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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Rust

let new_node = ListNode::new([1]);

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

Asum % 10

Bsum + carry

Ccarry

Dsum / 10

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum / 10 instead of sum % 10.

Adding carry directly without modulo.

We use sum % 10 to get the digit to store in the new node because the sum can be more than 9.

2fill in blank

medium

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

DSA Rust

carry = [1];

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Asum / 10

Bsum + carry

Csum % 10

Dcarry % 10

Attempts:

3 left

💡 Hint

Common Mistakes

Using modulo instead of division for carry.

Adding carry again instead of updating it.

Carry is the tens digit of the sum, so we divide sum by 10 to get it.

3fill in blank

hard

Fix the error in the loop condition to continue while there are digits or carry.

DSA Rust

while l1.is_some() || l2.is_some() || [1] > 0 {

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Anew_node

Bsum

Ccarry

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum instead of carry in the condition.

Checking new_node or result which are not loop conditions.

We continue the loop while carry is greater than 0 to add any leftover carry digit.

4fill in blank

hard

Fill both blanks to move to the next nodes in the lists if they exist.

DSA Rust

l1 = if let Some(node) = l1 { [1] } else { [2] };
l2 = if let Some(node) = l2 { [1] } else { [2] };

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Anode.next

BNone

Cnode

DSome(node)

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning node instead of node.next.

Using Some(node) instead of None when no next node.

We move to the next node using node.next, or None if no next node exists.

5fill in blank

hard

Fill all three blanks to build the result list and update the current pointer.

DSA Rust

sum = [2] + [3] + carry;
current.next = Some(Box::new(ListNode::new([1])));
current = current.next.as_mut().unwrap();

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

Asum % 10

Bl1_val

Cl2_val

Dcarry

Attempts:

3 left

💡 Hint

Common Mistakes

Using carry instead of sum % 10 for the new node value.

Mixing up l1_val and l2_val in sum calculation.

We first calculate sum by adding l1_val, l2_val, and carry, then create a new node with sum % 10.

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