Pop Using Linked List Node in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to remove the first node from the linked list.

DSA Typescript

function pop(head: ListNode | null): ListNode | null {
  if (head === null) return null;
  return head[1];
}

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

Ahead

Bnext

Cvalue

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Returning head.prev instead of head.next.

Returning head.value instead of the next node.

To pop the first node, we return the next node after head, which is head.next.

2fill in blank

medium

Complete the code to update the head pointer after popping the first node.

DSA Typescript

function pop(head: ListNode | null): ListNode | null {
  if (head === null) return null;
  const newHead = head[1];
  return newHead;
}

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

Anext

Bprev

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using head.prev instead of head.next.

Returning the old head instead of the new head.

After popping, the new head is the node after the current head, which is head.next.

3fill in blank

hard

Fix the error in the pop function to correctly remove the first node and avoid memory leaks.

DSA Typescript

function pop(head: ListNode | null): ListNode | null {
  if (head === null) return null;
  const newHead = head.next;
  head[1] = null;
  return newHead;
}

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

Anext

Bprev

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting head.prev to null instead of head.next.

Not disconnecting the popped node, causing memory leaks.

Setting head.next to null disconnects the popped node from the list, helping garbage collection.

4fill in blank

hard

Fill both blanks to correctly pop the first node and update the new head's previous pointer.

DSA Typescript

function pop(head: ListNode | null): ListNode | null {
  if (head === null) return null;
  const newHead = head[1];
  if (newHead !== null) {
    newHead[2] = null;
  }
  head.next = null;
  return newHead;
}

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

Anext

Bprev

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting newHead.next to null instead of newHead.prev.

Not updating newHead.prev, causing incorrect list links.

After popping, newHead is head.next, and its prev pointer should be set to null to mark it as the new head.

5fill in blank

hard

Fill all three blanks to pop the first node, update the new head's previous pointer, and return the new head.

DSA Typescript

function pop(head: ListNode | null): ListNode | null {
  if (head === null) return null;
  const newHead = head[1];
  if (newHead !== null) {
    newHead[2] = null;
  }
  head[3] = null;
  return newHead;
}

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

Anext

Bprev

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using head.prev or head.value instead of head.next.

Not disconnecting the old head's next pointer.

We get newHead as head.next, set newHead.prev to null, and disconnect old head's next pointer by setting head.next to null.

Practice

1. What does the pop() operation do in a linked list when implemented using nodes?

easy

A. Removes the last node and returns its value

B. Adds a new node at the beginning

C. Removes the first node and returns its value

D. Searches for a node with a specific value

Solution

1. Step 1: Understand pop operation in linked list context

   Pop means removing the last element from the list, similar to popping from a stack.

2. Step 2: Identify what pop returns

   Pop removes the last node and returns its stored value.

3. Final Answer:

   Removes the last node and returns its value -> Option A

4. Quick Check:

   Pop = remove last node [OK]

Hint: Pop removes last node and returns its value [OK]

Common Mistakes:

• Confusing pop with push operation
• Thinking pop removes the first node
• Assuming pop searches for a value

2. Which of the following is the correct TypeScript syntax to define a node class for a singly linked list?

easy

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

B. class Node { val: number; next: Node; constructor(val: number) { this.val = val; this.next = null; } }

C. class Node { value: number; next: Node; constructor(value: number) { this.value = value; this.next = undefined; } }

D. class Node { value: number; next: Node | undefined; constructor(value: number) { this.value = value; this.next = null; } }

Solution

1. Step 1: Check property types for linked list node

   Node should have a value and a next pointer which can be null if no next node.

2. Step 2: Verify constructor initialization

   Constructor sets value and initializes next to null (no next node initially).

3. Final Answer:

   class Node { value: number; next: Node | null; constructor(value: number) { this.value = value; this.next = null; } } -> Option A

4. Quick Check:

   Node next can be null [OK]

Hint: Next pointer must allow null for list end [OK]

Common Mistakes:

• Not allowing next to be null
• Using undefined instead of null
• Mismatching property names

3. Given the following linked list code, what will be the printed output after calling pop() once?

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

class LinkedList {
  head: Node | null = null;

  append(value: number) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }

  pop() {
    if (!this.head) return null;
    if (!this.head.next) {
      const val = this.head.value;
      this.head = null;
      return val;
    }
    let current = this.head;
    while (current.next && current.next.next) {
      current = current.next;
    }
    const val = current.next!.value;
    current.next = null;
    return val;
  }

  print() {
    let current = this.head;
    let result = '';
    while (current) {
      result += current.value + ' -> ';
      current = current.next;
    }
    result += 'null';
    console.log(result);
  }
}

const list = new LinkedList();
list.append(10);
list.append(20);
list.append(30);
list.pop();
list.print();

medium

A. null

B. 10 -> 20 -> null

C. 20 -> 30 -> null

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

Solution

1. Step 1: Understand list before pop

   List after appending 10, 20, 30 is: 10 -> 20 -> 30 -> null.

2. Step 2: Trace pop operation

   Pop removes last node (30), so list becomes 10 -> 20 -> null.

3. Final Answer:

   10 -> 20 -> null -> Option B

4. Quick Check:

   Pop removes last node [OK]

Hint: Pop removes last node, print shows remaining [OK]

Common Mistakes:

• Assuming pop removes first node
• Not updating next pointer to null
• Printing old list without pop effect

4. Identify the bug in the following pop() method for a singly linked list and select the correct fix.

pop() {
  if (!this.head) return null;
  let current = this.head;
  while (current.next) {
    current = current.next;
  }
  const val = current.value;
  current = null;
  return val;
}

medium

A. Return current.next.value instead of current.value

B. Initialize current as null before the loop

C. The loop should stop at the second last node; set current.next = null to remove last node

D. No bug; code works correctly

Solution

1. Step 1: Analyze loop behavior

   The loop moves current to the last node, but we need to stop at second last to remove last node.

2. Step 2: Fix removal of last node

   Set current.next = null at second last node to remove last node properly.

3. Final Answer:

   The loop should stop at the second last node; set current.next = null to remove last node -> Option C

4. Quick Check:

   Stop at second last node to remove last [OK]

Hint: Stop at second last node to remove last [OK]

Common Mistakes:

• Setting current = null does not remove node
• Loop goes to last node instead of second last
• Returning wrong node value

5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> 35 -> null. You want to implement a pop() method that removes the last node and returns its value. However, if the list has only one node, pop() should remove it and set the list to empty. What will be the state of the list after calling pop() twice?
Choose the correct printed output after these operations:

const list = new LinkedList();
list.append(5);
list.append(15);
list.append(25);
list.append(35);
list.pop();
list.pop();
list.print();

hard

A. null

B. 5 -> 15 -> 25 -> null

C. 25 -> 35 -> null

D. 5 -> 15 -> null

Solution

1. Step 1: Initial list and first pop

   List starts as 5 -> 15 -> 25 -> 35 -> null. First pop removes 35, list becomes 5 -> 15 -> 25 -> null.

2. Step 2: Second pop removes last node 25

   After second pop, list is 5 -> 15 -> null.

3. Final Answer:

   5 -> 15 -> null -> Option D

4. Quick Check:

   Two pops remove last two nodes [OK]

Hint: Each pop removes last node, so two pops remove last two nodes [OK]

Common Mistakes:

• Removing nodes from front instead of end
• Not updating list head when one node left
• Assuming pop removes all nodes