Reverse a Doubly Linked List in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap the next and prev pointers of the current node.

DSA Typescript

current.[1] = temp;

Drag options to blanks, or click blank then click option'

Anext

Bvalue

Cprev

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning temp to current.next instead of current.prev.

Trying to assign to current.value which is not a pointer.

We swap the next and prev pointers by assigning temp (original next) to current.prev.

2fill in blank

medium

Complete the code to move the current pointer to the next node in the original list.

DSA Typescript

current = current.[1];

Drag options to blanks, or click blank then click option'

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using current.next which now points backward after swapping.

Using current.head or current.tail which are not node pointers.

After swapping, current.prev points to the next node in the original list, so we move current to current.prev.

3fill in blank

hard

Complete the code to update the head pointer inside the loop after swapping pointers.

DSA Typescript

if (temp == null) {
  head = [1];
}

Drag options to blanks, or click blank then click option'

Ahead.prev

Btemp.next

Ctemp.prev

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Using if (temp != null) which skips the last node.

Assigning head = temp which is null.

Assigning head = temp.prev which is undefined.

Inside the loop, after swapping pointers for the last node, temp == null and current is the original tail (new head), so head = current.

4fill in blank

hard

Fill both blanks to complete the while loop condition and the pointer swap inside the loop.

DSA Typescript

while (current != [1]) {
  let temp = current.[2];
  current.[3] = current.[4];
  current.[5] = temp;
  current = current.[6];
}

Drag options to blanks, or click blank then click option'

Anull

Bnext

Cprev

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop condition like current != head.

Mixing up next and prev pointers during swap.

The loop runs until current is null. We store current.next in temp, swap current.next and current.prev, then move current to current.prev (original next).

5fill in blank

hard

Fill all three blanks to complete the pointer swap inside the loop of the reverse function.

DSA Typescript

function reverseDoublyLinkedList(head: Node | null): Node | null {
  let current = head;
  let temp: Node | null = null;
  while (current != null) {
    temp = current.next;
    current.[1] = current.[2];
    current.[3] = temp;
    if (temp == null) {
      head = current;
    }
    current = temp;
  }
  return head;
}

Drag options to blanks, or click blank then click option'

Anext

Bprev

Attempts:

3 left

💡 Hint

Common Mistakes

Not swapping pointers correctly (e.g., current.next = temp).

Confusing next and prev in the assignments.

temp = current.next (hardcoded), current.next = current.prev, current.prev = temp (pointers swapped), if(temp == null) head = current (update new head), current = temp (move forward).

Practice

1. What is the main purpose of reversing a doubly linked list?

easy

A. To sort the nodes in ascending order

B. To reverse the order of nodes so the last becomes first and vice versa

C. To delete all nodes from the list

D. To convert it into a singly linked list

Solution

1. Step 1: Understand the structure of a doubly linked list

   A doubly linked list has nodes with pointers to both previous and next nodes.

2. Step 2: Purpose of reversing

   Reversing swaps the direction so the last node becomes the first and the first becomes the last.

3. Final Answer:

   To reverse the order of nodes so the last becomes first and vice versa -> Option B

4. Quick Check:

   Reversing = changing node order [OK]

Hint: Reversing flips the list order, not deleting or sorting [OK]

Common Mistakes:

• Confusing reversing with deleting nodes
• Thinking reversing sorts the list
• Assuming it converts to singly linked list

2. Which of the following TypeScript code snippets correctly swaps the next and prev pointers of a node in a doubly linked list?

easy

A. let temp = node.next; node.next = node.prev; node.prev = temp;

B. node.next = node.next; node.prev = node.prev;

C. node.prev = node.next; node.next = node.prev;

D. let temp = node.prev; node.prev = node.next; node.next = temp.next;

Solution

1. Step 1: Identify correct swapping logic

   Swapping pointers requires a temporary variable to hold one pointer before assignment.

2. Step 2: Check each option

   let temp = node.next; node.next = node.prev; node.prev = temp; correctly uses a temp variable to swap next and prev. Others either assign incorrectly or lose references.

3. Final Answer:

   let temp = node.next; node.next = node.prev; node.prev = temp; -> Option A

4. Quick Check:

   Swap pointers using temp variable [OK]

Hint: Use a temp variable to swap pointers safely [OK]

Common Mistakes:

• Assigning pointers without temp causes data loss
• Mixing up next and prev assignments
• Using incorrect pointer references

3. What will be the output of the following TypeScript code that reverses a doubly linked list?

class Node {
  data: number;
  next: Node | null = null;
  prev: Node | null = null;
  constructor(data: number) { this.data = data; }
}

function reverse(head: Node | null): Node | null {
  let current = head;
  let temp: Node | null = null;
  while (current !== null) {
    temp = current.prev;
    current.prev = current.next;
    current.next = temp;
    current = current.prev;
  }
  if (temp !== null) {
    head = temp.prev;
  }
  return head;
}

// Create list: 1 <-> 2 <-> 3
const n1 = new Node(1);
const n2 = new Node(2);
n1.next = n2; n2.prev = n1;
const n3 = new Node(3);
n2.next = n3; n3.prev = n2;

const newHead = reverse(n1);

// Print reversed list
let result = [];
let node = newHead;
while (node !== null) {
  result.push(node.data);
  node = node.next;
}
console.log(result);

medium

A. []

B. [1, 2, 3]

C. [3, 2, 1]

D. [3, 1, 2]

Solution

1. Step 1: Trace pointer swaps in the loop

   Each node swaps its next and prev. The loop moves forward using the old next pointer now in prev.

2. Step 2: Update head to new first node

   After loop, head is updated to last node processed, which is node with data 3.

3. Final Answer:

   [3, 2, 1] -> Option C

4. Quick Check:

   Reversed list order = [3, 2, 1] [OK]

Hint: Reversing swaps pointers and updates head to last node [OK]

Common Mistakes:

• Not updating head after reversal
• Confusing next and prev in loop
• Printing original list order

4. Identify the bug in this TypeScript function that attempts to reverse a doubly linked list:

function reverse(head: Node | null): Node | null {
  let current = head;
  while (current !== null) {
    let temp = current.next;
    current.next = current.prev;
    current.prev = temp;
    current = current.next;
  }
  return head;
}

medium

A. The function should return current instead of head

B. The pointers next and prev are swapped incorrectly

C. The loop condition should be current.next !== null

D. The function does not update the head to the new first node after reversal

Solution

1. Step 1: Check pointer swapping logic

   The swapping of next and prev is correct using a temp variable.

2. Step 2: Verify head update after reversal

   The function returns original head without updating it to the new first node, so the reversed list head is incorrect.

3. Final Answer:

   The function does not update the head to the new first node after reversal -> Option D

4. Quick Check:

   Head must update to new first node [OK]

Hint: Always update head to last processed node after reversal [OK]

Common Mistakes:

• Forgetting to update head after reversal
• Returning old head instead of new head
• Incorrect loop condition causing infinite loop

5. Given a doubly linked list with nodes containing integer data, how would you modify the reversal function to also count how many nodes were reversed and return both the new head and the count?

hard

A. Modify the function to use an object return type like { head: Node | null, count: number }, increment count inside the loop, and update head after reversal.

B. Add a global variable to count nodes and return only the new head.

C. Count nodes before reversal and return the count without changing the function.

D. Return an array with the reversed list nodes and count without modifying the reversal logic.

Solution

1. Step 1: Add a counter variable inside the reversal function

   Initialize a count variable to zero and increment it each time a node is processed in the reversal loop.

2. Step 2: Return both new head and count

   After reversal, return an object containing the updated head and the count of nodes reversed.

3. Final Answer:

   Modify the function to use an object return type like { head: Node | null, count: number }, increment count inside the loop, and update head after reversal. -> Option A

4. Quick Check:

   Return head and count together after reversal [OK]

Hint: Use an object return type to return multiple values [OK]

Common Mistakes:

• Using global variables instead of local counters
• Not updating head before returning
• Returning count without reversing