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

Time Complexity: Dequeue Using Linked List

O(n)

Understanding Time Complexity

We want to understand how fast operations like adding or removing items work in a dequeue built with a linked list.

How does the time taken change as we add more items?

Scenario Under Consideration

Analyze the time complexity of these dequeue operations using a linked list.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class Dequeue:
    def __init__(self):
        self.front = None
        self.rear = None

    def enqueue_front(self, data):
        new_node = Node(data)
        if not self.front:
            self.front = self.rear = new_node
        else:
            new_node.next = self.front
            self.front = new_node

    def dequeue_rear(self):
        if not self.rear:
            return None
        if self.front == self.rear:
            temp = self.rear
            self.front = self.rear = None
            return temp.data
        current = self.front
        while current.next != self.rear:
            current = current.next
        temp = self.rear
        self.rear = current
        self.rear.next = None
        return temp.data

This code adds an item at the front and removes an item from the rear of the dequeue using a linked list.

Identify Repeating Operations

Look for loops or repeated steps in the code.

• Primary operation: The while loop in dequeue_rear that moves through nodes to find the one before the rear.
• How many times: It runs once for each node except the last, so up to n-1 times if there are n nodes.

How Execution Grows With Input

As the number of items grows, the time to remove from the rear grows too because we must find the node before the last.

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

Pattern observation: The steps increase roughly in a straight line with the number of items.

Final Time Complexity

Time Complexity: O(n)

This means removing from the rear takes longer as the list grows, roughly proportional to the number of items.

Common Mistake

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

[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 how linked list operations scale helps you explain trade-offs clearly and choose the right data structure for the job.

Self-Check

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

Practice

1. What is a dequeue (double-ended queue) implemented using a linked list?

easy

A. A queue where elements can only be added at the rear

B. A list that only allows insertion at the front

C. A stack implemented using arrays

D. A data structure where elements can be added or removed from both ends

Solution

1. Step 1: Understand dequeue operations

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

2. Step 2: Linked list implementation

   Using a linked list, nodes can be added or removed from both ends efficiently.

3. Final Answer:

   A data structure where elements can be added or removed from both ends -> Option D

4. Quick Check:

   Dequeue = double-ended queue [OK]

Hint: Dequeue means double-ended queue, both ends allowed [OK]

Common Mistakes:

• Confusing dequeue with stack or simple queue
• Thinking insertion is only at one end
• Assuming array implementation only

2. Which of the following is the correct way to initialize an empty dequeue using a linked list in Python?

easy

A. class Dequeue: def __init__(self): self.front = None; self.rear = None

B. class Dequeue: def __init__(self): self.list = []

C. class Dequeue: def __init__(self): self.head = 0; self.tail = 0

D. class Dequeue: def __init__(self): self.front = 0; self.rear = 0

Solution

1. Step 1: Identify linked list pointers

   For linked list, front and rear pointers are initialized to None to indicate empty.

2. Step 2: Check options

   class Dequeue: def __init__(self): self.front = None; self.rear = None correctly initializes front and rear as None; others use incorrect types or structures.

3. Final Answer:

   class Dequeue: def __init__(self): self.front = None; self.rear = None -> Option A

4. Quick Check:

   Linked list empty pointers = None [OK]

Hint: Empty linked list pointers are None, not zero or list [OK]

Common Mistakes:

• Using list instead of linked list
• Initializing pointers with 0 instead of None
• Confusing front/rear with indices

3. What will be the output after performing these operations on an empty dequeue implemented with linked list?
dq.insert_rear(10)
dq.insert_front(20)
dq.insert_rear(30)
dq.delete_front()
Print the dequeue from front to rear.

medium

A. 20 -> 10 -> 30 -> null

B. 20 -> 30 -> null

C. 10 -> 30 -> null

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

Solution

1. Step 1: Perform insert_rear(10) on empty dequeue

   Dequeue becomes: 10 -> null

2. Step 2: Perform insert_front(20)

   Dequeue becomes: 20 -> 10 -> null

3. Step 3: Perform insert_rear(30)

   Dequeue becomes: 20 -> 10 -> 30 -> null

4. Step 4: Perform delete_front()

   Remove 20 from front, dequeue becomes: 10 -> 30 -> null

5. Final Answer:

   10 -> 30 -> null -> Option C

6. Quick Check:

   After operations, front to rear = 10, 30 [OK]

Hint: Track each operation stepwise front to rear [OK]

Common Mistakes:

• Forgetting delete_front removes front element
• Mixing order of insert_front and insert_rear
• Printing rear to front instead of front to rear

4. Identify the error in this dequeue linked list method for deleting from rear:

def delete_rear(self):
    if self.rear is None:
        return None
    temp = self.rear
    self.rear = self.rear.next
    return temp.data

medium

A. Does not check if dequeue is empty

B. Incorrectly moves rear pointer forward instead of backward

C. Returns data before deleting node

D. Uses wrong variable name for rear

Solution

1. Step 1: Understand rear pointer movement

   In linked list, rear points to last node; to delete rear, rear must move backward (to previous node).

2. Step 2: Analyze code

   Code moves rear to rear.next (forward), which is incorrect for rear deletion.

3. Final Answer:

   Incorrectly moves rear pointer forward instead of backward -> Option B

4. Quick Check:

   Rear deletion moves pointer backward [OK]

Hint: Rear deletion moves pointer backward, not forward [OK]

Common Mistakes:

• Moving rear pointer forward instead of backward
• Not handling empty dequeue properly
• Confusing front and rear pointers

5. You want to implement a dequeue using a doubly linked list to support these operations efficiently:
1. Insert at front
2. Insert at rear
3. Delete from front
4. Delete from rear

Which of the following is the best approach to maintain pointers for O(1) time complexity for all operations?

hard

A. Maintain front and rear pointers; each node has next and prev links

B. Maintain only front pointer; traverse list for rear operations

C. Use singly linked list with only next pointers and rear pointer

D. Use array to simulate linked list with indices

Solution

1. Step 1: Understand operation requirements

   All operations must be O(1), so no traversal allowed.

2. Step 2: Analyze pointer needs

   Doubly linked list with front and rear pointers allows insert/delete at both ends in O(1).

3. Step 3: Check options

   Maintain front and rear pointers; each node has next and prev links matches this approach; others require traversal or use arrays which are not linked lists.

4. Final Answer:

   Maintain front and rear pointers; each node has next and prev links -> Option A

5. Quick Check:

   Doubly linked list + front/rear pointers = O(1) ops [OK]

Hint: Use doubly linked list with front and rear pointers [OK]

Common Mistakes:

• Using singly linked list causing O(n) rear operations
• Maintaining only one pointer causing traversal
• Confusing arrays with linked lists