Merge Two Sorted Linked Lists in DSA Rust - Time & Space Complexity

Time Complexity: Merge Two Sorted Linked Lists

O(n + m)

Understanding Time Complexity

We want to know how the time needed to merge two sorted linked lists changes as the lists get bigger.

Specifically, how does the merging process scale with the size of the input lists?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn merge_two_lists(
    mut list1: Option>, 
    mut list2: Option>
) -> Option> {
    let mut dummy = Box::new(ListNode::new(0));
    let mut tail = &mut dummy;

    while list1.is_some() && list2.is_some() {
        if list1.as_ref().unwrap().val <= list2.as_ref().unwrap().val {
            tail.next = list1;
            tail = tail.next.as_mut().unwrap();
            list1 = tail.next.take();
        } else {
            tail.next = list2;
            tail = tail.next.as_mut().unwrap();
            list2 = tail.next.take();
        }
    }
    tail.next = if list1.is_some() { list1 } else { list2 };
    dummy.next
}
    

This code merges two sorted linked lists into one sorted linked list by repeatedly choosing the smaller head node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that compares nodes and moves pointers.
• How many times: It runs until all nodes from both lists are processed, roughly the total number of nodes combined.

How Execution Grows With Input

As the total number of nodes increases, the number of comparisons and pointer moves grows roughly the same amount.

Input Size (n)	Approx. Operations
10	About 10 comparisons and moves
100	About 100 comparisons and moves
1000	About 1000 comparisons and moves

Pattern observation: The operations grow linearly with the total number of nodes.

Final Time Complexity

Time Complexity: O(n + m)

This means the time to merge grows directly with the total number of nodes in both lists combined.

Common Mistake

[X] Wrong: "Merging two lists takes quadratic time because of nested loops."

[OK] Correct: The code only uses one loop that moves through both lists once, so it does not repeat work inside another loop.

Interview Connect

Understanding this merging process helps you handle linked list problems efficiently, a common skill interviewers look for.

Self-Check

"What if the input lists were arrays instead of linked lists? How would the time complexity change?"

Practice

1. What is the main goal when merging two sorted linked lists?

easy

A. Reverse both linked lists

B. Remove duplicates from both lists

C. Create a new sorted linked list containing all elements from both lists

D. Sort each list separately without merging

Solution

1. Step 1: Understand the problem

   Merging two sorted linked lists means combining them into one list that remains sorted.

2. Step 2: Identify the goal

   The goal is to create a new sorted linked list that contains all elements from both input lists.

3. Final Answer:

   Create a new sorted linked list containing all elements from both lists -> Option C

4. Quick Check:

   Merging = Combined sorted list [OK]

Hint: Merging means combining sorted lists into one sorted list [OK]

Common Mistakes:

• Thinking merging means reversing lists
• Trying to sort lists separately without merging
• Confusing merging with removing duplicates

2. Which Rust code snippet correctly declares a singly linked list node struct for merging?

easy

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

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

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

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

Solution

1. Step 1: Understand Rust linked list node

   Rust needs heap allocation for recursive types, so next must be Option.

2. Step 2: Check each option

   struct ListNode { val: i32, next: Option } uses Option, which is correct. Others use references or direct types causing errors.

3. Final Answer:

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

4. Quick Check:

   Use Option for linked list nodes [OK]

Hint: Use Option for recursive linked list nodes in Rust [OK]

Common Mistakes:

• Using direct ListNode type for next causes infinite size error
• Using references without lifetimes causes borrow errors
• Using Option<ListNode> misses boxing needed for recursion

3. Given two sorted linked lists where one list is empty and the other is 0 -> 2 -> 4 -> null, what is the merged list? How does the merge function handle this edge case?

let l1: Option = None;
let l2 = Some(Box::new(ListNode { val: 0, next: Some(Box::new(ListNode { val: 2, next: Some(Box::new(ListNode { val: 4, next: None })) })) }));

let merged = merge_two_lists(l1, l2);

easy

A. 0 -> 2 -> 4 -> null

B. null

C. 0 -> null

D. 2 -> 4 -> null

Solution

1. Step 1: Check input lists

   l1 is None (empty), l2 is 0 -> 2 -> 4 -> null.

2. Step 2: Merge function returns non-empty list

   When one list is None, function returns the other list as is.

3. Final Answer:

   0 -> 2 -> 4 -> null -> Option A

4. Quick Check:

   Empty + list = list unchanged [OK]

Hint: If one list empty, merged list is the other list [OK]

Common Mistakes:

• Returning null when one list is empty
• Trying to merge empty list nodes incorrectly
• Losing nodes from non-empty list

4. What is the output linked list after merging these two sorted lists? List1: 1 -> 3 -> 5 -> null List2: 2 -> 4 -> 6 -> null

fn merge_two_lists(l1: Option, l2: Option) -> Option {
    match (l1, l2) {
        (None, None) => None,
        (Some(node), None) | (None, Some(node)) => Some(node),
        (Some(mut n1), Some(mut n2)) => {
            if n1.val < n2.val {
                n1.next = merge_two_lists(n1.next.take(), Some(n2));
                Some(n1)
            } else {
                n2.next = merge_two_lists(Some(n1), n2.next.take());
                Some(n2)
            }
        }
    }
}

medium

A. 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

B. 1 -> 3 -> 5 -> 2 -> 4 -> 6 -> null

C. 2 -> 4 -> 6 -> 1 -> 3 -> 5 -> null

D. 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> null

Solution

1. Step 1: Trace first comparison

   Compare 1 and 2: 1 < 2, so 1 is first node.

2. Step 2: Recursively merge remaining lists

   Next merge 3->5 and 2->4->6. 2 < 3, so 2 next, then 3, 4, 5, 6.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null -> Option A

4. Quick Check:

   Sorted merge = 1,2,3,4,5,6 [OK]

Hint: Pick smaller head each step to merge sorted lists [OK]

Common Mistakes:

• Appending lists instead of merging sorted
• Not updating next pointers correctly
• Mixing order of nodes in merged list

5. Identify the bug in this Rust merge function and select the fix:

fn merge_two_lists(l1: Option, l2: Option) -> Option {
    match (l1, l2) {
        (None, None) => None,
        (Some(node), None) => Some(node),
        (None, Some(node)) => Some(node),
        (Some(mut n1), Some(mut n2)) => {
            if n1.val <= n2.val {
                n1.next = merge_two_lists(n1.next, Some(n2));
                Some(n1)
            } else {
                n2.next = merge_two_lists(Some(n1), n2.next);
                Some(n2)
            }
        }
    }
}

medium

A. Remove match and use if-else only

B. Change if condition to n1.val < n2.val

C. Replace Some(mut n1) with &mut n1

D. Use n1.next.take() and n2.next.take() to avoid borrowing errors

Solution

1. Step 1: Identify ownership issue

   n1.next and n2.next are Options; assigning without take() causes borrowing errors.

2. Step 2: Fix by using take()

   Using n1.next.take() and n2.next.take() moves ownership out safely for recursion.

3. Final Answer:

   Use n1.next.take() and n2.next.take() to avoid borrowing errors -> Option D

4. Quick Check:

   Use take() to move Option ownership [OK]

Hint: Use .take() to move Option values safely in Rust [OK]

Common Mistakes:

• Not using take() causes borrow checker errors
• Changing comparison operator unnecessarily
• Using references incorrectly in pattern matching