Reverse a Singly Linked List Recursive in DSA Rust - Time & Space Complexity

Time Complexity: Reverse a Singly Linked List Recursive

O(n)

Understanding Time Complexity

We want to understand how the time needed to reverse a linked list grows as the list gets longer.

How does the number of steps change when the list size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn reverse_recursive(head: Option>) -> Option> {
    match head {
        None => None,
        Some(mut node) => {
            if node.next.is_none() {
                Some(node)
            } else {
                let next = node.next.take();
                let new_head = reverse_recursive(next);
                if let Some(mut next_node) = new_head.as_deref_mut() {
                    let mut tail = &mut next_node;
                    while let Some(ref mut next) = tail.next {
                        tail = next;
                    }
                    tail.next = Some(node);
                }
                node.next = None;
                new_head
            }
        }
    }
}

This function reverses a singly linked list by calling itself on the next node until it reaches the end, then rewires the links backward.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive call on the next node in the list.
• How many times: Once for each node in the list, so n times for a list of size n.

How Execution Grows With Input

Each node causes one recursive call, so the steps grow directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 recursive calls and rewiring steps
100	About 100 recursive calls and rewiring steps
1000	About 1000 recursive calls and rewiring steps

Pattern observation: The work grows linearly as the list gets longer.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows in direct proportion to the number of nodes.

Common Mistake

[X] Wrong: "Recursive reversal is faster because it uses recursion instead of loops."

[OK] Correct: Both recursion and loops visit each node once, so they take similar time. Recursion adds extra function call overhead but does not reduce the number of steps.

Interview Connect

Understanding this helps you explain how recursion works on linked lists and shows you can analyze how your code scales with input size.

Self-Check

"What if we changed the recursive reversal to an iterative approach? How would the time complexity change?"

Practice

1. What is the main idea behind reversing a singly linked list using recursion?

easy

A. Reverse the rest of the list first, then adjust the current node's next pointer

B. Swap the values of the first and last nodes only

C. Iterate through the list and reverse pointers in a loop

D. Create a new list and copy elements in reverse order

Solution

1. Step 1: Understand recursive reversal logic

   In recursion, we reverse the list starting from the end by calling the function on the next node first.

2. Step 2: Adjust pointers after recursion returns

   After the recursive call returns, we set the next node's next pointer to the current node, effectively reversing the link.

3. Final Answer:

   Reverse the rest of the list first, then adjust the current node's next pointer -> Option A

4. Quick Check:

   Recursive reversal = reverse rest first [OK]

Hint: Think: reverse tail first, then fix current link [OK]

Common Mistakes:

• Trying to swap node values instead of pointers
• Reversing pointers before recursive call
• Using iteration instead of recursion

2. Which of the following Rust function signatures correctly represents a recursive function to reverse a singly linked list?

easy

A. fn reverse_list(head: Node) -> Node

B. fn reverse_list(head: &mut Node) -> &mut Node

C. fn reverse_list(head: Option) -> Option

D. fn reverse_list(head: &Node) -> Node

Solution

1. Step 1: Identify correct ownership and return type

   In Rust, linked lists use Option to represent nullable pointers with ownership.

2. Step 2: Match function signature for recursive reversal

   The function should take ownership of the list head and return the new head after reversal, so Option input and output is correct.

3. Final Answer:

   fn reverse_list(head: Option) -> Option -> Option C

4. Quick Check:

   Ownership with Option = fn reverse_list(head: Option) -> Option [OK]

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

Common Mistakes:

• Using references without ownership causing borrow errors
• Returning plain Node instead of Option
• Passing mutable references incorrectly

3. Given the following Rust code snippet for recursive reversal, what will be the output list after reversal if the input list is 1 -> 2 -> 3 -> null?

fn reverse_list(head: Option) -> Option {
    match head {
        None => None,
        Some(mut node) => {
            if node.next.is_none() {
                Some(node)
            } else {
                let next = node.next.take();
                let new_head = reverse_list(next);
                let mut tail = new_head.as_ref();
                while let Some(t) = tail {
                    if t.next.is_none() {
                        break;
                    }
                    tail = t.next.as_ref();
                }
                tail.unwrap().next = Some(node);
                node.next = None;
                new_head
            }
        }
    }
}

medium

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. 2 -> 3 -> 1 -> null

D. Compile-time error due to borrowing

Solution

1. Step 1: Analyze pointer manipulation and borrowing

   The code tries to get a mutable reference to tail via as_ref() and then mutate tail.unwrap().next, which causes a borrowing error in Rust.

2. Step 2: Understand Rust's borrowing rules

   Rust does not allow mutable access through an immutable reference (as_ref()), so this code will cause a compile-time error.

3. Final Answer:

   Compile-time error due to borrowing -> Option D

4. Quick Check:

   Immutable ref then mutable write = error [OK]

Hint: Check mutable vs immutable references in Rust code [OK]

Common Mistakes:

• Ignoring Rust's strict borrowing rules
• Assuming as_ref() allows mutation
• Not using take() properly to avoid borrow conflicts

4. Identify the bug in this recursive Rust function to reverse a singly linked list:

fn reverse_list(head: Option) -> Option {
    match head {
        None => None,
        Some(mut node) => {
            if node.next.is_none() {
                Some(node)
            } else {
                let next = node.next.take();
                let new_head = reverse_list(next);
                node.next.as_mut().unwrap().next = Some(node);
                node.next = None;
                new_head
            }
        }
    }
}

medium

A. Function does not return the new head

B. Using node.next.as_mut().unwrap() after take() causes None unwrap panic

C. Missing base case for empty list

D. No bug, code is correct

Solution

1. Step 1: Understand effect of take() on node.next

   Calling node.next.take() sets node.next to None, so node.next.as_mut() returns None.

2. Step 2: Identify unsafe unwrap causing panic

   Calling unwrap() on None causes a runtime panic, so node.next.as_mut().unwrap() is invalid after take().

3. Final Answer:

   Using node.next.as_mut().unwrap() after take() causes None unwrap panic -> Option B

4. Quick Check:

   unwrap on None after take() = panic [OK]

Hint: Don't unwrap after take() without checking for None [OK]

Common Mistakes:

• Assuming node.next still exists after take()
• Ignoring unwrap panic possibility
• Not returning new_head properly

5. You want to reverse a singly linked list recursively in Rust, but also count how many nodes were reversed. Which approach correctly combines reversal and counting? Options: A) Modify the recursive function to return a tuple (new_head, count) and increment count after each recursion. B) Use a global mutable variable to count nodes during recursion. C) Reverse the list first, then iterate again to count nodes. D) Count nodes before reversal and ignore during recursion.

hard

A. Modify the recursive function to return a tuple (new_head, count) and increment count after each recursion.

B. Use a global mutable variable to count nodes during recursion.

C. Reverse the list first, then iterate again to count nodes.

D. Count nodes before reversal and ignore during recursion.

Solution

1. Step 1: Combine reversal and counting in recursion

   Modify the recursive function to return both the new head and the count of nodes reversed as a tuple.

2. Step 2: Increment count after each recursive call

   Each recursive call returns the count from deeper calls; add 1 for the current node before returning.

3. Final Answer:

   Modify the recursive function to return a tuple (new_head, count) and increment count after each recursion. -> Option A

4. Quick Check:

   Return tuple (head, count) in recursion = Modify the recursive function to return a tuple (new_head, count) and increment count after each recursion. [OK]

Hint: Return (head, count) tuple from recursion to track nodes [OK]

Common Mistakes:

• Using global variables causing unsafe code
• Counting nodes separately after reversal wastes time
• Ignoring counting during recursion