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

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after pushing nodes in this linked list?

Consider the following Python code that pushes nodes to the front of a linked list. What will be the printed linked list after all push operations?

DSA Python

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

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

    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    def printList(self):
        temp = self.head
        while temp:
            print(temp.data, end=' -> ')
            temp = temp.next
        print('null')

llist = LinkedList()
llist.push(10)
llist.push(20)
llist.push(30)
llist.printList()

A 30 -> 10 -> 20 -> null

B 10 -> 20 -> 30 -> null

C null -> 30 -> 20 -> 10

D 30 -> 20 -> 10 -> null

💡 Hint

Remember, push adds a new node at the front of the list.

Each push adds a new node at the beginning. So the last pushed node (30) becomes the head, followed by 20, then 10.

❓ Predict Output

intermediate

What is the linked list after these push operations?

Given this code snippet, what will be the final linked list printed?

DSA Python

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

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

    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    def printList(self):
        temp = self.head
        while temp:
            print(temp.data, end=' -> ')
            temp = temp.next
        print('null')

llist = LinkedList()
llist.push(5)
llist.push(15)
llist.push(25)
llist.push(35)
llist.printList()

A 5 -> 15 -> 25 -> 35 -> null

B 35 -> 25 -> 15 -> 5 -> null

C null -> 5 -> 15 -> 25 -> 35

D 15 -> 5 -> 25 -> 35 -> null

💡 Hint

New nodes are added at the front, so the last pushed node is the head.

Push adds nodes at the front, so the list order is reversed from the push order.

🔧 Debug

advanced

Why does this push method cause an error?

Look at this push method in a linked list class. It causes an error when called. What is the cause?

DSA Python

def push(self, new_data):
    new_node = Node(new_data)
    new_node.next = self.head.next
    self.head = new_node

A self.head might be None, so self.head.next causes AttributeError

B The method does not assign new_node.data correctly

C new_node.next should be None, not self.head.next

D self.head is not updated properly to new_node

💡 Hint

Think about what happens if the list is empty before push.

If the list is empty, self.head is None. Accessing self.head.next causes an error because None has no attribute 'next'.

❓ Predict Output

advanced

What is the linked list after pushing nodes with these values?

Given the following code, what will be the printed linked list?

DSA Python

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

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

    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    def printList(self):
        temp = self.head
        while temp:
            print(temp.data, end=' -> ')
            temp = temp.next
        print('null')

llist = LinkedList()
for value in [1, 2, 3, 4, 5]:
    llist.push(value)
llist.printList()

A 5 -> 4 -> 3 -> 2 -> 1 -> null

B 1 -> 2 -> 3 -> 4 -> 5 -> null

C null -> 1 -> 2 -> 3 -> 4 -> 5

D 5 -> 3 -> 1 -> 2 -> 4 -> null

💡 Hint

Push adds each new node at the front, reversing the order.

The last pushed value (5) becomes the head, so the list is reversed from the input order.

🧠 Conceptual

expert

What is the time complexity of pushing a node at the front of a linked list?

Consider a singly linked list. What is the time complexity of the push operation that adds a node at the front?

A O(n^2) - Quadratic time

B O(n) - Linear time

C O(1) - Constant time

D O(log n) - Logarithmic time

💡 Hint

Think about how many nodes you need to visit to add at the front.

Adding a node at the front only changes the head pointer and the new node's next pointer, which takes constant time.

Practice

1. What does the push operation do in a linked list when using a node?

easy

A. Removes the last node from the linked list

B. Searches for a node with a specific value

C. Deletes the first node of the linked list

D. Adds a new node at the beginning of the linked list

Solution

1. Step 1: Understand push operation meaning

   Push means adding a new element to the start of the linked list.

2. Step 2: Identify push effect on linked list

   The new node becomes the head, pointing to the old head node.

3. Final Answer:

   Adds a new node at the beginning of the linked list -> Option D

4. Quick Check:

   Push = Add at start [OK]

Hint: Push always adds node at the start of the list [OK]

Common Mistakes:

• Thinking push adds at the end
• Confusing push with pop (removal)
• Assuming push deletes nodes

2. Which of the following is the correct syntax to create a new node and push it to the front of a linked list in Python?

easy

A. head.next = Node(data); head = head.next

B. head = Node(data); head.next = new_node

C. new_node = Node(data); new_node.next = head; head = new_node

D. new_node = Node(data); head = new_node.next

Solution

1. Step 1: Create new node with data

   Use new_node = Node(data) to create a node.

2. Step 2: Link new node to current head and update head

   Set new_node.next = head and then head = new_node.

3. Final Answer:

   new_node = Node(data); new_node.next = head; head = new_node -> Option C

4. Quick Check:

   Correct push syntax = new_node = Node(data); new_node.next = head; head = new_node [OK]

Hint: New node points to head, then head updates to new node [OK]

Common Mistakes:

• Assigning head before linking new_node.next
• Confusing order of assignments
• Using undefined variables

3. What will be the linked list after executing the following code?

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

def push(head, data):
    new_node = Node(data)
    new_node.next = head
    head = new_node
    return head

head = None
head = push(head, 10)
head = push(head, 20)
head = push(head, 30)

medium

A. 30 -> 20 -> 10 -> null

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

C. 30 -> 10 -> 20 -> null

D. 20 -> 30 -> 10 -> null

Solution

1. Step 1: Push 10 to empty list

   List becomes 10 -> null.

2. Step 2: Push 20 to front

   New head is 20, pointing to 10: 20 -> 10 -> null.

3. Step 3: Push 30 to front

   New head is 30, pointing to 20: 30 -> 20 -> 10 -> null.

4. Final Answer:

   30 -> 20 -> 10 -> null -> Option A

5. Quick Check:

   Push adds at front, last pushed is first [OK]

Hint: Last pushed node is always the head [OK]

Common Mistakes:

• Assuming push adds at end
• Not updating head after push
• Mixing order of nodes

4. Find the error in this push function for a linked list:

def push(head, data):
    new_node = Node(data)
    head = new_node
    new_node.next = head
    return head

medium

A. new_node is not created properly

B. new_node.next is set after head is updated, causing a loop

C. head is not returned

D. Function does not accept head as parameter

Solution

1. Step 1: Analyze assignment order

   head is set to new_node before linking new_node.next.

2. Step 2: Identify pointer mistake

   new_node.next = head sets next to itself, creating a loop.

3. Final Answer:

   new_node.next is set after head is updated, causing a loop -> Option B

4. Quick Check:

   Link new_node.next before updating head [OK]

Hint: Set new_node.next before changing head [OK]

Common Mistakes:

• Assigning head before linking next
• Creating self-referential loops
• Forgetting to return head

5. Given a linked list with nodes 5 -> 15 -> 25 -> null, what will be the list after pushing nodes with values 35 and 45 in order using the push function?

head = Node(5)
head.next = Node(15)
head.next.next = Node(25)
head = push(head, 35)
head = push(head, 45)

hard

A. 45 -> 35 -> 5 -> 15 -> 25 -> null

B. 5 -> 15 -> 25 -> 35 -> 45 -> null

C. 35 -> 45 -> 5 -> 15 -> 25 -> null

D. 45 -> 35 -> 25 -> 15 -> 5 -> null

Solution

1. Step 1: Original list is 5 -> 15 -> 25 -> null

   Head points to 5, which points to 15, then 25.

2. Step 2: Push 35

   New node 35 points to head (5), head updates to 35: 35 -> 5 -> 15 -> 25 -> null.

3. Step 3: Push 45

   New node 45 points to head (35), head updates to 45: 45 -> 35 -> 5 -> 15 -> 25 -> null.

4. Final Answer:

   45 -> 35 -> 5 -> 15 -> 25 -> null -> Option A

5. Quick Check:

   Push adds nodes at front in order [OK]

Hint: New nodes become new head, pushing old nodes back [OK]

Common Mistakes:

• Appending instead of pushing at front
• Mixing node order after pushes
• Not updating head after push