Push Using Linked List Node in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C#

Node newNode = new Node([1]);

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

Adata

Bvalue

Cnode

Ditem

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not defined in the method.

Confusing the node variable with the data value.

The constructor of Node takes the data value to store in the node. Here, 'data' is the correct variable.

2fill in blank

medium

Complete the code to link the new node to the current head of the list.

DSA C#

newNode.[1] = head;

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

Alink

Bnext

Cprev

Dpointer

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

Using an undefined property name.

In a singly linked list, each node points to the next node using the 'next' reference.

3fill in blank

hard

Fix the error in updating the head pointer to the new node.

DSA C#

head = [1];

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

AnewNode

BnewNode.next

Chead.next

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning head to newNode.next which skips the new node.

Assigning head to null which empties the list.

After pushing, the head should point to the new node itself, not its next.

4fill in blank

hard

Fill both blanks to complete the push method that adds a new node at the start.

DSA C#

public void Push(int data) {
    Node newNode = new Node([1]);
    newNode.[2] = head;
    head = newNode;
}

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

Adata

Bvalue

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' for singly linked list.

Passing wrong variable to node constructor.

The new node is created with 'data' and its 'next' points to the current head.

5fill in blank

hard

Fill all three blanks to complete the push method with correct variable and pointer assignments.

DSA C#

public void Push(int [1]) {
    Node [2] = new Node([1]);
    [2].next = head;
    head = [2];
}

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

Adata

BnewNode

Cnode

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Confusing parameter name with node variable name.

The method parameter is 'data', the new node variable is 'newNode', and all references use 'newNode'.

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