Add Two Numbers Represented as Linked List in DSA Java - 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 work change when the lists have more digits?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class ListNode {
  int val;
  ListNode next;
  ListNode(int val) { this.val = val; }
}

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
  ListNode dummy = new ListNode(0);
  ListNode current = dummy;
  int carry = 0;
  while (l1 != null || l2 != null || carry != 0) {
    int sum = carry;
    if (l1 != null) { sum += l1.val; l1 = l1.next; }
    if (l2 != null) { 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, handling carry, and creates a new linked list for the result.

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 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 work grows directly with the number of digits; doubling digits 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.

Common Mistake

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

[OK] Correct: The code processes each digit once, so time depends on the number of digits, not the value of the numbers.

Interview Connect

Understanding this helps you explain how to handle linked lists and carry in addition, a common coding challenge skill.

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" mainly involve?

easy

A. Reversing a linked list

B. Sorting two linked lists in ascending order

C. Finding the middle node of a linked list

D. Adding digits node by node from two linked lists representing numbers

Solution

1. Step 1: Understand the problem context

   The problem involves two linked lists where each node contains a 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 digits node by node from two linked lists representing numbers -> Option D

4. Quick Check:

   Adding digits node by node = D [OK]

Hint: Focus on digit-wise addition with carry [OK]

Common Mistakes:

• Confusing addition with sorting
• Ignoring carry during addition
• Assuming numbers are stored in forward order

2. Which of the following Java code snippets correctly initializes a new node for a singly linked list representing a digit?

easy

A. ListNode node = ListNode(5);

B. ListNode node = new ListNode(); node.next = 5;

C. ListNode node = new ListNode(5);

D. ListNode node = new ListNode.val(5);

Solution

1. Step 1: Recall Java object creation syntax

   In Java, to create a new object, use the 'new' keyword followed by the constructor call with parameters.

2. Step 2: Check each option for correct syntax

   ListNode node = new ListNode(5); uses 'new ListNode(5);' which is correct. ListNode node = ListNode(5); misses 'new'. ListNode node = new ListNode(); node.next = 5; incorrectly sets next instead of val. ListNode node = new ListNode.val(5); is invalid syntax.

3. Final Answer:

   ListNode node = new ListNode(5); -> Option C

4. Quick Check:

   Java new object syntax = B [OK]

Hint: Use 'new' keyword with constructor for object creation [OK]

Common Mistakes:

• Omitting 'new' keyword
• Incorrect constructor call syntax
• Trying to call constructor like a method without 'new'
• Confusing node.val and node.next

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)

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummy;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummy.next;
}

medium

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

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

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

D. 2 -> 4 -> 3 -> 5 -> 6 -> 4 -> null

Solution

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

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

2. Step 2: Construct the resulting linked list

   Digits are 7 (units), 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 represented as 7->0->8 [OK]

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

Common Mistakes:

• Ignoring carry in addition
• Appending nodes incorrectly
• Mixing digit order in output

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

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode curr = dummy;
    int carry = 0;
    while (l1 != null && l2 != null) {
        int sum = carry + l1.val + l2.val;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        l1 = l1.next;
        l2 = l2.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummy.next;
}

medium

A. The carry is not calculated correctly

B. The while loop should use '||' instead of '&&' to handle different length lists

C. The dummy node is not initialized properly

D. The function does not return the correct node

Solution

1. Step 1: Analyze the loop condition

   The loop uses '&&' which stops when either list ends, missing remaining nodes of longer list.

2. Step 2: Correct loop condition to '||'

   Using '||' ensures the loop continues until both lists are fully processed, handling different lengths correctly.

3. Final Answer:

   The while loop should use '||' instead of '&&' to handle different length lists -> Option B

4. Quick Check:

   Loop condition must cover all nodes = A [OK]

Hint: Use '||' to process all nodes in both lists [OK]

Common Mistakes:

• Using '&&' causing early loop exit
• Not handling carry after loop
• 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 where the input lists are extremely long (millions of nodes) and memory is limited?

hard

A. Use iterative addition with constant extra space and process nodes one by one

B. Convert linked lists to integers, add, then convert back to linked list

C. Store all digits in arrays before adding to optimize speed

D. Reverse both lists first, add, then reverse the result

Solution

1. Step 1: Consider memory constraints with large inputs

   Converting to integers or arrays requires large memory, not feasible for millions of nodes.

2. Step 2: Use iterative addition with constant extra space

   Process nodes one by one, create result nodes on the fly, and avoid storing entire numbers in memory.

3. Final Answer:

   Use iterative addition with constant extra space and process nodes one by one -> Option A

4. Quick Check:

   Iterative node-wise addition saves memory = C [OK]

Hint: Add nodes one by one without full number conversion [OK]

Common Mistakes:

• Trying to convert entire list to integer causing overflow
• Using extra arrays increasing memory usage
• Reversing lists unnecessarily increasing complexity