Reverse a Doubly Linked List in DSA C - 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 gets bigger.

Specifically, how does the work grow when the number of nodes increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void reverseDoublyLinkedList(Node** head) {
    Node* current = *head;
    Node* temp = NULL;
    while (current != NULL) {
        temp = current->prev;
        current->prev = current->next;
        current->next = temp;
        current = current->prev;
    }
    if (temp != NULL) {
        *head = temp->prev;
    }
}
    

This code reverses the links of a doubly linked list so the last node becomes the first.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: It runs once for every node in the list, so n times if there are n nodes.

How Execution Grows With Input

As the list grows, the number of steps grows directly with the number of nodes.

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 number of nodes.

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 operation, so the total time is still linear.

Interview Connect

Understanding this helps you explain how pointer changes affect performance and shows you can analyze linked list operations clearly.

Self-Check

"What if we tried to reverse the list recursively instead of iteratively? How would the time complexity change?"

Practice

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

easy

A. To convert it into a singly linked list

B. To delete all nodes from the list

C. To make the last node the first and the first node the last

D. To sort the list in ascending order

Solution

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

   A doubly linked list has nodes with pointers to both previous and next nodes.

2. Step 2: Define reversing operation

   Reversing means swapping the direction so the last node becomes the first and vice versa.

3. Final Answer:

   To make the last node the first and the first node the last -> Option C

4. Quick Check:

   Reverse = last becomes first [OK]

Hint: Reversing flips start and end nodes [OK]

Common Mistakes:

• Thinking reversing deletes nodes
• Confusing reversing with sorting
• Assuming it changes list type

2. Which of the following is the correct way to swap the next and prev pointers of a node in C?

easy

A. temp = node->prev; node->prev = node->next; node->next = node->next;

B. node->next = node->prev; node->prev = node->next;

C. node->prev = node->next; node->next = node->prev;

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

Solution

1. Step 1: Use a temporary variable to hold one pointer

   Store node->next in temp to avoid losing its value.

2. Step 2: Swap pointers correctly

   Assign node->next to node->prev, then node->prev to temp.

3. Final Answer:

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

4. Quick Check:

   Swap pointers using temp variable [OK]

Hint: Always use temp variable to swap pointers safely [OK]

Common Mistakes:

• Swapping without temp loses pointer data
• Assigning both pointers to same value
• Mixing order of assignments

3. Given the following doubly linked list nodes and code snippet, what will be the new head node's data after reversal?

struct Node {
  int data;
  struct Node* next;
  struct Node* prev;
};

// List: 1 <-> 2 <-> 3 <-> NULL
struct Node* head = node1; // node1->data = 1

// Reversal code snippet:
struct Node* current = head;
struct Node* temp = NULL;
while (current != NULL) {
  temp = current->prev;
  current->prev = current->next;
  current->next = temp;
  current = current->prev;
}
if (temp != NULL) {
  head = temp->prev;
}

medium

A. 1

B. 3

C. 2

D. NULL

Solution

1. Step 1: Trace pointer swaps in the loop

   Each node swaps its next and prev pointers, moving current backward through original list.

2. Step 2: Update head after loop

   After loop, head is set to temp->prev, which points to original last node (data 3).

3. Final Answer:

   3 -> Option B

4. Quick Check:

   New head = original tail = 3 [OK]

Hint: New head is original tail after reversal [OK]

Common Mistakes:

• Not updating head pointer after reversal
• Confusing prev and next pointers
• Assuming head remains same

4. Identify the error in the following code snippet that attempts to reverse a doubly linked list:

void reverse(struct Node** head_ref) {
  struct Node* current = *head_ref;
  struct Node* temp = NULL;
  while (current != NULL) {
    temp = current->next;
    current->next = current->prev;
    current->prev = temp;
    current = current->prev;
  }
  if (temp != NULL) {
    *head_ref = temp->prev;
  }
}

medium

A. temp should store current->prev instead of current->next

B. current should move to current->next instead of current->prev

C. The head pointer update is incorrect

D. No error, code is correct

Solution

1. Step 1: Verify pointer swaps

   The next/prev swaps correctly reverse directions as current traverses the original next chain (now in prev).

2. Step 2: Analyze temp assignment

   Temp should store current->prev before swapping, not current->next, to avoid losing pointer.

3. Final Answer:

   temp should store current->prev instead of current->next -> Option A

4. Quick Check:

   Correct temp assignment preserves traversal [OK]

Hint: Temp must hold current->prev before swapping pointers [OK]

Common Mistakes:

• Incorrect head pointer update logic
• Moving current incorrectly after swap
• Not updating head pointer properly

5. You have a doubly linked list with nodes containing integer data. After reversing the list using the standard pointer swap method, which of the following statements is TRUE about the list's traversal?

hard

A. Traversing from the new head using next pointers visits nodes in original reverse order

B. Traversing from the new head using prev pointers visits nodes in original order

C. Traversing from the original head using next pointers visits nodes in reversed order

D. Traversing from the new head using next pointers visits nodes in original order

Solution

1. Step 1: Understand reversal effect on pointers

   Reversing swaps next and prev pointers, so next pointers now point backward in original list.

2. Step 2: Analyze traversal from new head

   Starting at new head, following next pointers visits nodes in original reverse order.

3. Final Answer:

   Traversing from the new head using next pointers visits nodes in original reverse order -> Option A

4. Quick Check:

   Next pointers after reverse = original prev [OK]

Hint: Next pointers after reverse follow original prev links [OK]

Common Mistakes:

• Assuming next pointers still go forward after reverse
• Confusing traversal directions
• Thinking original head remains start