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 grows.

Specifically, how does the work increase when the list has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void reverseDoublyLinkedList(Node*& head) {
    Node* current = head;
    Node* temp = nullptr;
    while (current != nullptr) {
        temp = current->prev;
        current->prev = current->next;
        current->next = temp;
        current = current->prev;
    }
    if (temp != nullptr) {
        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: Exactly once per node, so n times for n nodes.

How Execution Grows With Input

As the list gets longer, 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 list size; doubling the list doubles the work.

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 step, so the total time is still linear.

Interview Connect

Understanding this linear time reversal helps you handle linked list problems confidently, a common topic in interviews.

Self-Check

"What if we tried to reverse the list using recursion instead of a loop? How would the time complexity change?"

Practice

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

easy

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

B. To delete all nodes from the list

C. To sort the list in ascending order

D. To convert it into a singly linked list

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 A

4. Quick Check:

   Reverse = swap start and end [OK]

Hint: Reversing swaps head and tail pointers [OK]

Common Mistakes:

• Confusing reversing with deleting nodes
• Thinking reversing sorts the list
• Assuming it changes list type

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

easy

A. curr->next = curr->prev; curr->prev = curr->next;

B. std::swap(curr->next, curr->prev);

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

D. curr->next = curr->next; curr->prev = curr->prev;

Solution

1. Step 1: Understand pointer swapping

   Swapping two pointers requires a temporary storage or a swap function to avoid overwriting.

2. Step 2: Use std::swap for safe swapping

   std::swap exchanges values safely without losing data.

3. Final Answer:

   std::swap(curr->next, curr->prev); -> Option B

4. Quick Check:

   Use std::swap to exchange pointers [OK]

Hint: Use std::swap to exchange two pointers safely [OK]

Common Mistakes:

• Assigning one pointer to another without temp causes data loss
• Swapping pointers manually without temp variable
• Confusing assignment with swapping

3. What will be the output of the following code snippet after reversing the doubly linked list?

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

// List: 1 <-> 2 <-> 3
Node* head = new Node{1, nullptr, nullptr};
Node* second = new Node{2, nullptr, head};
head->next = second;
Node* third = new Node{3, nullptr, second};
second->next = third;

// Reverse function
Node* reverse(Node* head) {
  Node* curr = head;
  Node* temp = nullptr;
  while (curr != nullptr) {
    temp = curr->prev;
    curr->prev = curr->next;
    curr->next = temp;
    curr = curr->prev;
  }
  if (temp != nullptr) head = temp->prev;
  return head;
}

head = reverse(head);

// Print list
Node* tempPrint = head;
while (tempPrint != nullptr) {
  std::cout << tempPrint->data << " ";
  tempPrint = tempPrint->next;
}

medium

A. 2 3 1

B. 1 2 3

C. 3 2 1

D. 3 1 2

Solution

1. Step 1: Trace pointer swaps in reverse function

   Each node swaps its next and prev pointers, then moves to original next (now prev).

2. Step 2: Update head to new first node

   After loop, head is updated to last processed node's prev, which is the original tail.

3. Final Answer:

   3 2 1 -> Option C

4. Quick Check:

   Reversed list prints from tail to head [OK]

Hint: After reversing, head points to original tail [OK]

Common Mistakes:

• Not updating head after reversal
• Confusing next and prev pointers
• Printing from old head instead of new

4. Identify the error in the following reverse function for a doubly linked list and select the fix.

Node* reverse(Node* head) {
  Node* curr = head;
  Node* temp = nullptr;
  while (curr != nullptr) {
    temp = curr->prev;
    curr->prev = curr->next;
    curr->next = temp;
    curr = curr->next;  // Error here
  }
  if (temp != nullptr) head = temp->prev;
  return head;
}

medium

A. Change 'head = temp->prev;' to 'head = temp->next;'

B. Remove the line 'temp = curr->prev;'

C. Add 'curr = curr->prev;' before the loop

D. Change 'curr = curr->next;' to 'curr = curr->prev;'

Solution

1. Step 1: Analyze pointer movement in loop

   After swapping next and prev, moving curr to curr->next is wrong because next and prev are swapped.

2. Step 2: Correct pointer update

   We must move curr to curr->prev (original next) after swapping.

3. Final Answer:

   Change 'curr = curr->next;' to 'curr = curr->prev;' -> Option D

4. Quick Check:

   Move curr along swapped pointers [OK]

Hint: After swapping, move curr to prev pointer [OK]

Common Mistakes:

• Moving curr to next instead of prev after swap
• Not updating head correctly
• Ignoring pointer swap effects

5. Given a doubly linked list with nodes containing integers, how would you modify the reverse function to also count how many nodes were reversed and return that count along with the new head?

hard

A. Modify the function to return a pair, increment count inside the loop, and return both.

B. Add a global variable to count nodes and return only the head.

C. Count nodes before reversal and return the count without changing the function.

D. Use recursion to reverse and count nodes simultaneously without changing return type.

Solution

1. Step 1: Change function return type

   Use std::pair to return both new head and count.

2. Step 2: Count nodes during reversal

   Initialize count=0; increment count inside the reversal loop for each node processed.

3. Step 3: Return both values

   Return std::make_pair(new_head, count) after reversal completes.

4. Final Answer:

   Modify the function to return a pair, increment count inside the loop, and return both. -> Option A

5. Quick Check:

   Return both head and count using pair [OK]

Hint: Return pair with head and count, increment count in loop [OK]

Common Mistakes:

• Using global variables instead of return values
• Counting nodes before reversal only
• Trying recursion without changing return type