Pop Using Linked List Node in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

def pop_head(head):
    if head is None:
        return None
    new_head = head.[1]
    return new_head

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

Aprev

Bnext

Cvalue

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' to get the new head.

Trying to access 'value' or 'data' instead of the next node.

To remove the first node, we return the next node as the new head.

2fill in blank

medium

Complete the code to pop the last node from a singly linked list.

DSA Python

def pop_tail(head):
    if head is None or head.next is None:
        return None
    current = head
    while current.[1].next is not None:
        current = current.next
    current.next = None
    return head

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

Aprev

Bdata

Cvalue

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

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

Trying to access 'value' or 'data' instead of the next node.

We traverse using the 'next' pointer to find the second last node.

3fill in blank

hard

Fix the error in the code to pop the head node and return its value.

DSA Python

def pop_head_value(head):
    if head is None:
        return None, None
    value = head.[1]
    head = head.next
    return value, head

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

Avalue

Bnext

Cdata

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'value' which is not defined in the node.

Using 'next' or 'prev' which are pointers, not data.

The node's data is stored in the 'data' attribute, not 'value'.

4fill in blank

hard

Fill both blanks to pop the last node and return the updated list and popped value.

DSA Python

def pop_tail_value(head):
    if head is None:
        return None, None
    if head.next is None:
        return None, head.data
    current = head
    while current.[1].next is not None:
        current = current.next
    value = current.[2].data
    current.next = None
    return head, value

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

Anext

Bprev

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

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

Trying to access 'data' directly from current instead of current.next.

We use 'next' to traverse and access the last node's 'data'.

5fill in blank

hard

Fill all three blanks to pop the head node, return its value, and the updated list.

DSA Python

def pop_head_return(head):
    if head is None:
        return None, None
    value = head.[1]
    new_head = head.[2]
    head.[3] = None
    return value, new_head

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

Adata

Bnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is not present in singly linked list nodes.

Not disconnecting the old head's next pointer.

We get the data, update the head to next, and disconnect the old head's next.

Practice

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

easy

A. Removes the last node and returns its value

B. Adds a new node at the beginning

C. Searches for a node with a specific value

D. Counts the total number of nodes

Solution

1. Step 1: Understand pop operation in linked list

   Pop means removing the last element from the list and returning its value.

2. Step 2: Identify pop behavior with nodes

   In a linked list, pop removes the tail node and returns its data.

3. Final Answer:

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

4. Quick Check:

   Pop = Remove last node [OK]

Hint: Pop always removes the last node in a linked list [OK]

Common Mistakes:

• Confusing pop with push (adding nodes)
• Thinking pop removes the first node
• Assuming pop searches for a value

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

easy

A. def pop(self):\n if not self.head:\n return None\n current = self.head\n while current.next.next:\n current = current.next\n popped = current.next\n current.next = None\n return popped.data

B. def pop(self):\n popped = self.head\n self.head = self.head.next\n return popped.data

C. def pop(self):\n return self.head.data

D. def pop(self):\n self.head = None

Solution

1. Step 1: Check pop logic for singly linked list

   Pop removes the last node, so we must traverse to the second last node.

2. Step 2: Verify syntax correctness

   def pop(self):\n if not self.head:\n return None\n current = self.head\n while current.next.next:\n current = current.next\n popped = current.next\n current.next = None\n return popped.data correctly traverses until current.next.next is None, then removes last node and returns its data.

3. Final Answer:

   Code that traverses to second last node -> Option A

4. Quick Check:

   Traverse to second last node, remove last [OK]

Hint: Pop needs traversal to second last node before removal [OK]

Common Mistakes:

• Removing head instead of last node
• Not checking for empty list
• Returning node instead of node data

3. What will be the output after popping a node from the linked list 1 -> 2 -> 3 -> null using the pop method below?

def pop(self):
    if not self.head:
        return None
    if not self.head.next:
        val = self.head.data
        self.head = None
        return val
    current = self.head
    while current.next.next:
        current = current.next
    val = current.next.data
    current.next = None
    return val

medium

A. 2 -> 3 -> null

B. 1 -> 3 -> null

C. 1 -> 2 -> null

D. 3 -> null

Solution

1. Step 1: Identify the node popped

   Pop removes the last node, which is 3 in the list 1 -> 2 -> 3 -> null.

2. Step 2: Determine list after pop

   After removing 3, the list becomes 1 -> 2 -> null.

3. Final Answer:

   1 -> 2 -> null -> Option C

4. Quick Check:

   Last node removed, list ends at 2 [OK]

Hint: Pop removes last node, list shortens by one [OK]

Common Mistakes:

• Removing first node instead of last
• Not updating next pointer to None
• Returning wrong list after pop

4. Find the error in this pop method for a singly linked list with at least two nodes:

def pop(self):
    current = self.head
    while current.next:
        current = current.next
    val = current.data
    current = None
    return val

medium

A. It does not handle empty list case

B. It does not update the previous node's next pointer

C. It returns None instead of node data

D. It removes the head instead of last node

Solution

1. Step 1: Analyze traversal and removal

   The code moves current to last node but does not track previous node.

2. Step 2: Check pointer update

   Setting current = None only changes local variable, does not update previous node's next pointer to None, so last node remains linked.

3. Final Answer:

   It does not update the previous node's next pointer -> Option B

4. Quick Check:

   Must update previous.next to None to remove last node [OK]

Hint: Always update previous node's next to None when popping last node [OK]

Common Mistakes:

• Not handling empty list
• Not tracking previous node
• Setting local variable to None instead of pointer

5. Given a linked list with nodes 10 -> 20 -> 30 -> 40 -> null, write the sequence of node values after two consecutive pop operations using a correct pop method that removes the last node each time.

hard

A. 10 -> 30 -> null

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

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

D. 10 -> 20 -> null

Solution

1. Step 1: First pop removes last node 40

   List changes from 10 -> 20 -> 30 -> 40 -> null to 10 -> 20 -> 30 -> null.

2. Step 2: Second pop removes new last node 30

   List changes from 10 -> 20 -> 30 -> null to 10 -> 20 -> null.

3. Final Answer:

   10 -> 20 -> null -> Option D

4. Quick Check:

   Two pops remove last two nodes [OK]

Hint: Each pop removes last node, shorten list stepwise [OK]

Common Mistakes:

• Removing nodes from start instead of end
• Not updating list after first pop
• Confusing node values after pops