Pop Using Linked List Node in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after popping the last node from the linked list?

Given the linked list and the pop operation code below, what is the printed linked list after popping the last node?

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def append(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        last = self.head
        while last.next:
            last = last.next
        last.next = new_node

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

    def print_list(self):
        current = self.head
        result = []
        while current:
            result.append(str(current.data))
            current = current.next
        if result:
            print(' -> '.join(result) + ' -> null')
        else:
            print('null')

ll = LinkedList()
ll.append(10)
ll.append(20)
ll.append(30)
ll.pop()
ll.print_list()

A 10 -> 20 -> null

B 10 -> 20 -> 30 -> null

C 20 -> 30 -> null

D 10 -> null

💡 Hint

Think about what happens to the last node when you pop it from the list.

The pop method removes the last node (30). After popping, the list contains 10 and 20 nodes only.

❓ Predict Output

intermediate

What is the output after popping from a single-node linked list?

Consider the linked list with one node and the pop operation below. What is printed after popping the only node?

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def append(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        last = self.head
        while last.next:
            last = last.next
        last.next = new_node

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

    def print_list(self):
        current = self.head
        result = []
        while current:
            result.append(str(current.data))
            current = current.next
        if result:
            print(' -> '.join(result) + ' -> null')
        else:
            print('null')

ll = LinkedList()
ll.append(100)
ll.pop()
ll.print_list()

A 100

B 100 -> null

C Error: Cannot pop from empty list

D null

💡 Hint

What happens to the head when you pop the only node?

Popping the only node sets the head to None, so printing the list shows 'null' with no nodes.

❓ Predict Output

advanced

What is the output after popping twice from the linked list?

Given the linked list and pop method below, what is the printed linked list after popping twice?

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def append(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        last = self.head
        while last.next:
            last = last.next
        last.next = new_node

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

    def print_list(self):
        current = self.head
        result = []
        while current:
            result.append(str(current.data))
            current = current.next
        if result:
            print(' -> '.join(result) + ' -> null')
        else:
            print('null')

ll = LinkedList()
ll.append(1)
ll.append(2)
ll.append(3)
ll.append(4)
ll.pop()
ll.pop()
ll.print_list()

A 1 -> 2 -> 3 -> null

B 1 -> 2 -> null

C 2 -> 3 -> 4 -> null

D 1 -> 3 -> null

💡 Hint

Each pop removes the last node. Think about which nodes remain after two pops.

Popping twice removes nodes 4 and 3. Remaining nodes are 1 and 2.

🔧 Debug

advanced

Why does this pop method cause an error on a single-node list?

Examine the pop method below. It causes an error when popping from a list with only one node. What is the cause?

DSA Python

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

A It causes infinite loop when the list has two nodes.

B It returns None instead of the popped value when the list has two nodes.

C It does not handle the case when the list has only one node, causing AttributeError.

D It does not handle the case when the list has exactly two nodes, causing AttributeError.

💡 Hint

Check what happens when current.next.next is accessed on a single-node list.

When the list has two nodes, current is head. current.next.next is None, so the while loop condition fails immediately. But if the list has only one node, current.next is None, so current.next.next raises AttributeError. The code misses the single-node case.

🧠 Conceptual

expert

What is the time complexity of popping the last node in a singly linked list without tail pointer?

Consider a singly linked list with no tail pointer. What is the time complexity of the pop operation that removes the last node?

A O(n) because we must traverse the list to find the second last node

B O(1) because we can directly access the last node

C O(log n) because of binary search on nodes

D O(n^2) because of nested traversal

💡 Hint

Think about how you find the node before the last in a singly linked list.

Without a tail pointer, to pop the last node, you must traverse from head to find the second last node, which takes O(n) time.

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