Singly linked list structure in Data Structures Theory - Time & Space Complexity

Time Complexity: Singly linked list structure

O(n)

Understanding Time Complexity

When working with a singly linked list, it is important to understand how the time needed to perform operations changes as the list grows.

We want to know how the time to find or access elements changes as the list gets longer.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

    def find(self, target):
        current = self.head
        while current is not None:
            if current.value == target:
                return True
            current = current.next
        return False
    

This code defines a singly linked list and a method to find if a value exists by checking each node one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each node in the list one by one.
• How many times: Up to once for every node until the target is found or the list ends.

How Execution Grows With Input

As the list gets longer, the time to find a value grows roughly in direct proportion to the number of nodes.

Input Size (n)	Approx. Operations
10	Up to 10 checks
100	Up to 100 checks
1000	Up to 1000 checks

Pattern observation: The work grows steadily as the list grows longer, checking each node one after another.

Final Time Complexity

Time Complexity: O(n)

This means the time to find a value grows linearly with the number of nodes in the list.

Common Mistake

[X] Wrong: "Finding a value in a singly linked list is as fast as in an array because both store data."

[OK] Correct: Unlike arrays, singly linked lists do not allow direct access by position, so you must check nodes one by one, making it slower for searching.

Interview Connect

Understanding how linked lists work and their time costs helps you explain data structure choices clearly and solve problems efficiently.

Self-Check

"What if we added a pointer to the tail node? How would that affect the time complexity of finding a value?"

Practice

1. What does each node in a singly linked list contain?

easy

A. Only data without any pointers

B. Data and pointers to both next and previous nodes

C. Data and a pointer to the next node

D. Only a pointer to the previous node

Solution

1. Step 1: Understand node structure in singly linked list

   Each node stores some data and a pointer to the next node only, not to the previous one.

2. Step 2: Compare options with definition

   Data and a pointer to the next node matches the definition exactly. Data and pointers to both next and previous nodes describes a doubly linked list. Options C and D miss the pointer to the next node.

3. Final Answer:

   Data and a pointer to the next node -> Option C

4. Quick Check:

   Node = data + next pointer [OK]

Hint: Remember: singly linked means one pointer per node [OK]

Common Mistakes:

• Confusing singly linked with doubly linked list
• Thinking nodes have no pointers
• Assuming pointer to previous node exists

2. Which of the following is the correct way to represent the head of a singly linked list in code?

easy

A. head = None

B. head = head.next

C. head = head.prev

D. head = data

Solution

1. Step 1: Understand the head pointer initialization

   The head points to the first node or is None if the list is empty.

2. Step 2: Analyze options

   head = None correctly initializes head as None for an empty list. head = head.next tries to move head to next node, which is invalid if list is empty. head = head.prev uses prev pointer which does not exist in singly linked list. head = data assigns data to head, which is incorrect.

3. Final Answer:

   head = None -> Option A

4. Quick Check:

   Empty list head = None [OK]

Hint: Head starts as None when list is empty [OK]

Common Mistakes:

• Using prev pointer in singly linked list
• Assigning data directly to head
• Moving head before list has nodes

3. Given the following singly linked list nodes:

node1 = Node(data=10, next=node2)
node2 = Node(data=20, next=node3)
node3 = Node(data=30, next=None)
head = node1

What is the data value of head.next.next?

medium

A. 20

B. 30

C. 10

D. None

Solution

1. Step 1: Trace the linked nodes from head

   head points to node1 with data 10. head.next points to node2 with data 20. head.next.next points to node3 with data 30.

2. Step 2: Identify the data at head.next.next

   The data at node3 is 30.

3. Final Answer:

   30 -> Option B

4. Quick Check:

   head.next.next.data = 30 [OK]

Hint: Follow next pointers step-by-step [OK]

Common Mistakes:

• Confusing node positions
• Assuming next of last node is not None
• Mixing data values with node references

4. Identify the error in this singly linked list insertion code snippet:

new_node = Node(data=5)
new_node.next = head
head = new_node.next

medium

A. new_node.next should be None

B. No error, code is correct

C. Data assignment to new_node is missing

D. head should be assigned to new_node, not new_node.next

Solution

1. Step 1: Understand insertion at head

   To insert new_node at the start, new_node.next should point to current head, then head should be updated to new_node.

2. Step 2: Analyze the code error

   The code sets new_node.next = head correctly, but then sets head = new_node.next, which points to old head, not new_node. This loses the new node as head.

3. Final Answer:

   head should be assigned to new_node, not new_node.next -> Option D

4. Quick Check:

   head = new_node after insertion [OK]

Hint: After linking, head must point to new_node [OK]

Common Mistakes:

• Assigning head to new_node.next instead of new_node
• Forgetting to link new_node.next to old head
• Not updating head pointer after insertion

5. You want to reverse a singly linked list with nodes:

head -> node1 -> node2 -> node3 -> None

Which sequence of pointer changes correctly reverses the list?

hard

A. Change each node's next pointer to the previous node, then update head to last node

B. Swap data values between nodes without changing pointers

C. Set head to None and create a new list with reversed data

D. Change each node's next pointer to None

Solution

1. Step 1: Understand reversal of singly linked list

   Reversing means changing each node's next pointer to point to the previous node instead of the next.

2. Step 2: Identify correct pointer update sequence

   Change each node's next pointer to the previous node, then update head to last node correctly describes reversing pointers and updating head to the last node. Swap data values between nodes without changing pointers only swaps data, not pointers. Set head to None and create a new list with reversed data discards the list unnecessarily. Change each node's next pointer to None breaks links without reversing.

3. Final Answer:

   Change each node's next pointer to the previous node, then update head to last node -> Option A

4. Quick Check:

   Reverse = reverse pointers + update head [OK]

Hint: Reverse pointers, then reset head to last node [OK]

Common Mistakes:

• Swapping data instead of pointers
• Not updating head after reversal
• Breaking links without reconnecting