Pop Using Linked List Node in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to remove the top node from the stack and return its value.

DSA Java

public int pop() {
    if (top == null) return -1;
    int value = top.data;
    top = top.[1];
    return value;
}

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

Aprev

Bnext

Cchild

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes errors because the stack is singly linked.

Returning the wrong node's data.

The 'next' pointer of the top node points to the next node in the stack. Updating top to top.next removes the current top.

2fill in blank

medium

Complete the code to check if the stack is empty before popping.

DSA Java

public int pop() {
    if (top == [1]) return -1;
    int value = top.data;
    top = top.next;
    return value;
}

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

Afalse

B0

Cempty

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Checking top == 0 is incorrect because top is a node, not a number.

Using 'empty' or 'false' which are not valid node references.

The stack is empty if top is null, so we check if top == null before popping.

3fill in blank

hard

Fix the error in the pop method to correctly update the top pointer.

DSA Java

public int pop() {
    if (top == null) return -1;
    int value = top.data;
    top = [1];
    return value;
}

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

Atop.next

Btop.parent

Ctop.prev

Dtop.child

Attempts:

3 left

💡 Hint

Common Mistakes

Using top.prev causes null pointer exceptions because it's singly linked.

Using unrelated pointers like child or parent.

The top pointer should move to the next node in the stack, which is top.next.

4fill in blank

hard

Fill both blanks to pop the top node and update the stack size.

DSA Java

public int pop() {
    if (top == null) return -1;
    int value = top.data;
    top = top.[1];
    size[2];
    return value;
}

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

Anext

B++

C--

D+=

Attempts:

3 left

💡 Hint

Common Mistakes

Increasing size after popping is incorrect.

Using wrong pointer names like prev.

We update top to top.next to remove the node, and decrease size by 1 using size--.

5fill in blank

hard

Fill all three blanks to pop the top node, update size, and handle empty stack.

DSA Java

public int pop() {
    if (top == [1]) return -1;
    int value = top.[2];
    top = top.[3];
    size--;
    return value;
}

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

Anull

Bdata

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which does not exist in singly linked list.

Not checking for null before popping.

Check if top is null to detect empty stack, get value from top.data, and update top to top.next.

Practice

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

easy

A. Removes and returns the first node of the linked list

B. Adds a new node at the end of the linked list

C. Removes the last node without returning it

D. Returns the size of the linked list

Solution

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

   The pop operation removes the top element, which is the first node in a linked list acting like a stack.

2. Step 2: Identify what is returned

   Pop returns the value of the removed first node and updates the head pointer to the next node.

3. Final Answer:

   Removes and returns the first node of the linked list -> Option A

4. Quick Check:

   Pop = Remove first node [OK]

Hint: Pop removes the head node and returns it [OK]

Common Mistakes:

• Thinking pop removes the last node
• Confusing pop with push operation
• Assuming pop returns list size

2. Which of the following is the correct Java syntax to pop a node from the head of a singly linked list?

easy

A. head = head.next; return head.data;

B. Node temp = head; head = head.next; return temp.data;

C. Node temp = head.next; head = temp; return head.data;

D. return head.next.data; head = head.next;

Solution

1. Step 1: Check correct order of operations

   We must save the current head node in a temp variable before moving head to next.

2. Step 2: Return the data of the removed node

   Return the data from the saved temp node, not the new head.

3. Final Answer:

   Node temp = head; head = head.next; return temp.data; -> Option B

4. Quick Check:

   Save head, move head, return old data [OK]

Hint: Save head before moving it to next [OK]

Common Mistakes:

• Returning data from new head instead of old
• Not saving old head before moving
• Returning data before updating head

3. Given the linked list nodes: 10 -> 20 -> 30 -> null, what is the list after one pop operation using this code?

Node pop() {
  if (head == null) return null;
  Node temp = head;
  head = head.next;
  return temp;
}

medium

A. 30 -> null

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

C. 20 -> 30 -> null

D. null

Solution

1. Step 1: Identify the node removed by pop()

   Pop removes the head node which has value 10.

2. Step 2: Update head to next node

   Head moves from 10 to 20, so list becomes 20 -> 30 -> null.

3. Final Answer:

   20 -> 30 -> null -> Option C

4. Quick Check:

   Pop removes first node 10 [OK]

Hint: Pop removes first node, head moves next [OK]

Common Mistakes:

• Assuming pop removes last node
• Not updating head pointer
• Returning old list unchanged

4. What is the error in this pop method for a singly linked list?

Node pop() {
  if (head == null) return null;
  Node temp = head.next;
  head = temp;
  return temp;
}

medium

A. It skips the first node and removes the second node instead

B. It causes a null pointer exception when list is empty

C. It returns the old head node instead of new head

D. It does not update the head pointer

Solution

1. Step 1: Analyze which node is assigned to temp

   Temp is assigned head.next, so the first node is skipped.

2. Step 2: Understand effect on head and return

   Head moves to second node, so first node is never removed; second node is removed instead.

3. Final Answer:

   It skips the first node and removes the second node instead -> Option A

4. Quick Check:

   Pop must remove first node, not second [OK]

Hint: Pop must save head, not head.next [OK]

Common Mistakes:

• Assigning temp to head.next instead of head
• Returning the new head instead of the old head
• Returning wrong node

5. You have a linked list: 5 -> 15 -> 25 -> 35 -> null. You perform two pop operations using this method:

Node pop() {
  if (head == null) return null;
  Node temp = head;
  head = head.next;
  return temp;
}

What is the state of the list after these two pops?

hard

A. null

B. 15 -> 25 -> 35 -> null

C. 35 -> null

D. 25 -> 35 -> null

Solution

1. Step 1: First pop removes 5

   After first pop, head moves from 5 to 15, list is 15 -> 25 -> 35 -> null.

2. Step 2: Second pop removes 15

   After second pop, head moves from 15 to 25, list is 25 -> 35 -> null.

3. Final Answer:

   25 -> 35 -> null -> Option D

4. Quick Check:

   Two pops remove first two nodes [OK]

Hint: Each pop removes head, moves head next [OK]

Common Mistakes:

• Removing last nodes instead of first
• Not updating head after each pop
• Confusing pop with push