Add Two Numbers Represented as Linked List in DSA Rust - 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 (ones place 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 node contains the least significant digit, so addition starts from there.

intermediate

What is the purpose of the carry variable in the addition of two linked lists?

The carry variable stores the overflow digit when the sum of two digits exceeds 9, to be added to the next digit sum.

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 exhausted, treating missing digits as zero, and include any remaining carry.

intermediate

What is the final step after processing all nodes and carry in the addition of two linked lists?

If there is a leftover carry after processing all nodes, add a new node with that carry value at the end of the result list.

In the problem 'Add Two Numbers Represented as Linked List', how are the digits stored in the linked list?

A In reverse order, least significant digit first

B In normal order, most significant digit first

C Random order

D Sorted order

Digits are stored in reverse order so addition starts from the head node representing the ones place.

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

A Stop when the shorter list ends

B Continue adding with zero for missing digits

C Ignore the extra digits

D Reverse the longer list

Continue adding digits, treating missing digits as zero, to correctly sum all digits.

What does the carry represent in the addition process?

A The sum of two digits

B The difference between digits

C The leftover value after sum exceeds 9

D The product of digits

Carry stores the leftover value when sum of digits is greater than 9, to add to the next digit.

What is the output format of the sum in this problem?

A A linked list with digits in normal order

B An integer number

C A string of digits

D A linked list with digits in reverse order

The sum is returned as a linked list with digits stored in reverse order, same as input.

If after adding all digits, carry is still 1, what should be done?

A Add a new node with value 1 at the end

B Ignore the carry

C Subtract 1 from the last node

D Restart the addition

A new node with the carry value should be added to represent the extra digit.

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 carry and different length linked lists when adding two numbers represented as linked lists.

Carry is like the extra you keep in mind when adding digits.

Practice

1. What does the problem "Add Two Numbers Represented as Linked List" typically involve?

easy

A. Sorting two linked lists in ascending order

B. Adding two numbers where each digit is stored in a node of a linked list

C. Merging two linked lists into one

D. Finding the middle node of 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 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 two numbers where each digit is stored in a node of a linked list -> Option B

4. Quick Check:

   Adding digits node-wise = Adding two numbers where each digit is stored in a node of a linked list [OK]

Hint: Think of adding digit by digit like elementary school addition [OK]

Common Mistakes:

• Confusing addition with merging or sorting
• Ignoring the carry between digits
• Assuming digits are stored in forward order

2. Which Rust code snippet correctly defines a singly linked list node for this problem?

easy

A. struct ListNode { val: i32, next: &ListNode }

B. struct ListNode { val: i32, next: ListNode }

C. struct ListNode { val: i32, next: Option }

D. struct ListNode { val: i32, next: Box<ListNode> }

Solution

1. Step 1: Recall Rust linked list node structure

   Each node holds a value and an optional pointer to the next node using Box wrapped in Option for possible None.

2. Step 2: Check each option

   struct ListNode { val: i32, next: Option } uses Option which is correct. struct ListNode { val: i32, next: ListNode } and D lack Option, causing infinite size. struct ListNode { val: i32, next: &ListNode } uses reference which is invalid for ownership.

3. Final Answer:

   struct ListNode { val: i32, next: Option } -> Option C

4. Quick Check:

   Use Option for linked list nodes in Rust [OK]

Hint: Use Option for next pointer in Rust linked lists [OK]

Common Mistakes:

• Using direct ListNode type for next causing infinite size
• Using references without lifetime annotations
• Omitting Option causing no None termination

3. What is the output linked list when adding (2 -> 4 -> 3) and (5 -> 6 -> 4) representing numbers 342 and 465 respectively?

medium

A. 7 -> 1 -> 8

B. 8 -> 0 -> 7

C. 7 -> 10 -> 7

D. 7 -> 0 -> 8

Solution

1. Step 1: Add digits 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 -> 0 -> 8 representing 807 in reverse order.

3. Final Answer:

   7 -> 0 -> 8 -> Option D

4. Quick Check:

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

Hint: Add digits and carry from left to right in linked list order [OK]

Common Mistakes:

• Not carrying over when sum >= 10
• Mixing digit order in output
• Adding digits without carry

4. Identify the bug in this Rust code snippet for adding two linked lists:

while l1.is_some() || l2.is_some() {
    let sum = l1.unwrap().val + l2.unwrap().val + carry;
    carry = sum / 10;
    let node = ListNode::new(sum % 10);
    // ...
}

medium

A. Calling unwrap() without checking if Option is Some causes panic

B. Carry is not updated correctly

C. ListNode::new does not create a node

D. Loop condition should use && instead of ||

Solution

1. Step 1: Analyze unwrap usage

   unwrap() on Option without checking Some causes panic if None occurs.

2. Step 2: Check loop condition and unwrap safety

   Loop runs while either list is Some, but unwrap is called on both without checking, causing runtime error.

3. Final Answer:

   Calling unwrap() without checking if Option is Some causes panic -> Option A

4. Quick Check:

   Always check Option before unwrap to avoid panic [OK]

Hint: Check Option is Some before unwrap to prevent panic [OK]

Common Mistakes:

• Assuming both lists have same length without checks
• Ignoring Option safety causing runtime panic
• Incorrect loop condition logic

5. Given two linked lists representing numbers in reverse order, how would you handle adding them if one list is longer than the other?

hard

A. Continue adding remaining digits of longer list with carry until all nodes processed

B. Stop addition when shorter list ends and ignore remaining digits

C. Reverse both lists first then add digits

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

Solution

1. Step 1: Understand unequal length addition

   When one list ends, continue adding remaining digits of longer list plus carry.

2. Step 2: Maintain carry and process all nodes

   Carry from previous additions must be added until no nodes or carry remain.

3. Final Answer:

   Continue adding remaining digits of longer list with carry until all nodes processed -> Option A

4. Quick Check:

   Process all digits and carry regardless of list lengths [OK]

Hint: Add all digits and carry until both lists and carry are empty [OK]

Common Mistakes:

• Stopping addition early ignoring leftover digits
• Not adding carry after lists end
• Reversing lists unnecessarily