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DSA C#programming

Insert at Beginning Head Insert in DSA C#

Choose your learning style10 modes available
Mental Model
Adding a new item at the start of a list means making it the first item and linking the old first item after it.
Analogy: Imagine a line of people waiting. To add a new person at the front, they step in front and the rest follow behind them.
null
↑ head
Dry Run Walkthrough
Input: list: null (empty), insert value 5 at beginning
Goal: Add a new node with value 5 at the start so it becomes the first node
Step 1: Create new node with value 5
5 -> null
↑ new node
Why: We need a new node to insert
Step 2: Set new node's next to current head (which is null)
5 -> null
↑ new node
Why: Link new node to the old first node (none here)
Step 3: Update head to point to new node
5 -> null
↑ head
Why: Head must point to the new first node
Result:
5 -> null
↑ head
Annotated Code
DSA C#
using System;

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 LinkedList()
    {
        head = null;
    }

    public void InsertAtBeginning(int value)
    {
        Node newNode = new Node(value); // create new node
        newNode.next = head; // link new node to current head
        head = newNode; // update head to new node
    }

    public void PrintList()
    {
        Node current = head;
        while (current != null)
        {
            Console.Write(current.data + " -> ");
            current = current.next;
        }
        Console.WriteLine("null");
    }
}

public class Program
{
    public static void Main()
    {
        LinkedList list = new LinkedList();
        list.InsertAtBeginning(5);
        list.PrintList();
    }
}
Node newNode = new Node(value); // create new node
create a new node with the given value
newNode.next = head; // link new node to current head
point new node's next to current head to keep the list connected
head = newNode; // update head to new node
move head pointer to new node to make it the first node
OutputSuccess
5 -> null
Complexity Analysis
Time: O(1) because insertion at the beginning does not require traversal
Space: O(1) because only one new node is created
vs Alternative: Compared to inserting at the end which requires O(n) traversal, this is faster and simpler
Edge Cases
empty list
new node becomes the head and points to null
DSA C#
newNode.next = head; // link new node to current head
list with existing nodes
new node points to old head, head updates to new node
DSA C#
head = newNode; // update head to new node
When to Use This Pattern
When you need to add an item quickly at the start of a list, use head insertion because it avoids traversing the list.
Common Mistakes
Mistake: Not updating the head pointer after creating the new node
Fix: Assign head = newNode to make the new node the first element
Mistake: Setting newNode.next to null instead of current head
Fix: Set newNode.next = head to keep the existing list connected
Summary
Adds a new node at the start of a linked list.
Use when you want to quickly insert at the beginning without traversing.
The new node points to the old head, then head moves to the new node.

Practice

(1/5)
1. What does the "Insert at Beginning" operation do in a singly linked list?
easy
A. Deletes the last node of the list
B. Adds a new node at the start of the list
C. Deletes the first node of the list
D. Adds a new node at the end of the list

Solution

  1. Step 1: Understand the operation

    Inserting at the beginning means placing a new node before the current first node.
  2. Step 2: Effect on the list

    The new node becomes the head, pushing existing nodes after it.
  3. Final Answer:

    Adds a new node at the start of the list -> Option B
  4. Quick Check:

    Insert at beginning = Add new head [OK]
Hint: New node becomes head, old head moves next [OK]
Common Mistakes:
  • Confusing insert at beginning with insert at end
  • Thinking it deletes nodes
  • Assuming it changes the tail node
2. Which of the following is the correct syntax to insert a new node at the beginning of a singly linked list in C#?
easy
A. new Node(data).next = head; head = new Node(data);
B. head = new Node(data); head.next = oldHead;
C. head.next = new Node(data);
D. Node newNode = new Node(data); newNode.next = head; head = newNode;

Solution

  1. Step 1: Create new node and link

    Create a new node and set its next pointer to current head.
  2. Step 2: Update head pointer

    Assign head to the new node to make it the first node.
  3. Final Answer:

    Node newNode = new Node(data); newNode.next = head; head = newNode; -> Option D
  4. Quick Check:

    New node points to old head, head updated [OK]
Hint: New node next = head; head = new node [OK]
Common Mistakes:
  • Not updating head after creating new node
  • Assigning head before setting newNode.next
  • Creating multiple new nodes unnecessarily
3. What will be the output of the following code snippet?
class Node {
  public int data;
  public Node next;
  public Node(int d) { data = d; next = null; }
}

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

Node temp = head;
while(temp != null) {
  Console.Write(temp.data + " -> ");
  temp = temp.next;
}
Console.WriteLine("null");
medium
A. 5 -> 10 -> 20 -> null
B. 20 -> 10 -> 5 -> null
C. 10 -> 20 -> 5 -> null
D. 5 -> 20 -> 10 -> null

Solution

  1. Step 1: Initial list creation

    Head points to 10, which points to 20 (10 -> 20 -> null).
  2. Step 2: Insert new node at beginning

    New node with data 5 points to old head (10), then head updated to new node (5).
  3. Step 3: Traverse and print list

    Traversal prints 5 -> 10 -> 20 -> null.
  4. Final Answer:

    5 -> 10 -> 20 -> null -> Option A
  5. Quick Check:

    New head 5, followed by 10 and 20 [OK]
Hint: New node points to old head, becomes new head [OK]
Common Mistakes:
  • Confusing order of nodes after insertion
  • Not updating head pointer
  • Printing nodes before insertion
4. Identify the error in the following code that tries to insert a node at the beginning of a singly linked list:
Node newNode = new Node(15);
newNode.next = newNode;
head = newNode;
medium
A. head should be assigned before setting newNode.next
B. Node constructor is missing data parameter
C. newNode.next should point to head, not newNode
D. No error, code is correct

Solution

  1. Step 1: Analyze newNode.next assignment

    newNode.next points to itself, creating a cycle instead of linking to the list.
  2. Step 2: Correct pointer assignment

    newNode.next should point to current head to insert at beginning properly.
  3. Final Answer:

    newNode.next should point to head, not newNode -> Option C
  4. Quick Check:

    Link new node to old head, not itself [OK]
Hint: newNode.next = head, not newNode [OK]
Common Mistakes:
  • Setting next pointer to newNode itself
  • Not updating head after insertion
  • Confusing order of assignments
5. You have a singly linked list with nodes containing values [30, 40, 50]. You want to insert nodes with values [10, 20] at the beginning in that order so the final list is [10, 20, 30, 40, 50]. Which sequence of insertions achieves this?
hard
A. Insert 20 at beginning, then insert 10 at beginning
B. Insert 10 at beginning, then insert 20 at beginning
C. Insert 10 at end, then insert 20 at beginning
D. Insert 20 at end, then insert 10 at beginning

Solution

  1. Step 1: Understand insertion order effect

    Each insert at beginning places new node before current head.
  2. Step 2: Insert 10 first, then 20

    Inserting 10 first results in list [10, 30, 40, 50]. Then inserting 20 at beginning results in [20, 10, 30, 40, 50].
  3. Step 3: Reverse the order to get desired list

    To get [10, 20, 30, 40, 50], insert 20 first, then 10.
  4. Final Answer:

    Insert 20 at beginning, then insert 10 at beginning -> Option A
  5. Quick Check:

    Last inserted node becomes new head [OK]
Hint: Insert last value first to keep order at head [OK]
Common Mistakes:
  • Inserting in given order causes reversed nodes
  • Confusing insert at beginning with insert at end
  • Not updating head after each insertion
Recommended

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10 questions~5 mineasy hard