Add Two Numbers Represented as Linked List in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does each node in the linked list represent in the problem 'Add Two Numbers Represented as Linked List'?

Each node represents a single digit of a number, stored in reverse order (the 1's digit is at the head).

beginner

Why do we add digits starting from the head of the linked lists in this problem?

Because the digits are stored in reverse order, the head contains the least significant digit, so addition starts from the units place.

beginner

What is the role of the carry in adding two numbers represented as linked lists?

The carry stores the overflow digit when the sum of two digits exceeds 9, and it is added to the next pair of digits.

intermediate

How do you handle linked lists of different lengths when adding two numbers represented as linked lists?

Continue adding digits until both lists are fully traversed, treating missing digits as 0, and include any remaining carry.

intermediate

What is the time complexity of adding two numbers represented as linked lists?

The time complexity is O(max(m, n)) where m and n are the lengths of the two linked lists, since each node is processed once.

In the problem 'Add Two Numbers Represented as Linked List', what does the head node represent?

A The carry digit

B The most significant digit

C The middle digit

D The least significant digit

The head node stores the least significant digit because the number is stored in reverse order.

What should you do if one linked list is longer than the other during addition?

A Continue adding with the longer list's digits and carry

B Reverse the longer list

C Ignore the extra digits

D Stop when the shorter list ends

You continue adding the remaining digits of the longer list along with any carry.

What is the purpose of the carry variable in the addition process?

A To store the sum of digits

B To count the number of nodes

C To store the overflow digit from sum > 9

D To reverse the linked list

Carry holds the overflow digit when the sum of two digits is greater than 9.

If the sum of the last digits plus carry is 15, what digit is stored in the new node?

A 5

B 1

C 0

D 15

Only the last digit (5) is stored; the carry (1) is passed to the next addition.

What is the output linked list for adding (2 -> 4 -> 3) and (5 -> 6 -> 4)?

A 8 -> 0 -> 7

B 7 -> 0 -> 8

C 7 -> 10 -> 7

D 7 -> 6 -> 7

342 + 465 = 807, stored as 7 -> 0 -> 8 in reverse order.

Explain step-by-step how to add two numbers represented as linked lists.

Think about how you add numbers on paper from right to left.

Describe how to handle the carry when adding two linked lists digit by digit.

Carry is like the extra you keep for the next column in addition.

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