Doubly linked list in Data Structures Theory - Time & Space Complexity

Time Complexity: Doubly linked list

O(n)

Understanding Time Complexity

When working with doubly linked lists, it is important to understand how the time to perform operations changes as the list grows.

We want to know how fast we can add, remove, or find items as the list gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

class DoublyLinkedList:
    def __init__(self):
        self.head = None

    def append(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        current = self.head
        while current.next:
            current = current.next
        current.next = new_node
        new_node.prev = current
    

This code adds a new item to the end of a doubly linked list by walking through the list to find the last node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves through each node until the end.
• How many times: It runs once for each node in the list until the last one is found.

How Execution Grows With Input

As the list gets longer, the time to find the end grows roughly in direct proportion to the number of items.

Input Size (n)	Approx. Operations
10	About 10 steps to reach the end
100	About 100 steps to reach the end
1000	About 1000 steps to reach the end

Pattern observation: The time grows linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to add an item at the end grows directly with the number of items in the list.

Common Mistake

[X] Wrong: "Adding to the end of a doubly linked list always takes constant time because we can go backwards."

[OK] Correct: Without keeping a reference to the last node, we must walk from the start to find the end, which takes time proportional to the list size.

Interview Connect

Understanding how linked list operations scale helps you explain your choices clearly and shows you know how data structures behave in real situations.

Self-Check

"What if we kept a reference to the tail node? How would the time complexity of appending change?"

Practice

1. What is a key feature of a doubly linked list compared to a singly linked list?

easy

A. It stores data in a fixed-size array

B. Nodes only link to the next node

C. Each node has links to both the previous and next nodes

D. It uses a stack structure internally

Solution

1. Step 1: Understand node links in doubly linked lists

   A doubly linked list node contains two links: one to the previous node and one to the next node.

2. Step 2: Compare with singly linked list

   A singly linked list node only has a link to the next node, not the previous one.

3. Final Answer:

   Each node has links to both the previous and next nodes -> Option C

4. Quick Check:

   Doubly linked list = two links per node [OK]

Hint: Remember: doubly means two links per node [OK]

Common Mistakes:

• Thinking nodes only link forward
• Confusing with arrays or stacks
• Assuming fixed size storage

2. Which of the following is the correct way to update links when inserting a new node newNode after a node current in a doubly linked list?

easy

A. newNode.next = current.next; newNode.prev = current; current.next.prev = newNode; current.next = newNode;

B. newNode.prev = current.next; newNode.next = current; current.prev = newNode; current.next = newNode;

C. current.next = newNode; newNode.next = current.prev; current.prev = newNode;

D. newNode.next = current.prev; newNode.prev = current.next; current.next = newNode;

Solution

1. Step 1: Set newNode's next and prev pointers

   newNode.next should point to current.next, and newNode.prev should point to current.

2. Step 2: Update surrounding nodes to link to newNode

   current.next.prev must point back to newNode, and current.next must point to newNode.

3. Final Answer:

   newNode.next = current.next; newNode.prev = current; current.next.prev = newNode; current.next = newNode; -> Option A

4. Quick Check:

   Update newNode links, then fix neighbors [OK]

Hint: Update new node links before neighbors [OK]

Common Mistakes:

• Mixing prev and next pointers
• Not updating current.next.prev
• Assigning pointers in wrong order

3. Consider the following doubly linked list nodes and code snippet:

Node1: data=10, prev=None, next=Node2
Node2: data=20, prev=Node1, next=Node3
Node3: data=30, prev=Node2, next=None

current = Node2
print(current.prev.data, current.data, current.next.data)

What will be the output?

medium

A. None 20 30

B. 30 20 10

C. 20 20 20

D. 10 20 30

Solution

1. Step 1: Identify current node and its neighbors

   current is Node2 with data 20; prev is Node1 with data 10; next is Node3 with data 30.

2. Step 2: Print data values of prev, current, and next nodes

   Output will be 10 (prev), 20 (current), 30 (next).

3. Final Answer:

   10 20 30 -> Option D

4. Quick Check:

   Prev, current, next data = 10, 20, 30 [OK]

Hint: Prev and next data surround current data [OK]

Common Mistakes:

• Confusing prev and next nodes
• Assuming None has data
• Printing node references instead of data

4. The following code attempts to delete a node current from a doubly linked list. What is the error?

current.prev.next = current.next
current.next.prev = current.prev
current = None

medium

A. It should only update current.prev.next, not current.next.prev

B. It does not handle the case when current is the head node

C. It should update current.next.next instead of current.prev.next

D. It incorrectly sets current to None instead of deleting

Solution

1. Step 1: Analyze pointer updates for deletion

   The code updates the previous node's next pointer and the next node's prev pointer to bypass current.

2. Step 2: Check edge cases for head or tail nodes

   If current is the head (no prev), current.prev is None, so current.prev.next causes an error.

3. Final Answer:

   It does not handle the case when current is the head node -> Option B

4. Quick Check:

   Deleting head needs special handling [OK]

Hint: Check for None before updating neighbors [OK]

Common Mistakes:

• Ignoring head or tail node cases
• Assuming current is always in middle
• Confusing which pointers to update

5. You want to reverse a doubly linked list by swapping the next and prev pointers of each node. Which of the following steps is correct to achieve this?

hard

A. For each node, swap its next and prev pointers, then move to the new prev node

B. For each node, swap its next and prev pointers, then move to the new next node

C. For each node, only swap next pointers and ignore prev pointers

D. For each node, set next to None and prev to head

Solution

1. Step 1: Swap next and prev pointers for each node

   Swapping next and prev reverses the direction of links.

2. Step 2: Move to the new prev node after swapping

   After swapping, the original next is now prev, so move to node.prev to continue.

3. Final Answer:

   For each node, swap its next and prev pointers, then move to the new prev node -> Option A

4. Quick Check:

   Swap links, then follow new prev to reverse [OK]

Hint: After swapping, move to node.prev to continue [OK]

Common Mistakes:

• Moving to node.next after swapping
• Swapping only one pointer
• Setting pointers to None incorrectly