Why Doubly linked list in Data Structures Theory? - Purpose & Use Cases

The Big Idea

What if you could move both forward and backward through your data with ease, just like flipping pages in a book?

The Scenario

Imagine you have a long chain of paper clips linked together. You want to find a specific clip, but you can only move forward from the first clip, one by one, until you find it.

Now, what if you want to go back to the previous clip? You have no way to do that easily because you only linked them forward.

The Problem

Using a simple chain that only moves forward means you must start from the beginning every time you want to go backward or find something near the end.

This makes searching slow and frustrating, especially if the chain is very long.

Also, removing or adding clips in the middle is tricky because you have to remember the clip before it, which is not directly linked.

The Solution

A doubly linked list solves this by linking each clip both forward and backward.

This way, you can easily move in both directions, making searching, adding, or removing clips much faster and simpler.

Before vs After

✗ Before

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

# Can only move forward

✓ After

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

# Can move forward and backward

What It Enables

It enables quick and flexible navigation through data in both directions, making many operations efficient and easy to implement.

Real Life Example

Think of a music playlist where you can go to the next song or easily go back to the previous one without restarting the list.

Key Takeaways

Doubly linked lists link elements both forward and backward.

This allows easy movement in both directions.

It makes adding, removing, and searching more efficient than single-direction lists.

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