Reverse a Singly Linked List Iterative in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the previous pointer to null.

DSA Typescript

let prev = [1];

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

Anull

Bhead

Cundefined

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Setting prev to head causes incorrect reversal.

Using undefined instead of null breaks pointer logic.

We start reversing by setting prev to null because the new tail will point to null.

2fill in blank

medium

Complete the code to store the next node before changing links.

DSA Typescript

let next = current.[1];

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

Aprev

Bhead

Cnext

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev instead of next loses the list.

Assigning null breaks the chain.

We save current.next in next to not lose the rest of the list when reversing.

3fill in blank

hard

Fix the error in reversing the link by assigning the correct pointer.

DSA Typescript

current.next = [1];

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

Anext

Bhead

Ccurrent

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning next causes no reversal.

Assigning current or head causes loops or errors.

To reverse the link, current.next must point to prev, the previous node.

4fill in blank

hard

Fill both blanks to update pointers for the next iteration.

DSA Typescript

prev = [1];
current = [2];

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

Acurrent

Bnext

Cprev

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping these pointers breaks the loop.

Assigning prev to next causes loss of reference.

After reversing the link, move prev to current and current to next to continue.

5fill in blank

hard

Fill all three blanks to complete the iterative reverse function.

DSA Typescript

function reverseList(head: ListNode | null): ListNode | null {
  let prev = [1];
  let current = head;
  while (current !== null) {
    let next = current.[2];
    current.[3] = prev;
    prev = current;
    current = next;
  }
  return prev;
}

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

Anull

Bnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up next and prev pointers.

Not returning prev at the end.

The function starts with prev = null. Inside the loop, it saves current.next in next, then reverses the link by setting current.next = prev. Finally, it moves prev and current forward.

Practice

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

easy

A. To sort the linked list in ascending order

B. To change the direction of the links between nodes without using recursion

C. To delete all nodes from the linked list

D. To find the middle node of the linked list

Solution

1. Step 1: Understand the operation of reversing

   Reversing a singly linked list means changing the direction of the next pointers so the last node becomes the first.

2. Step 2: Identify iterative approach

   The iterative method changes links step-by-step using a loop without recursion.

3. Final Answer:

   To change the direction of the links between nodes without using recursion -> Option B

4. Quick Check:

   Reversing iteratively = changing links direction without recursion [OK]

Hint: Reversing means flipping links, not sorting or deleting [OK]

Common Mistakes:

• Confusing reversing with sorting
• Thinking recursion is required
• Assuming nodes are deleted

2. Which of the following TypeScript code snippets correctly initializes pointers for iterative reversal of a singly linked list?

easy

A. let prev = null; let current = head;

B. let prev = head; let current = null;

C. let prev = head; let current = head.next;

D. let prev = null; let current = null;

Solution

1. Step 1: Understand pointer roles

   Prev starts as null because the new tail points to null; current starts at head to traverse.

2. Step 2: Match initialization

   Only let prev = null; let current = head; sets prev to null and current to head correctly for iteration.

3. Final Answer:

   let prev = null; let current = head; -> Option A

4. Quick Check:

   Prev null, current head = correct start [OK]

Hint: Prev starts null, current starts at head for iteration [OK]

Common Mistakes:

• Setting prev to head initially
• Starting current as null
• Both pointers null at start

3. Given the linked list: 1 -> 2 -> 3 -> null, what will be the output after running this iterative reversal code?

function reverseList(head: ListNode | null): ListNode | null {
  let prev = null;
  let current = head;
  while (current !== null) {
    let nextTemp = current.next;
    current.next = prev;
    prev = current;
    current = nextTemp;
  }
  return prev;
}

medium

A. null

B. 1 -> 2 -> 3 -> null

C. 3 -> 2 -> 1 -> null

D. 1 -> null

Solution

1. Step 1: Trace the reversal loop

   Initially prev=null, current=1. Each iteration reverses current.next to prev and moves pointers forward.

2. Step 2: Final reversed list

   After loop, prev points to node 3, which links to 2, then 1, then null.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option C

4. Quick Check:

   Reversed list = 3 -> 2 -> 1 -> null [OK]

Hint: Follow pointers step-by-step to see reversed links [OK]

Common Mistakes:

• Returning head instead of prev
• Not updating current correctly
• Confusing next pointer updates

4. Identify the error in this iterative reversal code snippet:

function reverseList(head: ListNode | null): ListNode | null {
  let prev = head;
  let current = head.next;
  while (current !== null) {
    let nextTemp = current.next;
    current.next = prev;
    prev = current;
    current = nextTemp;
  }
  return prev;
}

medium

A. current should start at head, not head.next

B. The function should return current instead of prev

C. nextTemp should be assigned after updating current.next

D. prev should be initialized to null, not head

Solution

1. Step 1: Check pointer initialization

   Prev must start as null because the new tail points to null, not head.

2. Step 2: Understand impact of wrong initialization

   Starting prev at head causes a cycle or incorrect reversal.

3. Final Answer:

   prev should be initialized to null, not head -> Option D

4. Quick Check:

   Prev must be null at start [OK]

Hint: Prev always null at start to avoid cycles [OK]

Common Mistakes:

• Starting prev at head causing loops
• Starting current at wrong node
• Returning wrong pointer

5. You have a singly linked list with nodes containing values: 10 -> 20 -> 30 -> 40 -> null. After reversing it iteratively, you want to insert a new node with value 25 between 20 and 30 in the reversed list. What will be the correct order of nodes after insertion?

hard

A. 40 -> 30 -> 25 -> 20 -> 10 -> null

B. 40 -> 30 -> 20 -> 25 -> 10 -> null

C. 40 -> 30 -> 20 -> 10 -> 25 -> null

D. 40 -> 25 -> 30 -> 20 -> 10 -> null

Solution

1. Step 1: Reverse the original list

   Original: 10 -> 20 -> 30 -> 40 -> null reversed becomes 40 -> 30 -> 20 -> 10 -> null.

2. Step 2: Insert 25 between 30 and 20

   In reversed list, 30 points to 20, so insert 25 after 30: 30 -> 25 -> 20.

3. Step 3: Final list order

   40 -> 30 -> 25 -> 20 -> 10 -> null.

4. Final Answer:

   40 -> 30 -> 25 -> 20 -> 10 -> null -> Option A

5. Quick Check:

   Insert between 30 and 20 in reversed list = 40->30->25->20->10 [OK]

Hint: Reverse first, then insert at correct reversed position [OK]

Common Mistakes:

• Inserting 25 between wrong nodes after reversal
• Confusing original and reversed order
• Not updating next pointers correctly