Dequeue Using Linked List in DSA C++ - Time & Space Complexity

Time Complexity: Dequeue Using Linked List

O(1)

Understanding Time Complexity

We want to understand how fast operations run in a dequeue built with a linked list.

Specifically, how the time to add or remove items changes as the number of items grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

class Dequeue {
  Node* front;
  Node* rear;
public:
  void insertFront(int x) {
    Node* temp = new Node{x, front, nullptr};
    if (front) front->prev = temp;
    front = temp;
    if (!rear) rear = temp;
  }
  void deleteRear() {
    if (!rear) return;
    Node* temp = rear;
    rear = rear->prev;
    if (rear) rear->next = nullptr;
    else front = nullptr;
    delete temp;
  }
};
    

This code shows how to add an item at the front and remove an item from the rear in a dequeue using a doubly linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding or removing a node involves only a few pointer changes.
• How many times: Each operation runs once per call, no loops or recursion inside.

How Execution Grows With Input

Each insert or delete operation takes the same small number of steps no matter how many items are in the dequeue.

Input Size (n)	Approx. Operations
10	3-5 pointer changes
100	3-5 pointer changes
1000	3-5 pointer changes

Pattern observation: The time stays about the same no matter how many items are in the dequeue.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing items from either end takes a constant amount of time regardless of size.

Common Mistake

[X] Wrong: "Removing from the rear takes longer because we have to find the last node."

[OK] Correct: The linked list keeps a direct pointer to the rear node, so no searching is needed.

Interview Connect

Knowing that dequeue operations run in constant time shows you understand efficient data structure design.

This skill helps you explain and write fast code for real problems.

Self-Check

"What if we used a singly linked list without a rear pointer? How would the time complexity of deleting from the rear change?"

Practice

1. What is the main advantage of implementing a Dequeue using a linked list over an array?

easy

A. It allows insertion and deletion from both ends in O(1) time without shifting elements.

B. It uses less memory than an array.

C. It automatically sorts the elements.

D. It restricts insertion to only one end.

Solution

1. Step 1: Understand Dequeue operations

   A dequeue allows insertion and deletion from both front and rear ends.

2. Step 2: Compare linked list and array

   Linked list supports O(1) insertion/deletion at both ends without shifting, unlike arrays.

3. Final Answer:

   It allows insertion and deletion from both ends in O(1) time without shifting elements. -> Option A

4. Quick Check:

   Linked list dequeue = O(1) insert/delete both ends [OK]

Hint: Linked list avoids shifting, so both ends are fast [OK]

Common Mistakes:

• Thinking arrays allow O(1) insert/delete at both ends
• Assuming linked list sorts elements automatically
• Confusing memory usage advantages

2. Which of the following is the correct syntax to insert an element at the front of a dequeue implemented using a linked list in C++?

easy

A. void insertFront(int data) { front = data; }

B. void insertFront(int data) { Node* newNode = new Node(data); newNode->next = front; front = newNode; }

C. void insertFront(int data) { front->data = data; }

D. void insertFront(int data) { delete front; front = new Node(data); }

Solution

1. Step 1: Create new node with data

   We allocate a new node and assign the data to it.

2. Step 2: Link new node to current front and update front pointer

   Set newNode->next to current front, then update front to newNode.

3. Final Answer:

   void insertFront(int data) { Node* newNode = new Node(data); newNode->next = front; front = newNode; } -> Option B

4. Quick Check:

   New node points to old front, front updated [OK]

Hint: New node next points to old front, then front updated [OK]

Common Mistakes:

• Assigning data directly to front pointer
• Deleting front before inserting
• Not linking new node to old front

3. Consider the following code snippet for a dequeue using linked list:

Dequeue dq;
dq.insertRear(10);
dq.insertFront(20);
dq.insertRear(30);
dq.deleteFront();
dq.insertFront(40);
dq.deleteRear();
dq.display();

What will be the output of dq.display()?

medium

A. 40 -> 10 -> null

B. 40 -> 20 -> 10 -> null

C. 20 -> 40 -> null

D. 10 -> 20 -> 30 -> null

Solution

1. Step 1: Trace insertions and deletions step-by-step

   insertRear(10): 10 insertFront(20): 20 -> 10 insertRear(30): 20 -> 10 -> 30 deleteFront(): removes 20, now 10 -> 30 insertFront(40): 40 -> 10 -> 30 deleteRear(): removes 30, now 40 -> 10

2. Step 2: Final list after all operations

   40 -> 10 -> null

3. Final Answer:

   40 -> 10 -> null -> Option A

4. Quick Check:

   Trace insert/delete carefully for final list [OK]

Hint: Trace each operation stepwise to find final list [OK]

Common Mistakes:

• Forgetting to update front or rear after deletion
• Mixing order of insertFront and insertRear
• Assuming deleted elements remain

4. The following code snippet for deleting from rear in a dequeue using linked list has a bug:

void deleteRear() {
  if (rear == nullptr) return;
  Node* temp = front;
  while (temp->next != rear) {
    temp = temp->next;
  }
  delete rear;
  rear = temp;
  rear->next = nullptr;
}

What is the bug in this code?

medium

A. It sets rear->next to non-null value.

B. It deletes the front node instead of rear.

C. It does not handle the case when there is only one node in the dequeue.

D. It causes memory leak by not deleting any node.

Solution

1. Step 1: Analyze the loop for single node case

   If only one node exists, front == rear, so while loop condition temp->next != rear fails immediately, temp = front.

2. Step 2: Check what happens after loop

   Deleting rear node, then setting rear = temp (which is front). But if only one node, temp == rear, so deleting rear deletes the only node, then rear points to deleted memory, causing dangling pointer.

3. Final Answer:

   It does not handle the case when there is only one node in the dequeue. -> Option C

4. Quick Check:

   Single node case needs special handling [OK]

Hint: Check single-node case separately before loop [OK]

Common Mistakes:

• Ignoring single-node edge case
• Assuming loop always runs at least once
• Not updating rear pointer correctly

5. You want to implement a function reverseDequeue() that reverses the order of elements in a dequeue implemented using a linked list. Which approach is best to reverse the dequeue efficiently?

hard

A. Delete all nodes and re-insert them in reverse order.

B. Create a new linked list and insert elements from the original dequeue at the rear one by one.

C. Use an auxiliary stack to push all elements, then pop and insert back into the dequeue.

D. Traverse the linked list once, reversing the next pointers of each node, then swap front and rear pointers.

Solution

1. Step 1: Reverse linked list pointers in one traversal

   Reversing next pointers of each node in one pass efficiently reverses the list.

2. Step 2: Swap front and rear pointers

   After reversal, front and rear pointers must be swapped to reflect new ends.

3. Final Answer:

   Traverse the linked list once, reversing the next pointers of each node, then swap front and rear pointers. -> Option D

4. Quick Check:

   In-place reversal with pointer swap is most efficient [OK]

Hint: Reverse pointers in one pass, then swap front and rear [OK]

Common Mistakes:

• Using extra memory unnecessarily
• Re-inserting nodes instead of reversing pointers
• Not swapping front and rear after reversal