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

Time Complexity: Reverse a Doubly Linked List

O(n)

Understanding Time Complexity

We want to understand how the time needed to reverse a doubly linked list changes as the list grows.

How does the work increase when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct Node {
    data: i32,
    prev: Option<*mut Node>,
    next: Option>,
}

fn reverse(head: &mut Option>) {
    let mut current = head.take();
    let mut prev = None;
    while let Some(mut node) = current {
        let next = node.next.take();
        node.next = prev.take();
        node.prev = next.as_deref_mut().map(|n| &mut **n as *mut _);
        prev = Some(node);
        current = next;
    }
    *head = prev;
}
    

This code reverses the links of a doubly linked list by swapping the next and prev pointers for each node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once per node, so as many times as there are nodes in the list.

How Execution Grows With Input

As the list size grows, the number of steps grows directly with it.

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

Pattern observation: The work grows in a straight line with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows directly with the list size; doubling the list doubles the work.

Common Mistake

[X] Wrong: "Reversing a doubly linked list takes more than linear time because of the two pointers per node."

[OK] Correct: Each node is visited once, and swapping pointers is a constant-time action, so the total time is still linear.

Interview Connect

Understanding this helps you show you can work with linked lists efficiently, a common skill interviewers look for.

Self-Check

"What if we changed the list to a singly linked list? How would the time complexity of reversing it change?"

Practice

1. What is the main purpose of reversing a doubly linked list?

easy

A. To change the order of nodes so the last becomes first and vice versa

B. To delete all nodes from the list

C. To add new nodes at the end of the list

D. To sort the nodes in ascending order

Solution

1. Step 1: Understand the structure of a doubly linked list

   A doubly linked list has nodes connected both forward and backward.

2. Step 2: Purpose of reversing

   Reversing swaps the direction so the last node becomes the first and the first becomes the last.

3. Final Answer:

   To change the order of nodes so the last becomes first and vice versa -> Option A

4. Quick Check:

   Reversing means flipping order = C [OK]

Hint: Reversing flips the list order front to back [OK]

Common Mistakes:

• Confusing reversing with deleting nodes
• Thinking reversing sorts the list
• Mixing up adding nodes with reversing

2. Which Rust code snippet correctly swaps the next and prev pointers of a node in a doubly linked list?

easy

A. let temp = node.prev; node.prev = node.next; node.next = temp.next;

B. node.next = node.next; node.prev = node.prev;

C. node.prev = node.next; node.next = node.prev;

D. let temp = node.next; node.next = node.prev; node.prev = temp;

Solution

1. Step 1: Identify correct swapping logic

   Swapping means storing one pointer temporarily, then exchanging values.

2. Step 2: Check each option

   let temp = node.next; node.next = node.prev; node.prev = temp; correctly uses a temp variable to swap next and prev. Others either assign incorrectly or use wrong variables.

3. Final Answer:

   let temp = node.next; node.next = node.prev; node.prev = temp; -> Option D

4. Quick Check:

   Swap pointers with temp variable = D [OK]

Hint: Use a temp variable to swap pointers safely [OK]

Common Mistakes:

• Assigning pointers without temp causes data loss
• Mixing up next and prev in assignments
• Using wrong pointer in swap

3. Given the Rust code below that reverses a doubly linked list, what will be the printed output?

struct Node {
    data: i32,
    prev: Option,
    next: Option,
}

// Assume list is 1 <-> 2 <-> 3
// After reversal, print from head

// Output format: data values separated by spaces

// Output?

medium

A. 3 2 1

B. 1 2 3

C. 2 3 1

D. 3 1 2

Solution

1. Step 1: Understand initial list order

   The list is 1 <-> 2 <-> 3, so printing from head gives 1 2 3.

2. Step 2: Effect of reversal

   Reversing swaps next and prev pointers, so the new head is 3, and order is 3 2 1.

3. Final Answer:

   3 2 1 -> Option A

4. Quick Check:

   Reversed list prints backwards = 3 2 1 [OK]

Hint: Reversed doubly linked list prints nodes backward [OK]

Common Mistakes:

• Printing original order after reversal
• Confusing node data order
• Assuming reversal sorts nodes

4. Identify the bug in this Rust function that attempts to reverse a doubly linked list:

fn reverse(mut head: Option) -> Option {
    let mut current = head;
    while let Some(mut node) = current {
        let temp = node.next;
        node.next = node.prev;
        node.prev = temp;
        current = node.next;
    }
    head
}

medium

A. The loop condition should be while current.is_some()

B. Swapping next and prev pointers is done incorrectly

C. The function does not update head to the new first node after reversal

D. The function should return current instead of head

Solution

1. Step 1: Analyze pointer swapping

   Swapping next and prev is done correctly inside the loop.

2. Step 2: Check head update

   The function returns original head without updating it to the new first node after reversal, so head remains old start.

3. Final Answer:

   The function does not update head to the new first node after reversal -> Option C

4. Quick Check:

   Head must point to new first node after reverse = A [OK]

Hint: Update head to last processed node after reversal [OK]

Common Mistakes:

• Not updating head pointer after reversal
• Returning old head instead of new head
• Incorrect loop condition causing infinite loop

5. You have a doubly linked list with nodes containing integers. After reversing it, you want to find the sum of all nodes that were originally at even positions (2nd, 4th, etc.). Which approach correctly solves this in Rust?

hard

A. Reverse the list, then traverse and sum nodes at odd positions in the reversed list

B. Traverse original list, sum nodes at even positions, then reverse the list

C. Reverse the list, then sum all nodes without position checks

D. Traverse reversed list and sum nodes at even positions

Solution

1. Step 1: Understand position change after reversal

   Reversing changes node positions, so original even positions become different in reversed list.

2. Step 2: Correct approach

   Sum nodes at even positions before reversal to get original even-position sum, then reverse list if needed.

3. Final Answer:

   Traverse original list, sum nodes at even positions, then reverse the list -> Option B

4. Quick Check:

   Sum original even positions before reversing = A [OK]

Hint: Sum before reversing to keep original positions intact [OK]

Common Mistakes:

• Summing positions after reversal which changes order
• Ignoring position changes due to reversal
• Summing all nodes without filtering positions