Add Two Numbers Represented as Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the sum digit.

DSA Python

new_node = ListNode([1])

Drag options to blanks, or click blank then click option'

Asum % 10

Bsum + 10

Csum // 10

Dsum * 10

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum // 10 instead of sum % 10

Adding 10 to sum instead of modulo

Multiplying sum by 10

The digit to store in the node is the remainder when sum is divided by 10.

2fill in blank

medium

Complete the code to update the carry after adding two digits.

DSA Python

carry = [1]

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Asum // 10

Bsum % 10

Csum + 1

Dsum - 10

Attempts:

3 left

💡 Hint

Common Mistakes

Using modulo instead of integer division

Adding 1 to sum instead of dividing

Subtracting 10 from sum

Carry is the quotient when sum is divided by 10.

3fill in blank

hard

Fix the error in the loop condition to continue while nodes or carry exist.

DSA Python

while l1 or l2 or [1]:

Drag options to blanks, or click blank then click option'

Anew_node

Bsum

CNone

Dcarry

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum which is undefined here

Using new_node which is a node, not a condition

Using None which stops the loop

The loop must continue if there is a carry left to add.

4fill in blank

hard

Fill both blanks to correctly move to the next nodes in the lists.

DSA Python

l1 = l1[1] if l1 else None
l2 = l2[2] if l2 else None

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A.next

B.prev

C.head

D.tail

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev which is for doubly linked lists backwards

Using .head or .tail which are not node pointers

We move to the next node in each linked list using the .next attribute.

5fill in blank

hard

Fill both blanks to initialize the dummy node, set current pointer, and return the result list.

DSA Python

dummy = ListNode([1])
current = dummy
...
return dummy[2]

Drag options to blanks, or click blank then click option'

A0

B.next

C.prev

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Returning dummy instead of dummy.next

Using None instead of 0 for dummy node value

Using .prev which is not correct here

Dummy node starts with 0, current points to dummy, and we return dummy.next as the head of the result list.

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