Creating a Singly Linked List from Scratch in DSA Python - Debugging & Practice

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Inserting Nodes in a Singly Linked List

What is the printed state of the linked list after inserting nodes with values 10, 20, and 30 in that order?

DSA Python

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

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

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

    def print_list(self):
        current = self.head
        while current:
            print(current.data, end=' -> ' if current.next else ' -> null\n')
            current = current.next

sll = SinglyLinkedList()
sll.insert_at_end(10)
sll.insert_at_end(20)
sll.insert_at_end(30)
sll.print_list()

A 30 -> 20 -> 10 -> null

B 10 -> 20 -> 30 -> null

C 10 -> 30 -> 20 -> null

D null

💡 Hint

Think about how nodes are added at the end and how the list is printed from head to tail.

Nodes are inserted at the end, so the order is the same as insertion: 10, then 20, then 30. The print_list method prints from head to tail, showing 10 -> 20 -> 30 -> null.

🧠 Conceptual

intermediate

Understanding the Head Pointer in a Singly Linked List

What does the 'head' pointer in a singly linked list represent?

A It points to the last node in the list.

B It points to the middle node in the list.

C It points to the first node in the list.

D It points to a random node in the list.

💡 Hint

Think about where the list starts when you want to traverse it.

The head pointer always points to the first node of the list, which is the starting point for traversing the list.

🔧 Debug

advanced

Identify the Error in Node Insertion Code

What error will occur when running this code to insert a node at the end of the list?

DSA Python

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

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

    def insert_at_end(self, data):
        new_node = Node(data)
        if self.head is None:
            self.head = new_node
        else:
            current = self.head
            while current.next:
                current = current.next
            current.next = new_node

sll = SinglyLinkedList()
sll.insert_at_end(5)
sll.insert_at_end(15)
sll.insert_at_end(25)
print(sll.head.next.next.next.data)

A AttributeError: 'NoneType' object has no attribute 'data'

B No error, prints 25

C IndexError: list index out of range

D TypeError: 'int' object is not callable

💡 Hint

Check how many nodes are in the list and what happens when accessing next pointers.

The list has three nodes, so sll.head.next.next.next is None. Trying to access .data on None causes AttributeError.

❓ Predict Output

advanced

Output After Deleting a Node from Singly Linked List

What is the printed state of the linked list after deleting the node with value 20?

DSA Python

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

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

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

    def delete_node(self, key):
        current = self.head
        prev = None
        while current and current.data != key:
            prev = current
            current = current.next
        if not current:
            return
        if prev is None:
            self.head = current.next
        else:
            prev.next = current.next

    def print_list(self):
        current = self.head
        while current:
            print(current.data, end=' -> ' if current.next else ' -> null\n')
            current = current.next

sll = SinglyLinkedList()
sll.insert_at_end(10)
sll.insert_at_end(20)
sll.insert_at_end(30)
sll.delete_node(20)
sll.print_list()

A 30 -> 10 -> null

B 10 -> 20 -> 30 -> null

C 20 -> 30 -> null

D 10 -> 30 -> null

💡 Hint

Deleting a node removes it from the chain, so it won't appear in the printed list.

The node with value 20 is removed, so the list contains 10 followed by 30.

🧠 Conceptual

expert

Time Complexity of Searching in a Singly Linked List

What is the time complexity of searching for a value in a singly linked list with n nodes?

A O(n)

B O(1)

C O(n log n)

D O(log n)

💡 Hint

You may need to check each node one by one in the worst case.

Searching requires checking nodes one by one until the value is found or list ends, so it takes linear time O(n).