Pop Using Linked List Node in DSA Typescript - Execution Trace

Concept Flow - Pop Using Linked List Node

Start with head node

↓

Check if list is empty?

↓

Return null

↓

Move head to next node

↓

Disconnect popped node

↓

Return popped node's value

This flow shows how to remove the first node from a linked list by updating the head pointer and returning the removed node's value.

Execution Sample

DSA Typescript

class Node {
  constructor(public value: number, public next: Node | null = null) {}
}

let head = new Node(10, new Node(20, new Node(30)));

// Pop operation
function pop(): number | null {
  if (!head) return null;
  const popped = head;
  head = head.next;
  popped.next = null;
  return popped.value;
}

This code removes the first node from the linked list and returns its value.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Start with list	10 -> 20 -> 30 -> null	head -> Node(10)	head -> 10 -> 20 -> 30 -> null
2	Check if list is empty	10 -> 20 -> 30 -> null	head not null	head -> 10 -> 20 -> 30 -> null
3	Store head node to pop	10 -> 20 -> 30 -> null	popped -> Node(10)	head -> 10 -> 20 -> 30 -> null
4	Move head to next node	10 -> 20 -> 30 -> null	head -> Node(20)	head -> 20 -> 30 -> null
5	Disconnect popped node	10 (popped) 20 -> 30 -> null	popped.next = null	popped: 10 -> null, head: 20 -> 30 -> null
6	Return popped node's value	20 -> 30 -> null	head unchanged	head -> 20 -> 30 -> null
7	End	20 -> 30 -> null	head -> Node(20)	head -> 20 -> 30 -> null

💡 Pop operation ends after removing the first node and updating head pointer.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	Final
head	Node(10)	Node(10)	Node(20)	Node(20)	Node(20)
popped	null	Node(10)	Node(10)	Node(10)	Node(10)
popped.next	undefined	Node(20)	Node(20)	null	null

Key Moments - 3 Insights

Why do we set popped.next = null after moving head?

Setting popped.next = null disconnects the popped node from the list, preventing accidental links. See step 5 in execution_table where this change happens.

What happens if the list is empty when pop is called?

If head is null, pop returns null immediately without changes. This is shown in step 2 where the check for empty list occurs.

Why do we update head to head.next instead of head = null?

Because we want to remove only the first node and keep the rest of the list intact. Updating head to head.next moves the start pointer forward. See step 4 in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of head after step 4?

A Node with value 20

B Node with value 10

C null

D Node with value 30

💡 Hint

Check the 'Pointer Changes' column at step 4 in execution_table.

At which step is the popped node disconnected from the list?

A Step 2

B Step 3

C Step 5

D Step 6

💡 Hint

Look for 'Disconnect popped node' operation in execution_table.

If the list was empty initially, what would pop() return?

A The value of the first node

B null

C An error

D The last node's value

💡 Hint

Refer to step 2 where the empty list check happens.

Concept Snapshot

Pop Using Linked List Node:
- Check if list is empty (head == null)
- Store current head node as popped
- Move head to next node (head = head.next)
- Disconnect popped node (popped.next = null)
- Return popped node's value
- List shrinks by one node at front

Full Transcript

This visual execution shows how to remove the first node from a linked list. We start with the head node and check if the list is empty. If not empty, we store the head node as the popped node. Then we move the head pointer to the next node, effectively removing the first node from the list. We disconnect the popped node's next pointer to avoid accidental links. Finally, we return the popped node's value. The list now starts from the second node. If the list was empty, pop returns null immediately. This process is shown step-by-step in the execution table and variable tracker.

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