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

Time Complexity: Add Two Numbers Represented as Linked List

O(n)

Understanding Time Complexity

We want to understand how the time needed to add two numbers stored as linked lists grows as the numbers get longer.

How does the work increase when the linked lists have more digits?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn add_two_numbers(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    let mut p = l1;
    let mut q = l2;
    let mut carry = 0;
    let mut dummy_head = Box::new(ListNode::new(0));
    let mut current = &mut dummy_head;

    while p.is_some() || q.is_some() {
        let x = p.as_ref().map_or(0, |node| node.val);
        let y = q.as_ref().map_or(0, |node| node.val);
        let sum = carry + x + y;
        carry = sum / 10;
        current.next = Some(Box::new(ListNode::new(sum % 10)));
        current = current.next.as_mut().unwrap();
        if let Some(node) = p.as_mut() { p = node.next.take(); }
        if let Some(node) = q.as_mut() { q = node.next.take(); }
    }

    if carry > 0 {
        current.next = Some(Box::new(ListNode::new(carry)));
    }

    dummy_head.next
}

This code adds two numbers digit by digit, moving through each linked list once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single while loop that moves through both linked lists.
• How many times: The loop runs once for each digit in the longer linked list.

How Execution Grows With Input

As the number of digits (nodes) in the linked lists grows, the number of steps grows roughly the same.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The work grows linearly with the number of digits.

Final Time Complexity

Time Complexity: O(n)

This means the time to add the numbers grows directly in proportion to the number of digits.

Common Mistake

[X] Wrong: "The time complexity is O(n²) because we add digits multiple times."

[OK] Correct: Each digit is processed only once in a single pass, so the work grows linearly, not quadratically.

Interview Connect

Understanding this linear time complexity helps you explain how to efficiently add numbers stored in linked lists, a common problem that shows your grasp of list traversal and simple arithmetic operations.

Self-Check

"What if the digits 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" 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