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

Time Complexity: Dequeue Using Linked List

O(1)

Understanding Time Complexity

We want to understand how fast operations on a dequeue using a linked list run.

Specifically, how the time changes when we add or remove items from either end.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Add item at front
func (dq *Dequeue) AddFront(value int) {
  newNode := &Node{value: value}
  newNode.next = dq.front
  if dq.front != nil {
    dq.front.prev = newNode
  }
  dq.front = newNode
  if dq.rear == nil {
    dq.rear = newNode
  }
}
    

This code adds a new item at the front of the dequeue linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding or removing a node at the front or rear pointer.
• How many times: Each operation runs once per call, no loops inside.

How Execution Grows With Input

Each add or remove operation touches only a few nodes, no matter how big the dequeue is.

Input Size (n)	Approx. Operations
10	5
100	5
1000	5

Pattern observation: The number of steps stays about the same even if the dequeue grows larger.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing items at either end takes a constant amount of time, no matter the size.

Common Mistake

[X] Wrong: "Adding or removing from a dequeue takes longer as it gets bigger because we have to move through the list."

[OK] Correct: We keep direct pointers to front and rear, so we never walk through the list. Operations happen instantly at those ends.

Interview Connect

Knowing that dequeue operations run in constant time shows you understand efficient data structures, a key skill in coding interviews.

Self-Check

"What if we used a singly linked list without a rear pointer? How would the time complexity for adding at the rear change?"

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