Why Pop Using Linked List Node in DSA Typescript?

The Big Idea

What if you could remove the top item instantly without any searching or mistakes?

The Scenario

Imagine you have a stack of books on your desk. To remove the top book, you have to pick it up carefully without disturbing the rest. Now, imagine if you had to remove the top book but you only had a messy pile without any order or way to find the top quickly.

The Problem

Without a proper way to track the top book, you might have to search through the entire pile to find it. This is slow and can cause mistakes like dropping books or removing the wrong one. Manually managing this pile is confusing and error-prone.

The Solution

Using a linked list node to pop means you always know where the top book is. You can quickly remove it by updating a simple pointer. This keeps the stack organized and makes removing the top item fast and safe.

Before vs After

✗ Before

let books = ['book1', 'book2', 'book3'];
// To remove top, search and remove last manually
books.splice(books.length - 1, 1);

✓ After

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

// Pop by moving top pointer
let top: Node | null = new Node('book3');
top = top.next;

What It Enables

This lets you remove the top item instantly without searching, making your program faster and simpler.

Real Life Example

When you undo typing in a text editor, the program pops the last action from a stack quickly using linked list nodes.

Key Takeaways

Manual removal is slow and error-prone.

Linked list nodes keep track of the top item clearly.

Pop operation becomes fast and safe.

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