Dequeue Using Linked List in DSA Go - Build from Scratch

Dequeue Using Linked List

📖 Scenario: You are building a simple system to manage a line of customers waiting for service. The line can add customers at the front or back, and remove customers from the front or back as well. This is called a dequeue (double-ended queue).We will use a linked list to build this dequeue, where each customer is a node connected to the next and previous customers.

🎯 Goal: Create a dequeue using a linked list in Go. You will add customers to the front and back, remove customers from front and back, and finally print the current line of customers.

📋 What You'll Learn

Create a linked list node struct with integer data and pointers to next and previous nodes

Create a dequeue struct with pointers to the front and rear nodes

Add methods to insert at front and rear

Add methods to remove from front and rear

Print the dequeue from front to rear showing all elements

💡 Why This Matters

🌍 Real World

Dequeue structures are used in real-world systems like task scheduling, undo-redo operations, and managing customer service lines where adding and removing from both ends is needed.

💼 Career

Understanding linked lists and dequeue operations is important for software engineers working on system design, performance optimization, and implementing custom data structures.

1

Create Node and Dequeue structs

Create a struct called Node with fields data int, next *Node, and prev *Node. Then create a struct called Dequeue with fields front *Node and rear *Node.

DSA Go

// Your code here

Hint

Think of Node as a customer with links to the next and previous customers. Dequeue keeps track of the first and last customers.

2

Add InsertFront and InsertRear methods

Add a method InsertFront to Dequeue that takes an int and adds a new node at the front. Add another method InsertRear that adds a new node at the rear.

DSA Go

package main

type Node struct {
    data int
    next *Node
    prev *Node
}

type Dequeue struct {
    front *Node
    rear *Node
}

// Add InsertFront method here

// Add InsertRear method here

func main() {
    // Your code here
}

Hint

When the dequeue is empty, both front and rear point to the new node. Otherwise, link the new node properly at front or rear.

3

Add RemoveFront and RemoveRear methods

Add a method RemoveFront to Dequeue that removes the front node if it exists. Add a method RemoveRear that removes the rear node if it exists.

DSA Go

package main

type Node struct {
    data int
    next *Node
    prev *Node
}

type Dequeue struct {
    front *Node
    rear *Node
}

func (dq *Dequeue) InsertFront(value int) {
    newNode := &Node{data: value}
    if dq.front == nil {
        dq.front = newNode
        dq.rear = newNode
        return
    }
    newNode.next = dq.front
    dq.front.prev = newNode
    dq.front = newNode
}

func (dq *Dequeue) InsertRear(value int) {
    newNode := &Node{data: value}
    if dq.rear == nil {
        dq.front = newNode
        dq.rear = newNode
        return
    }
    dq.rear.next = newNode
    newNode.prev = dq.rear
    dq.rear = newNode
}

// Add RemoveFront method here

// Add RemoveRear method here

func main() {
    // Your code here
}

Hint

When removing, check if the dequeue is empty or has only one node. Update front and rear pointers carefully.

4

Print the Dequeue from front to rear

In main, create a Dequeue variable called dq. Insert 10 at front, 20 at rear, 5 at front, and 25 at rear. Then remove one element from front and one from rear. Finally, print the elements from front to rear separated by -> and ending with null.

DSA Go

package main

type Node struct {
    data int
    next *Node
    prev *Node
}

type Dequeue struct {
    front *Node
    rear *Node
}

func (dq *Dequeue) InsertFront(value int) {
    newNode := &Node{data: value}
    if dq.front == nil {
        dq.front = newNode
        dq.rear = newNode
        return
    }
    newNode.next = dq.front
    dq.front.prev = newNode
    dq.front = newNode
}

func (dq *Dequeue) InsertRear(value int) {
    newNode := &Node{data: value}
    if dq.rear == nil {
        dq.front = newNode
        dq.rear = newNode
        return
    }
    dq.rear.next = newNode
    newNode.prev = dq.rear
    dq.rear = newNode
}

func (dq *Dequeue) RemoveFront() {
    if dq.front == nil {
        return
    }
    if dq.front == dq.rear {
        dq.front = nil
        dq.rear = nil
        return
    }
    dq.front = dq.front.next
    dq.front.prev = nil
}

func (dq *Dequeue) RemoveRear() {
    if dq.rear == nil {
        return
    }
    if dq.front == dq.rear {
        dq.front = nil
        dq.rear = nil
        return
    }
    dq.rear = dq.rear.prev
    dq.rear.next = nil
}

func main() {
    var dq Dequeue
    // Your code here
}

Hint

Insert and remove nodes as instructed. Then loop from front to rear printing each node's data followed by ' -> ', ending with 'null'.

Practice

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

easy

A. It automatically sorts the elements.

B. It uses less memory than an array.

C. It allows insertion and deletion from both ends in constant time.

D. It prevents duplicate elements.

Solution

1. Step 1: Understand dequeue operations

   A dequeue supports insertion and deletion at both front and rear ends.

2. Step 2: Compare linked list vs array

   Linked list allows these operations in constant time without shifting elements, unlike arrays.

3. Final Answer:

   It allows insertion and deletion from both ends in constant time. -> Option C

4. Quick Check:

   Linked list dequeue = O(1) insert/delete both ends [OK]

Hint: Linked list avoids shifting, so both ends operate fast [OK]

Common Mistakes:

• Thinking arrays allow O(1) insert/delete at both ends
• Confusing memory usage advantage
• Assuming automatic sorting

2. Which of the following is the correct way to declare a node struct for a dequeue using a linked list in Go?

easy

A. type Node struct { data int; next Node; prev Node }

B. type Node struct { data int; next *Node; prev *Node }

C. type Node struct { data int; next *Node }

D. type Node struct { data int; prev *Node }

Solution

1. Step 1: Identify node fields for dequeue

   A dequeue node needs data, pointer to next and previous nodes for double linking.

2. Step 2: Check pointer usage in Go

   Pointers to Node must be *Node, not Node, to avoid recursive struct size.

3. Final Answer:

   type Node struct { data int; next *Node; prev *Node } -> Option B

4. Quick Check:

   Double linked node uses *Node pointers [OK]

Hint: Use *Node for next and prev to avoid infinite struct size [OK]

Common Mistakes:

• Using Node instead of *Node for next/prev
• Missing prev pointer for double linking
• Declaring only next pointer

3. Given the following Go code snippet for a dequeue implemented with a linked list, what will be printed after the operations?

dq := NewDequeue()
dq.InsertFront(10)
dq.InsertRear(20)
dq.InsertFront(5)
dq.DeleteRear()
dq.Display()

Assuming Display prints elements from front to rear separated by ' -> ' and ends with ' -> null'.

medium

A. 5 -> 10 -> null

B. 10 -> 5 -> null

C. 20 -> 10 -> 5 -> null

D. 5 -> 20 -> null

Solution

1. Step 1: Trace insertions and deletions

   InsertFront(10) => [10], InsertRear(20) => [10, 20], InsertFront(5) => [5, 10, 20], DeleteRear() removes 20 => [5, 10]

2. Step 2: Determine display output

   Display prints front to rear: 5 -> 10 -> null

3. Final Answer:

   5 -> 10 -> null -> Option A

4. Quick Check:

   After operations, dequeue = 5, 10 [OK]

Hint: Track front and rear changes step-by-step [OK]

Common Mistakes:

• Forgetting DeleteRear removes last element
• Mixing order of InsertFront and InsertRear
• Printing rear to front instead of front to rear

4. Consider this buggy Go code snippet for deleting from the front of a dequeue implemented with a linked list:

func (dq *Dequeue) DeleteFront() {
    if dq.front == nil {
        fmt.Println("Dequeue is empty")
        return
    }
    dq.front = dq.front.next
    dq.front.prev = nil
}

What is the error in this code?

medium

A. It does not check if dq.front.next is nil before setting dq.front.prev.

B. It should not set dq.front.prev to nil.

C. It should update dq.rear.prev instead of dq.front.prev.

D. It should set dq.rear to nil when the last element is deleted.

Solution

1. Step 1: Analyze deletion when only one element exists

   If only one node exists, after dq.front = dq.front.next, dq.front becomes nil.

2. Step 2: Check rear pointer update

   Rear pointer still points to old node; it must be set to nil to avoid dangling pointer.

3. Final Answer:

   It should set dq.rear to nil when the last element is deleted. -> Option D

4. Quick Check:

   Update rear on last element deletion [OK]

Hint: Update rear pointer when dequeue becomes empty [OK]

Common Mistakes:

• Not updating rear when front becomes nil
• Assuming front.prev always exists
• Confusing prev and next pointers

5. You want to implement a function in Go that reverses the elements of a dequeue implemented using a doubly linked list. Which approach correctly reverses the dequeue in-place?

hard

A. Traverse each node, swap its next and prev pointers, then swap front and rear pointers of the dequeue.

B. Create a new dequeue and insert elements from the old dequeue's front to rear into the new dequeue's rear.

C. Swap only the data values of front and rear nodes repeatedly until the middle is reached.

D. Delete all elements from front and insert them at rear repeatedly until original dequeue is empty.

Solution

1. Step 1: Understand in-place reversal of doubly linked list

   Swapping next and prev pointers of each node reverses the links.

2. Step 2: Update dequeue front and rear pointers

   After swapping pointers, front and rear pointers must be swapped to reflect new ends.

3. Final Answer:

   Traverse each node, swap its next and prev pointers, then swap front and rear pointers of the dequeue. -> Option A

4. Quick Check:

   Swap pointers in nodes + swap front/rear [OK]

Hint: Swap next/prev in nodes, then swap front and rear [OK]

Common Mistakes:

• Creating new dequeue instead of in-place reversal
• Swapping only data values without pointers
• Deleting and reinserting nodes unnecessarily