Insert at End Tail Insert in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the given value.

DSA C#

Node newNode = new Node([1]);

Drag options to blanks, or click blank then click option'

Avalue

Bnull

Chead

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'null' instead of the value.

Using 'head' or 'next' which are references, not values.

We create a new node by passing the value to the Node constructor.

2fill in blank

medium

Complete the code to check if the list is empty before inserting.

DSA C#

if (head == [1]) {
    head = newNode;
}

Drag options to blanks, or click blank then click option'

Anext

Bhead

CnewNode

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing head to newNode instead of null.

Using 'next' which is a pointer, not the list head.

If head is null, the list is empty, so we set head to the new node.

3fill in blank

hard

Fix the error in the loop to find the last node.

DSA C#

Node current = head;
while (current.[1] != null) {
    current = current.next;
}

Drag options to blanks, or click blank then click option'

Avalue

Bnext

Chead

DnewNode

Attempts:

3 left

💡 Hint

Common Mistakes

Checking current.head or current.value which are not pointers.

Using newNode which is unrelated here.

We check current.next to find the last node whose next is null.

4fill in blank

hard

Fill both blanks to link the new node at the end of the list.

DSA C#

current.[1] = [2];

Drag options to blanks, or click blank then click option'

Anext

Bhead

CnewNode

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning head or value instead of newNode.

Using wrong pointer names.

Set current.next to newNode to link the new node at the end.

5fill in blank

hard

Fill all three blanks to complete the tail insert method.

DSA C#

public void InsertAtEnd(int [1]) {
    Node newNode = new Node([2]);
    if (head == null) {
        head = newNode;
        return;
    }
    Node current = head;
    while (current.[3] != null) {
        current = current.next;
    }
    current.next = newNode;
}

Drag options to blanks, or click blank then click option'

Avalue

Cnext

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for parameter and newNode.

Checking or assigning wrong pointers.

The method parameter is 'value', newNode is created with 'value', and we traverse using 'next'.

Practice

1. What does the InsertAtEnd method do in a singly linked list?

easy

A. Adds a new node at the end (tail) of the list

B. Removes the first node of the list

C. Inserts a new node at the beginning (head) of the list

D. Deletes the last node of the list

Solution

1. Step 1: Understand the method name

   The method name InsertAtEnd suggests adding a node at the end of the list.

2. Step 2: Recall linked list operations

   Inserting at the end means adding a new node after the current last node (tail).

3. Final Answer:

   Adds a new node at the end (tail) of the list -> Option A

4. Quick Check:

   InsertAtEnd = Add node at tail [OK]

Hint: InsertAtEnd means add node at tail, not head or remove [OK]

Common Mistakes:

• Confusing insert at end with insert at beginning
• Thinking it removes nodes instead of adding
• Mixing up tail and head operations

2. Which of the following is the correct syntax to create a new node and insert it at the end of a singly linked list in C#?

easy

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

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

C. Node newNode = new Node(data); newNode.prev = tail; tail = newNode;

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

Solution

1. Step 1: Create a new node with data

   We create a new node using new Node(data).

2. Step 2: Link new node at the end

   To insert at end, set tail.next = newNode and update tail = newNode.

3. Final Answer:

   Node newNode = new Node(data); tail.next = newNode; tail = newNode; -> Option A

4. Quick Check:

   Insert at tail = tail.next = newNode; tail = newNode [OK]

Hint: Tail insert needs tail.next and tail update [OK]

Common Mistakes:

• Assigning newNode.next to head instead of tail.next
• Forgetting to update tail pointer
• Using prev pointer in singly linked list

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

LinkedList list = new LinkedList();
list.InsertAtEnd(10);
list.InsertAtEnd(20);
list.InsertAtEnd(30);
list.PrintList();

Assuming PrintList() prints nodes as value -> value -> ... -> null.

medium

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

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

C. 10 -> 20 -> null

D. null

Solution

1. Step 1: Insert nodes at end in order

   Nodes 10, 20, 30 are inserted at the tail one by one, so list order is 10, then 20, then 30.

2. Step 2: Print list in order

   Printing shows nodes in insertion order ending with null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option B

4. Quick Check:

   InsertAtEnd preserves order: 10,20,30 [OK]

Hint: InsertAtEnd keeps order; print shows same sequence [OK]

Common Mistakes:

• Assuming insert at end reverses order
• Missing last node in print
• Confusing head and tail positions

4. The following code tries to insert a node at the end of a singly linked list but causes a runtime error. What is the error?

public void InsertAtEnd(int data) {
    Node newNode = new Node(data);
    if (head == null) {
        head = newNode;
    } else {
        Node temp = head;
        while (temp.next != null) {
            temp = temp.next;
        }
        temp = newNode;
    }
}

medium

A. Node constructor is missing

B. The head is not updated when list is empty

C. The while loop condition is incorrect

D. The line temp = newNode; does not link the new node to the list

Solution

1. Step 1: Analyze the while loop

   The loop finds the last node where temp.next == null.

2. Step 2: Check how new node is linked

   Assigning temp = newNode; only changes local variable, not the last node's next pointer.

3. Step 3: Correct linking

   Should assign temp.next = newNode; to link new node at end.

4. Final Answer:

   The line temp = newNode; does not link the new node to the list -> Option D

5. Quick Check:

   Must set temp.next = newNode, not temp = newNode [OK]

Hint: Assign newNode to temp.next, not temp variable [OK]

Common Mistakes:

• Assigning newNode to temp instead of temp.next
• Not updating head when list empty
• Incorrect loop condition causing infinite loop

5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> null. You want to insert a new node with value 35 at the end. After insertion, what will be the correct sequence of nodes? Also, what changes must be made if the list is empty initially?

hard

A. Sequence: 5 -> 15 -> 25 -> null; No changes needed for empty list

B. Sequence: 35 -> 5 -> 15 -> 25 -> null; If empty, set tail = newNode only

C. Sequence: 5 -> 15 -> 25 -> 35 -> null; If empty, set head = newNode and tail = newNode

D. Sequence: 5 -> 15 -> 25 -> 35 -> null; If empty, set tail.next = newNode only

Solution

1. Step 1: Insert new node at end

   Adding 35 at end appends it after 25, so list becomes 5 -> 15 -> 25 -> 35 -> null.

2. Step 2: Handle empty list case

   If list is empty, both head and tail must point to the new node to start the list properly.

3. Final Answer:

   Sequence: 5 -> 15 -> 25 -> 35 -> null; If empty, set head = newNode and tail = newNode -> Option C

4. Quick Check:

   Insert at end appends; empty list needs head and tail set [OK]

Hint: Empty list: set head and tail to new node; else tail.next = newNode [OK]

Common Mistakes:

• Not updating both head and tail when list is empty
• Inserting new node at head instead of tail
• Only updating tail without linking new node