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

Time Complexity: Add Two Numbers Represented as Linked List

O(n)

Understanding Time Complexity

We want to understand 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.

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def addTwoNumbers(l1, l2):
    carry = 0
    dummy = ListNode()
    current = dummy
    while l1 or l2 or carry:
        val1 = l1.val if l1 else 0
        val2 = l2.val if l2 else 0
        total = val1 + val2 + carry
        carry = total // 10
        current.next = ListNode(total % 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, handling carry, and returns the sum as a 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 steps
100	About 100 steps
1000	About 1000 steps

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 number of digits.

Common Mistake

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

[OK] Correct: The code processes each digit once, so the length of the list matters, not the number's size.

Interview Connect

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

Self-Check

"What if the numbers 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 main 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 node-wise = Adding numbers represented as linked lists [OK]

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

Common Mistakes:

• Confusing addition with sorting
• Thinking the digits are stored in normal order
• Ignoring carry during addition

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

easy

A. class Node:
def __init__(self, val=0, next=None):
self.val = val
self.next = next

B. class Node:
def __init__(self, val, next):
val = self.val
next = self.next

C. class Node:
def __init__(self):
self.value = 0
self.next = None

D. class Node:
def __init__(self, val):
self.val = val
self.next = val

Solution

1. Step 1: Check correct attribute assignment

   class Node: def __init__(self, val=0, next=None): self.val = val self.next = next correctly assigns the value and next pointer using self.val and self.next.

2. Step 2: Verify default parameters and syntax

   class Node: def __init__(self, val=0, next=None): self.val = val self.next = next uses default values and proper syntax for linked list node creation.

3. Final Answer:

   class Node: def __init__(self, val=0, next=None): self.val = val self.next = next -> Option A

4. Quick Check:

   Correct node class syntax = class Node: def __init__(self, val=0, next=None): self.val = val self.next = next [OK]

Hint: Use self to assign attributes inside __init__ [OK]

Common Mistakes:

• Reversing assignment order (val = self.val)
• Forgetting to use self for attributes
• Incorrect default parameters

3. What will be the output linked list after adding these two numbers represented as linked lists? List 1: 2 -> 4 -> 3 -> null (represents 342) List 2: 5 -> 6 -> 4 -> null (represents 465) Code snippet:

def addTwoNumbers(l1, l2):
    carry = 0
    dummy = Node(0)
    current = dummy
    while l1 or l2 or carry:
        val1 = l1.val if l1 else 0
        val2 = l2.val if l2 else 0
        total = val1 + val2 + carry
        carry = total // 10
        current.next = Node(total % 10)
        current = current.next
        if l1: l1 = l1.next
        if l2: l2 = l2.next
    return dummy.next

medium

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

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

C. 2 -> 0 -> 8 -> null

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

Solution

1. Step 1: Add digits with carry

   Adding 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 -> 8 representing 807, which is 342+465.

3. Final Answer:

   7 -> 0 -> 8 -> null -> Option D

4. Quick Check:

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

Hint: Add digits and carry, build new nodes for sum digits [OK]

Common Mistakes:

• Not handling carry correctly
• Mixing digit order in output
• Adding digits without carry

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

def addTwoNumbers(l1, l2):
    carry = 0
    dummy = Node(0)
    current = dummy
    while l1 or l2:
        val1 = l1.val if l1 else 0
        val2 = l2.val if l2 else 0
        total = val1 + val2 + carry
        carry = total // 10
        current.next = Node(total % 10)
        current = current.next
        if l1: l1 = l1.next
        if l2: l2 = l2.next
    return dummy.next

medium

A. The initial dummy node is not created

B. The loop misses processing carry after both lists end

C. The nodes are not linked properly

D. The carry is not calculated correctly

Solution

1. Step 1: Analyze loop condition

   The loop runs only while l1 or l2 exist, but if carry remains after both end, it is ignored.

2. Step 2: Identify missing carry handling

   Carry should be checked in loop condition to add an extra node if needed.

3. Final Answer:

   The loop misses processing carry after both lists end -> Option B

4. Quick Check:

   Loop must include carry to avoid missing last digit [OK]

Hint: Include carry in loop condition to handle last digit [OK]

Common Mistakes:

• Ignoring carry after lists end
• Wrong loop condition
• Not updating current pointer

5. Given two linked lists representing numbers in reverse order, how would you modify the addition function to handle numbers stored in forward order (most significant digit first)?

hard

A. Use recursion to add digits from head to tail without reversing

B. Add digits directly without reversing, starting from head

C. Reverse both lists, add as usual, then reverse the result

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

Solution

1. Step 1: Understand digit order difference

   Forward order means most significant digit is at head, so addition must start from tail.

2. Step 2: Reverse lists to reuse existing logic

   Reversing lists makes digits least significant first, allowing normal addition and then reversing result restores order.

3. Final Answer:

   Reverse both lists, add as usual, then reverse the result -> Option C

4. Quick Check:

   Reverse to handle forward order addition [OK]

Hint: Reverse lists to reuse reverse-order addition logic [OK]

Common Mistakes:

• Adding digits from head without reversing
• Ignoring digit order difference
• Converting to int may cause overflow