Dequeue Using Linked List in DSA C++ - Execution Trace

Concept Flow - Dequeue Using Linked List

Create new node

↓

Is Dequeue empty?

Yes→Set head and tail to new node

|No↓

Insert at front or rear?

↓

Add node before head

↓

Update head or tail pointer

↓

Done

↓

Is Dequeue empty?

Yes→Underflow - cannot delete

|No↓

Delete from front or rear?

↓

Remove head node

↓

Update head or tail pointer

↓

Done

Shows how nodes are added or removed at both ends of the linked list, updating head and tail pointers accordingly.

Execution Sample

DSA C++

class Node {
public:
  int data;
  Node* next;
  Node* prev;
};

// InsertFront, InsertRear, DeleteFront, DeleteRear methods

Defines a doubly linked list node and basic dequeue operations to add or remove nodes at both ends.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	InsertFront(10)	10	head=Node(10), tail=Node(10)	10 <-> null
2	InsertRear(20)	10 -> 20	tail=Node(20), Node(10).next=Node(20), Node(20).prev=Node(10)	10 <-> 20 <-> null
3	InsertFront(5)	5 -> 10 -> 20	head=Node(5), Node(5).next=Node(10), Node(10).prev=Node(5)	5 <-> 10 <-> 20 <-> null
4	DeleteRear()	5 -> 10	tail=Node(10), Node(10).next=null	5 <-> 10 <-> null
5	DeleteFront()	10	head=Node(10), Node(10).prev=null	10 <-> null
6	DeleteFront()	empty	head=null, tail=null	null
7	DeleteFront()	empty	No change (underflow)	null

💡 Step 7 stops because dequeue is empty; cannot delete front.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7
head	null	Node(10)	Node(10)	Node(5)	Node(5)	Node(10)	null	null
tail	null	Node(10)	Node(20)	Node(20)	Node(10)	Node(10)	null	null
List Size	0	1	2	3	2	1	0	0

Key Moments - 3 Insights

Why do we update both head and tail pointers when the list is empty during insertion?

When the list is empty (Step 1), both head and tail must point to the new node because it is the only node. This is shown in execution_table row 1.

Why does deleting from an empty dequeue not change pointers?

At Step 7, the dequeue is empty, so deletion cannot happen. Pointers remain null to avoid errors, as shown in execution_table row 7.

Why do we update the prev pointer of the next node during insertion at front?

In Step 3, when inserting at front, the new node's next points to old head, and old head's prev must point back to new node to keep links correct, as shown in pointer changes.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the visual state of the list?

A 10 <-> 5 <-> 20 <-> null

B 5 <-> 10 <-> 20 <-> null

C 5 -> 10 -> 20 -> null

D 10 <-> 20 <-> 5 <-> null

💡 Hint

Check the Visual State column at Step 3 in execution_table.

At which step does the dequeue become empty?

A Step 7

B Step 5

C Step 6

D Step 4

💡 Hint

Look at List Size and Visual State columns in execution_table.

If we insertRear(30) after Step 6, what will be the new tail?

A Node(30)

B null

C Node(10)

D Node(5)

💡 Hint

After Step 6, list is empty, so new node becomes both head and tail.

Concept Snapshot

Dequeue using doubly linked list:
- InsertFront: add node before head, update head
- InsertRear: add node after tail, update tail
- DeleteFront: remove head node, update head
- DeleteRear: remove tail node, update tail
- Handle empty list by setting head and tail to null
- Maintain prev and next pointers for bidirectional links

Full Transcript

This visualization shows how a dequeue works using a doubly linked list. Nodes can be added or removed from both front and rear. When inserting into an empty dequeue, both head and tail point to the new node. When deleting, pointers update to remove nodes safely. The execution table tracks each step, showing nodes in the list, pointer changes, and the visual linked structure. Key moments clarify why pointers update in certain ways and how empty conditions are handled. The quiz tests understanding of list states and pointer updates.

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