Dequeue Using Linked List in DSA Java - Time & Space Complexity

Time Complexity: Dequeue Using Linked List

O(n)

Understanding Time Complexity

We want to understand how fast operations on a dequeue using a linked list run.

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 dequeue operations using a linked list.

class Dequeue {
  Node front, rear;

  void addFront(int val) {
    Node newNode = new Node(val);
    newNode.next = front;
    front = newNode;
    if (rear == null) rear = front;
  }

  void removeRear() {
    if (front == null) return;
    if (front == rear) {
      front = rear = null;
      return;
    }
    Node temp = front;
    while (temp.next != rear) {
      temp = temp.next;
    }
    temp.next = null;
    rear = temp;
  }
}
    

This code adds an item at the front and removes an item from the rear of the dequeue.

Identify Repeating Operations

Look for loops or repeated steps in the code.

• Primary operation: The while loop in removeRear() that finds the node before the rear.
• How many times: It runs once for each node from front until the one before rear, so up to n-1 times.

How Execution Grows With Input

As the number of items (n) grows, the time to remove from rear grows too because of the loop.

Input Size (n)	Approx. Operations
10	About 9 steps to find node before rear
100	About 99 steps
1000	About 999 steps

Pattern observation: The steps increase roughly one by one as n grows.

Final Time Complexity

Time Complexity: O(n)

This means removing from the rear takes longer as the list gets longer, growing in a straight line with n.

Common Mistake

[X] Wrong: "Removing from rear is always fast because we have a pointer to rear."

[OK] Correct: Even with a rear pointer, we must find the node before rear to update its next link, which takes time proportional to the list size.

Interview Connect

Understanding these time costs helps you explain trade-offs in data structure choices clearly and confidently.

Self-Check

"What if we also kept a pointer to the node before rear? How would that change the time complexity of removing from rear?"

Practice

1. What is the main advantage of using a linked list to implement a Dequeue compared to using an array?

easy

A. It uses less memory than arrays in all cases.

B. It allows dynamic memory allocation and easy insertion/deletion at both ends.

C. It provides faster access to elements by index.

D. It automatically sorts elements when inserted.

Solution

1. Step 1: Understand linked list properties

   A linked list can grow or shrink dynamically without fixed size limits.

2. Step 2: Compare with array limitations

   Arrays have fixed size and inserting/deleting at ends may require shifting elements, which is costly.

3. Final Answer:

   It allows dynamic memory allocation and easy insertion/deletion at both ends. -> Option B

4. Quick Check:

   Linked list dynamic size = Advantage [OK]

Hint: Linked list grows dynamically, arrays have fixed size [OK]

Common Mistakes:

• Thinking arrays always use less memory
• Confusing access speed with insertion speed
• Assuming linked list sorts elements automatically

2. Which of the following is the correct way to declare a node class for a doubly linked list used in a dequeue in Java?

easy

A. class Node { int data; Node next; }

B. class Node { int data; Node prev; }

C. class Node { int data; Node left; Node right; }

D. class Node { int data; Node next; Node prev; }

Solution

1. Step 1: Identify node requirements for dequeue

   A dequeue needs to insert/delete at both ends, so nodes must link forward and backward.

2. Step 2: Check node class fields

   Only class Node { int data; Node next; Node prev; } has both next and prev pointers for doubly linked list.

3. Final Answer:

   class Node { int data; Node next; Node prev; } -> Option D

4. Quick Check:

   Doubly linked list node = next + prev [OK]

Hint: Doubly linked list nodes have next and prev pointers [OK]

Common Mistakes:

• Using single pointer nodes for dequeue
• Confusing binary tree node pointers with linked list
• Omitting prev pointer needed for backward traversal

3. What will be the output after performing these operations on an empty dequeue implemented using a linked list?

dequeue.insertFront(10);
dequeue.insertRear(20);
dequeue.insertFront(5);
dequeue.deleteRear();
dequeue.display();

medium

A. 5 -> 10 -> null

B. 10 -> 5 -> null

C. 5 -> 20 -> null

D. 10 -> null

Solution

1. Step 1: Perform insertFront(10) and insertRear(20)

   List becomes: 10 -> 20 -> null

2. Step 2: Perform insertFront(5)

   List becomes: 5 -> 10 -> 20 -> null

3. Step 3: Perform deleteRear()

   Removes 20, list becomes: 5 -> 10 -> null

4. Step 4: Display list

   Output is 5 -> 10 -> null

5. Final Answer:

   5 -> 10 -> null -> Option A

6. Quick Check:

   Front insertions add left, rear deletion removes right [OK]

Hint: Track insertFront and insertRear positions carefully [OK]

Common Mistakes:

• Deleting front instead of rear
• Mixing order of insertions
• Forgetting to update pointers after deletion

4. Consider this buggy method to delete from the front of a dequeue implemented using a linked list:

public void deleteFront() {
  if (front == null) {
    System.out.println("Dequeue is empty");
    return;
  }
  front = front.next;
  front.prev = null;
}

What error can occur when this method is called on a dequeue with only one element?

medium

A. NullPointerException when setting front.prev = null

B. ArrayIndexOutOfBoundsException

C. No error, method works fine

D. Infinite loop in linked list

Solution

1. Step 1: Analyze deleteFront on single element list

   If only one node, after front = front.next, front becomes null.

2. Step 2: Check next line front.prev = null

   Since front is null, accessing front.prev causes NullPointerException.

3. Final Answer:

   NullPointerException when setting front.prev = null -> Option A

4. Quick Check:

   Accessing null pointer causes NullPointerException [OK]

Hint: Check for null before accessing front.prev [OK]

Common Mistakes:

• Assuming front is never null after deletion
• Ignoring edge case of single element
• Confusing exception types

5. You want to implement a dequeue using a linked list that supports these operations efficiently:
1. Insert at front
2. Insert at rear
3. Delete from front
4. Delete from rear
5. Get size of dequeue

Which data members should your dequeue class maintain to achieve all operations in O(1) time?

hard

A. Pointer to rear node only, no size counter

B. Pointer to front node only, and an integer size counter

C. Pointers to front and rear nodes, and an integer size counter

D. Array to store elements and a size counter

Solution

1. Step 1: Identify pointers needed for O(1) insert/delete at both ends

   We need both front and rear pointers to access both ends directly.

2. Step 2: Identify how to get size in O(1)

   Maintain an integer size counter updated on each insert/delete.

3. Final Answer:

   Pointers to front and rear nodes, and an integer size counter -> Option C

4. Quick Check:

   Front + rear pointers + size counter = O(1) ops [OK]

Hint: Keep front, rear pointers and size counter for O(1) operations [OK]

Common Mistakes:

• Using only one pointer for both ends
• Not maintaining size counter for quick size retrieval
• Using array instead of linked list for dequeue