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DSA Typescriptprogramming~10 mins

Dequeue Using Linked List in DSA Typescript - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to add a new node at the front of the dequeue.

DSA Typescript
addFront(data: number): void {
  const newNode = new Node(data);
  if (this.front === null) {
    this.front = this.rear = newNode;
  } else {
    newNode.next = this.[1];
    this.front.prev = newNode;
    this.front = newNode;
  }
}
Drag options to blanks, or click blank then click option'
Aprev
Brear
Cnext
Dfront
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'rear' instead of 'front' for the next pointer.
2fill in blank
medium

Complete the code to remove a node from the rear of the dequeue.

DSA Typescript
removeRear(): number | null {
  if (this.rear === null) return null;
  const data = this.rear.data;
  this.rear = this.rear.[1];
  if (this.rear !== null) {
    this.rear.next = null;
  } else {
    this.front = null;
  }
  return data;
}
Drag options to blanks, or click blank then click option'
Aprev
Bnext
Crear
Dfront
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'next' instead of 'prev' to move rear pointer.
3fill in blank
hard

Fix the error in the code to correctly check if the dequeue is empty.

DSA Typescript
isEmpty(): boolean {
  return this.[1] === null && this.rear === null;
}
Drag options to blanks, or click blank then click option'
Afront
Bnext
Cprev
Ddata
Attempts:
3 left
💡 Hint
Common Mistakes
Checking 'next' or 'prev' instead of 'front'.
4fill in blank
hard

Fill both blanks to correctly add a node at the rear of the dequeue.

DSA Typescript
addRear(data: number): void {
  const newNode = new Node(data);
  if (this.rear === null) {
    this.front = this.rear = newNode;
  } else {
    this.rear.[1] = newNode;
    newNode.[2] = this.rear;
    this.rear = newNode;
  }
}
Drag options to blanks, or click blank then click option'
Anext
Bprev
Cfront
Drear
Attempts:
3 left
💡 Hint
Common Mistakes
Swapping 'next' and 'prev' assignments.
5fill in blank
hard

Fill all three blanks to correctly remove a node from the front of the dequeue.

DSA Typescript
removeFront(): number | null {
  if (this.front === null) return null;
  const data = this.front.data;
  this.front = this.front.[1];
  if (this.front !== null) {
    this.front.[2] = null;
  } else {
    this.rear = null;
  }
  // Update rear if list becomes empty
  this.rear = this.front === null ? [3] : this.rear;
  return data;
}
Drag options to blanks, or click blank then click option'
Anext
Bprev
Cnull
Dthis.front
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'prev' instead of 'next' to move front pointer.
Not setting rear to null when list is empty.

Practice

(1/5)
1. What is the main advantage of using a linked list to implement a dequeue compared to an array?
easy
A. It allows insertion and deletion from both ends in constant time.
B. It uses less memory than arrays.
C. It automatically sorts the elements.
D. It prevents duplicate elements.

Solution

  1. Step 1: Understand dequeue operations

    A dequeue supports insertion and deletion from both front and rear ends.
  2. Step 2: Compare linked list and array

    Linked lists allow adding/removing nodes at both ends in constant time, unlike arrays which may require shifting elements.
  3. Final Answer:

    It allows insertion and deletion from both ends in constant time. -> Option A
  4. Quick Check:

    Linked list dequeue = O(1) insert/delete both ends [OK]
Hint: Linked list supports fast add/remove at both ends [OK]
Common Mistakes:
  • Thinking arrays allow constant time insert/delete at front
  • Confusing memory usage advantage
  • Assuming automatic sorting
2. Which of the following is the correct TypeScript method signature to insert an element at the front of a dequeue implemented using a linked list?
easy
A. insertFront(value: number): void
B. addRear(value: number): number
C. pushFront(): void
D. enqueueFront(value: string): string

Solution

  1. Step 1: Identify method purpose and parameter

    Inserting at front requires a method that takes a value (number) and returns nothing (void).
  2. Step 2: Match signature with correct naming and types

    insertFront(value: number): void matches: method name 'insertFront', parameter 'value: number', return type 'void'. Others have wrong names, return types, or parameter types.
  3. Final Answer:

    insertFront(value: number): void -> Option A
  4. Quick Check:

    Insert front method = insertFront(value: number): void [OK]
Hint: Method to add front takes value and returns void [OK]
Common Mistakes:
  • Using wrong method names like pushFront
  • Wrong return types like number or string
  • Incorrect parameter types
3. Consider the following TypeScript code snippet for a dequeue using linked list:
const dq = new Dequeue();
dq.insertRear(10);
dq.insertFront(20);
dq.insertRear(30);
dq.deleteFront();
dq.insertFront(40);
dq.deleteRear();
dq.display();
What will be the printed output?
medium
A. 10 -> 20 -> 30 -> null
B. 40 -> 10 -> null
C. 20 -> 10 -> 30 -> null
D. 40 -> 20 -> null

Solution

  1. Step 1: Trace each operation on dequeue

    Start empty. 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: Display final dequeue state

    Prints elements from front to rear: 40 -> 10 -> null
  3. Final Answer:

    40 -> 10 -> null -> Option B
  4. Quick Check:

    After operations, dequeue = 40 -> 10 -> null [OK]
Hint: Trace each insert/delete step carefully [OK]
Common Mistakes:
  • Forgetting deleteFront removes front element
  • Mixing order of insertFront and insertRear
  • Not updating links after deleteRear
4. The following TypeScript method for deleting the rear node in a dequeue implemented using a singly linked list has a bug. What is the bug?
deleteRear(): number | null {
  if (!this.rear) return null;
  let temp = this.front;
  while (temp.next !== this.rear) {
    temp = temp.next;
  }
  const value = this.rear.value;
  this.rear = temp;
  this.rear.next = null;
  return value;
}
medium
A. It causes an infinite loop when the list is empty.
B. It incorrectly updates the front pointer.
C. It returns the wrong value after deletion.
D. It does not handle the case when there is only one node.

Solution

  1. Step 1: Analyze loop and pointers

    The loop finds the node before rear by checking temp.next !== rear. If only one node exists, temp.next is null, so loop condition fails immediately.
  2. Step 2: Check single node case

    If only one node, front and rear point to same node. The code does not update front or rear to null, causing incorrect state after deletion.
  3. Final Answer:

    It does not handle the case when there is only one node. -> Option D
  4. Quick Check:

    Single node deleteRear needs special handling [OK]
Hint: Check single-node edge case in deleteRear [OK]
Common Mistakes:
  • Ignoring single-node list case
  • Assuming front pointer updates automatically
  • Overlooking null pointer exceptions
5. You want to implement a dequeue using a doubly linked list in TypeScript. Which of the following changes will improve the efficiency of deleting the rear node compared to a singly linked list implementation?
hard
A. Traverse from front to find the node before rear each time.
B. Use an array instead of linked list for rear deletion.
C. Maintain a rear pointer and use the previous pointer to delete rear in O(1) time.
D. Delete rear by shifting all elements forward.

Solution

  1. Step 1: Understand doubly linked list structure

    Each node has pointers to next and previous nodes, allowing backward traversal.
  2. Step 2: Use rear pointer and previous link

    With rear pointer and previous link, deleting rear node is direct and takes constant time without traversal.
  3. Final Answer:

    Maintain a rear pointer and use the previous pointer to delete rear in O(1) time. -> Option C
  4. Quick Check:

    Doubly linked list rear delete = O(1) using prev pointer [OK]
Hint: Use prev pointer in doubly linked list for fast rear delete [OK]
Common Mistakes:
  • Traversing from front wastes time
  • Using array shifts is inefficient
  • Ignoring doubly linked list advantages
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