };

Solution

1. Step 1: Understand C struct syntax for linked list nodes

   In C, to define a node with an integer value and pointer to next node, use 'struct Node* next;'.

2. Step 2: Identify correct pointer usage

   struct Node { int val; struct Node* next; }; correctly uses 'struct Node* next;' which is the pointer to the next node. struct Node { int val; Node* next; }; misses 'struct' keyword, C uses struct Node instead of pointer, D uses 'prev' which is for doubly linked list.

3. Final Answer:

   struct Node { int val; struct Node* next; }; -> Option D

4. Quick Check:

   Pointer to next node = struct Node* next [OK]

Hint: Use 'struct Node* next;' for next pointer in C struct [OK]

Common Mistakes:

• Omitting 'struct' keyword before Node*
• Using struct Node instead of pointer for next
• Confusing singly linked list with doubly linked list

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)
Code snippet adds digits with carry:

medium

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

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

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

D. 7 -> 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

   Digits are 7 (ones), 0 (tens), 8 (hundreds), so linked list is 7 -> 0 -> 8 -> null.

3. Final Answer:

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

4. Quick Check:

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

Hint: Add digits with carry, store remainder, move forward [OK]

Common Mistakes:

• Not handling carry correctly
• Adding digits without carry from previous sum
• Confusing order of digits in output

4. Given this buggy code snippet for adding two linked lists, what is the main error?

while(l1 != NULL || l2 != NULL) {
  int sum = carry;
  if(l1 != NULL) { sum += l1->val; l1 = l1->next; }
  if(l2 != NULL) { sum += l2->val; l2 = l2->next; }
  carry = sum / 10;
  sum = sum % 10;
  newNode->val = sum;
  newNode = newNode->next;
}

medium

A. newNode->next is not initialized before assignment, causing undefined behavior.

B. Carry is not updated correctly after sum calculation.

C. Loop condition should be && instead of ||.

D. Sum should be reset to zero inside the loop.

Solution

1. Step 1: Analyze pointer usage for newNode

   The code assigns newNode->val but then moves newNode to newNode->next without initializing newNode->next, so newNode->next is likely NULL or garbage.

2. Step 2: Identify missing node creation

   Before moving newNode = newNode->next, a new node must be created and linked to newNode->next to avoid undefined behavior.

3. Final Answer:

   newNode->next is not initialized before assignment, causing undefined behavior. -> Option A

4. Quick Check:

   Always create new node before moving pointer [OK]

Hint: Always allocate new node before moving next pointer [OK]

Common Mistakes:

• Not creating new nodes for the result list
• Incorrect loop condition causing missed nodes
• Incorrect carry update logic

5. You have two linked lists representing numbers in reverse order:
List1: 9 -> 9 -> 9 -> 9 -> null (represents 9999)
List2: 1 -> null (represents 1)
What is the correct resulting linked list after adding these two numbers?

hard

A. 9 -> 9 -> 9 -> 9 -> 1 -> null

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

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

D. 0 -> 0 -> 0 -> 0 -> null

Solution

1. Step 1: Add digits with carry step-by-step

   9+1=10 (digit 0, carry 1), next 9+0+1=10 (digit 0, carry 1), repeat for next two 9s similarly.

2. Step 2: Handle final carry

   After last digit, carry is 1, so add new node with value 1 at the end.

3. Final Answer:

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

4. Quick Check:

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

Hint: Add carry after last nodes if non-zero [OK]

Common Mistakes:

• Forgetting to add final carry node
• Stopping addition early when one list ends
• Misplacing the carry digit in the result