Push Using Linked List Node in DSA C# - Time & Space Complexity

Time Complexity: Push Using Linked List Node

O(1)

Understanding Time Complexity

We want to understand how long it takes to add a new item to the start of a linked list.

How does the time needed change as the list grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class Node {
    public int data;
    public Node next;
    public Node(int data) {
        this.data = data;
        this.next = null;
    }
}

public class LinkedList {
    public Node head;

    public void Push(int newData) {
        Node newNode = new Node(newData);
        newNode.next = head;
        head = newNode;
    }
}
    

This code adds a new node at the beginning of a linked list by adjusting pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Creating a new node and updating pointers.
• How many times: This happens once per push call, no loops or recursion involved.

How Execution Grows With Input

Each push adds one node by changing a few pointers, no matter how big the list is.

Input Size (n)	Approx. Operations
10	3 (create node, set next, update head)
100	3 (same steps, list size doesn't matter)
1000	3 (still just these steps)

Pattern observation: The number of steps stays the same no matter how large the list grows.

Final Time Complexity

Time Complexity: O(1)

This means adding a new node at the start takes the same small amount of time no matter how big the list is.

Common Mistake

[X] Wrong: "Adding a node takes longer as the list gets bigger because we have to move through the list."

[OK] Correct: We only change the first node's pointer, so we never walk through the list. The size does not affect the time.

Interview Connect

Knowing that adding at the start of a linked list is quick helps you explain efficient data structure choices in interviews.

Self-Check

"What if we had to add a node at the end of the linked list instead? How would the time complexity change?"

Practice

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

easy

A. Adds a new node at the beginning of the list

B. Removes the last node from the list

C. Searches for a node with a specific value

D. Deletes the first node from the list

Solution

1. Step 1: Understand push operation in linked list

   The push operation inserts a new node at the start of the linked list.

2. Step 2: Effect of push on list structure

   The new node becomes the head, and its next pointer points to the old head.

3. Final Answer:

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

4. Quick Check:

   Push means insert at head [OK]

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

Common Mistakes:

• Confusing push with pop operation
• Thinking push adds at the end
• Assuming push deletes nodes

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

easy

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

B. Node newNode = new Node(data); newNode.next = head; head = newNode;

C. head = new Node(data); head.next = null;

D. Node newNode = new Node(data); head = newNode.next; newNode = head;

Solution

1. Step 1: Create new node with data

   Use new Node(data) to create a node with the given data.

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

   Set newNode.next = head and then head = newNode to push at front.

3. Final Answer:

   Node newNode = new Node(data); newNode.next = head; head = newNode; -> Option B

4. Quick Check:

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

Hint: New node next points to head, then head updated [OK]

Common Mistakes:

• Assigning head before setting newNode.next
• Not initializing new node with data
• Swapping head and newNode incorrectly

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

Node head = null;
head = new Node(10);
head.next = new Node(20);
Node newNode = new Node(5);
newNode.next = head;
head = newNode;

Print the list from head to end.

medium

A. 10 -> 20 -> 5 -> null

B. 5 -> 20 -> 10 -> null

C. 20 -> 10 -> 5 -> null

D. 5 -> 10 -> 20 -> null

Solution

1. Step 1: Initial list creation

   Head points to node 10, which points to node 20, then null.

2. Step 2: Push new node with value 5

   New node 5 points to old head (10), then head updated to new node 5.

3. Final Answer:

   5 -> 10 -> 20 -> null -> Option D

4. Quick Check:

   New node added at front [OK]

Hint: New node points to old head, becomes new head [OK]

Common Mistakes:

• Appending new node at end instead of front
• Not updating head to new node
• Confusing node order after push

4. Identify the error in this push method for a singly linked list:

public void Push(int data) {
    Node newNode = new Node(data);
    newNode.next = null;
    head = newNode;
}

medium

A. newNode.next should point to current head before updating head

B. Node constructor is missing data parameter

C. head should be set to null before assigning newNode

D. No error, code is correct

Solution

1. Step 1: Analyze newNode.next assignment

   Setting newNode.next = null disconnects new node from existing list.

2. Step 2: Correct linking for push operation

   New node's next should point to current head to keep list intact before updating head.

3. Final Answer:

   newNode.next should point to current head before updating head -> Option A

4. Quick Check:

   Link new node to old head before head update [OK]

Hint: Link new node to old head before changing head [OK]

Common Mistakes:

• Setting newNode.next to null breaks list
• Not updating head after push
• Confusing order of assignments

5. Given a linked list with nodes 30 -> 20 -> 10 -> null, you want to push nodes 40 and 50 in that order using the push method. What will be the final list?

hard

A. 30 -> 20 -> 10 -> 50 -> 40 -> null

B. 40 -> 50 -> 30 -> 20 -> 10 -> null

C. 50 -> 40 -> 30 -> 20 -> 10 -> null

D. 30 -> 20 -> 10 -> 40 -> 50 -> null

Solution

1. Step 1: Initial list and first push

   Start with 30 -> 20 -> 10 -> null. Push 40: new head is 40, list is 40 -> 30 -> 20 -> 10 -> null.

2. Step 2: Second push with 50

   Push 50: new head is 50, list becomes 50 -> 40 -> 30 -> 20 -> 10 -> null.

3. Final Answer:

   50 -> 40 -> 30 -> 20 -> 10 -> null -> Option C

4. Quick Check:

   Push adds nodes at front in reverse order [OK]

Hint: Last pushed node becomes new head [OK]

Common Mistakes:

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