Creating a Singly Linked List from Scratch in DSA Python - Complexity Walkthrough

Time Complexity: Creating a Singly Linked List from Scratch

O(n^2)

Understanding Time Complexity

When we create a singly linked list from scratch, we add nodes one by one. Understanding how the time needed grows as we add more nodes helps us write better code.

We want to know: How does the time to build the list change as the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

    def append(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

This code creates a singly linked list and adds new nodes at the end one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop inside the append method that moves through the list to find the last node.
• How many times: For each new node added, the loop runs once for each existing node in the list until it reaches the end.

How Execution Grows With Input

As the list grows, adding a new node takes longer because we must walk through all existing nodes to find the end.

Input Size (n)	Approx. Operations
10	About 1 + 2 + ... + 10 = 55 steps
100	About 1 + 2 + ... + 100 = 5,050 steps
1000	About 1 + 2 + ... + 1000 = 500,500 steps

Pattern observation: The total steps grow roughly with the square of the number of nodes, because each append walks through the list.

Final Time Complexity

Time Complexity: O(n2)

This means that adding n nodes one by one like this takes time proportional to n squared, because each new node requires walking through the whole list so far.

Common Mistake

[X] Wrong: "Appending a node always takes the same time, so building the whole list is O(n)."

[OK] Correct: Without keeping track of the last node, each append must walk through the entire list, making later appends slower and total time grow faster than just n.

Interview Connect

Knowing how linked list operations scale helps you explain trade-offs clearly. This skill shows you understand how data structures work under the hood, which is valuable in real coding tasks.

Self-Check

"What if we kept a pointer to the last node to avoid walking the list each time? How would the time complexity change?"

Practice

1. What is the main purpose of a Singly Linked List in data structures?

easy

A. To store elements with keys and values for fast lookup

B. To store elements in a sequence where each element points to the next

C. To store elements in a sorted array for binary search

D. To store elements in a tree structure with parent and child nodes

Solution

1. Step 1: Understand the structure of a singly linked list

   A singly linked list stores elements in nodes where each node points to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike arrays or trees, singly linked lists do not store keys or have parent-child relationships; they only link sequentially.

3. Final Answer:

   To store elements in a sequence where each element points to the next -> Option B

4. Quick Check:

   Singly Linked List = sequence with next pointers [OK]

Hint: Remember: singly linked list nodes point only forward [OK]

Common Mistakes:

• Confusing linked lists with arrays or hash maps
• Thinking nodes have backward pointers (that's doubly linked list)
• Assuming linked lists store key-value pairs

2. Which of the following is the correct way to define a node class for a singly linked list in Python?

easy

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

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

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

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

Solution

1. Step 1: Identify the correct attributes for a singly linked list node

   A singly linked list node stores data and a pointer to the next node only, so it needs data and next.

2. Step 2: Check the constructor parameters and attributes

   class Node: def __init__(self, data): self.data = data self.next = None correctly initializes data and sets next to None by default. Other options include prev which is for doubly linked lists or incorrect parameters.

3. Final Answer:

   class Node: def __init__(self, data): self.data = data self.next = None -> Option C

4. Quick Check:

   Node has data and next only [OK]

Hint: Singly linked list nodes have no 'prev' pointer [OK]

Common Mistakes:

• Adding a 'prev' pointer by mistake
• Not initializing 'next' to None
• Using extra parameters unnecessarily

3. What will be the output of the following code that creates a singly linked list with nodes 10, 20, 30 and prints the list?

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 print_list(self):
        current = self.head
        while current:
            print(current.data, end=' -> ')
            current = current.next
        print('None')

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

medium

A. 10 -> 20 -> None

B. 30 -> 20 -> 10 -> None

C. None

D. 10 -> 20 -> 30 -> None

Solution

1. Step 1: Trace the append method calls

   Appending 10 sets head to node with data 10. Then 20 and 30 are added at the end by traversing the list.

2. Step 2: Trace the print_list method

   Prints nodes in order: 10, then 20, then 30, each followed by ' -> ', ending with 'None'.

3. Final Answer:

   10 -> 20 -> 30 -> None -> Option D

4. Quick Check:

   Append adds at end, print shows order [OK]

Hint: Append adds nodes at list end; print shows order [OK]

Common Mistakes:

• Assuming append adds at front
• Printing without 'None' at end
• Confusing order of nodes

4. The following code tries to append nodes to a singly linked list but does not work as expected. What is the error?

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 self.head is None:
            self.head = new_node
        else:
            last = self.head
            while last.next:
                last = last.next
            last = new_node

ll = LinkedList()
ll.append(1)
ll.append(2)
ll.append(3)

current = ll.head
while current:
    print(current.data, end=' -> ')
    current = current.next
print('None')

medium

A. The line 'last = new_node' does not link the new node to the list

B. The head is never initialized

C. The Node class is missing the 'prev' pointer

D. The while loop condition should be 'while last is not None'

Solution

1. Step 1: Analyze the append method's else block

   The code finds the last node correctly but then does 'last = new_node' which only changes the local variable, not the last node's next pointer.

2. Step 2: Identify the fix

   To link the new node, it should be 'last.next = new_node' to update the last node's next pointer.

3. Final Answer:

   The line 'last = new_node' does not link the new node to the list -> Option A

4. Quick Check:

   Assign to last.next to link nodes [OK]

Hint: Update last.next, not last variable, to link nodes [OK]

Common Mistakes:

• Assigning new_node to last instead of last.next
• Not updating the head when list is empty
• Confusing variable assignment with pointer update

5. You want to create a singly linked list from scratch that supports inserting a new node at the beginning and printing the list. Which code snippet correctly implements this?

hard

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

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

def insert_at_beginning(self, data):
new_node = Node(data)
new_node.next = self.head
self.head = new_node

def print_list(self):
current = self.head
while current:
print(current.data, end=' -> ')
current = current.next
print('None')

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

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

def insert_at_beginning(self, data):
new_node = Node(data)
new_node.prev = self.head
self.head = new_node

def print_list(self):
current = self.head
while current:
print(current.data, end=' -> ')
current = current.next
print('None')

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

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

def insert_at_beginning(self, data):
new_node = Node(data)
self.head = new_node

def print_list(self):
current = self.head
while current:
print(current.data, end=' -> ')
current = current.next
print('None')

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

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

def insert_at_beginning(self, data):
new_node = Node(data)
new_node.next = None
self.head = new_node

def print_list(self):
current = self.head
while current:
print(current.data, end=' -> ')
current = current.next
print('None')

Solution

1. Step 1: Understand insert at beginning logic

   To insert at the beginning, new node's next should point to current head, then head updated to new node.

2. Step 2: Check each option's insert_at_beginning method

   class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def insert_at_beginning(self, data): new_node = Node(data) new_node.next = self.head self.head = new_node def print_list(self): current = self.head while current: print(current.data, end=' -> ') current = current.next print('None') correctly sets new_node.next = self.head and updates head. class Node: def __init__(self, data): self.data = data self.prev = None self.next = None class LinkedList: def __init__(self): self.head = None def insert_at_beginning(self, data): new_node = Node(data) new_node.prev = self.head self.head = new_node def print_list(self): current = self.head while current: print(current.data, end=' -> ') current = current.next print('None') wrongly uses prev pointer. class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def insert_at_beginning(self, data): new_node = Node(data) self.head = new_node def print_list(self): current = self.head while current: print(current.data, end=' -> ') current = current.next print('None') misses linking new_node.next. class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def insert_at_beginning(self, data): new_node = Node(data) new_node.next = None self.head = new_node def print_list(self): current = self.head while current: print(current.data, end=' -> ') current = current.next print('None') sets new_node.next to None, losing old list.

3. Final Answer:

   Option A code snippet correctly inserts at beginning and prints list -> Option A

4. Quick Check:

   New node points to old head, then head updated [OK]

Hint: New node's next = old head; update head to new node [OK]

Common Mistakes:

• Not linking new node to old head
• Using prev pointer in singly linked list
• Setting new_node.next to None incorrectly