Delete Node at Specific Position in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the position is valid before deletion.

DSA C#

if (position < 1 || position > [1]) {
    return head;
}

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

ACountNodes(head)

Bhead.Length

Chead.Count()

Dhead.Size

Attempts:

3 left

💡 Hint

Common Mistakes

Using properties like Length or Count() which do not exist for linked lists.

Not checking if position is less than 1.

We use a helper function CountNodes(head) to count the total nodes and check if the position is valid.

2fill in blank

medium

Complete the code to delete the head node when position is 1.

DSA C#

if (position == 1) {
    Node temp = head;
    head = [1];
    temp = null;
    return head;
}

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

Ahead.next

Bhead.next.next

Chead

Dhead.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning head to itself or to head.prev which is null.

Not updating the head pointer.

To delete the head node, we move the head pointer to the next node.

3fill in blank

hard

Fix the error in the loop that moves to the node before the one to delete.

DSA C#

Node current = head;
for (int i = 1; i < [1] - 1; i++) {
    current = current.next;
}

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

Aposition + 1

Bposition

Cposition - 1

Dposition - 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using position - 2 or position + 1 which causes off-by-one errors.

Starting loop counter at 0 instead of 1.

The loop should run until the node before the position, so it runs position - 1 times. Using i < position - 1 is correct.

4fill in blank

hard

Fill both blanks to unlink the node at the given position.

DSA C#

Node temp = current.next;
current.next = temp.[1];
if (temp.[2] != null) {
    temp.next.prev = current;
}

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

Anext

Bprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev instead of next in the first blank.

Not checking if temp.next is null before accessing prev.

We set current.next to temp.next to unlink the node. Then check if temp.next is not null before updating its prev pointer.

5fill in blank

hard

Fill all three blanks to complete the delete function for a doubly linked list.

DSA C#

public Node DeleteAtPosition(Node head, int position) {
    if (position < 1 || position > [1]) return head;
    if (position == 1) {
        Node temp = head;
        head = head.[2];
        if (head != null) head.[3] = null;
        temp = null;
        return head;
    }
    Node current = head;
    for (int i = 1; i < position - 1; i++) {
        current = current.next;
    }
    Node tempNode = current.next;
    current.next = tempNode.next;
    if (tempNode.next != null) {
        tempNode.next.prev = current;
    }
    tempNode = null;
    return head;
}

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

ACountNodes(head)

Bnext

Cprev

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating the previous pointer of the new head to null.

Using incorrect function or property to count nodes.

The first blank checks the list length using CountNodes(head). The second blank moves head to the next node. The third blank sets the new head's previous pointer to null.

Practice

1. What does deleting a node at a specific position in a singly linked list mean?

easy

A. Adding a new node at that position

B. Removing the node at that position and linking its previous node to the next node

C. Swapping the node at that position with the last node

D. Reversing the entire list up to that position

Solution

1. Step 1: Understand node deletion in linked list

   Deleting a node means removing it and connecting the previous node to the next node to keep the list intact.

2. Step 2: Identify the effect of deletion at a specific position

   The node at the given position is removed, and the previous node's next pointer is updated to skip the deleted node.

3. Final Answer:

   Removing the node at that position and linking its previous node to the next node -> Option B

4. Quick Check:

   Delete node = Remove and link previous to next [OK]

Hint: Delete means remove node and link previous to next [OK]

Common Mistakes:

• Thinking deletion adds a node
• Confusing deletion with swapping nodes
• Assuming deletion reverses the list

2. Which of the following is the correct C# syntax to delete a node at position pos in a singly linked list starting with head?

easy

A. head.next = null;

B. head = head.next[pos];

C. if (pos == 0) head = head.next;

D. head.deleteNode(pos);

Solution

1. Step 1: Check deletion at head position

   If position is 0, deleting means moving head to next node: head = head.next;.

2. Step 2: Validate other options

   head.next = null;: head.next = null; deletes second node regardless of pos. head = head.next[pos];: head = head.next[pos]; invalid syntax (no array indexing on nodes). head.deleteNode(pos);: head.deleteNode(pos); calls non-existent method.

3. Final Answer:

   if (pos == 0) head = head.next; -> Option C

4. Quick Check:

   Delete head = head.next [OK]

Hint: To delete head, assign head to head.next [OK]

Common Mistakes:

• Using array-like indexing on linked list
• Calling undefined methods
• Setting next to null incorrectly

3. What will be the linked list after deleting the node at position 2 from the list: 1 -> 3 -> 5 -> 7 -> null?

medium

A. 1 -> 3 -> 7 -> null

B. 1 -> 5 -> 7 -> null

C. 3 -> 5 -> 7 -> null

D. 1 -> 3 -> 5 -> null

Solution

1. Step 1: Identify node at position 2

   Positions start at 0: 0->1, 1->3, 2->5, 3->7. Node at position 2 is 5.

2. Step 2: Delete node 5 and link previous to next

   Remove 5, link node with value 3 to node with value 7.

3. Final Answer:

   1 -> 3 -> 7 -> null -> Option A

4. Quick Check:

   Delete node 5 at pos 2 = 1->3->7 [OK]

Hint: Count from zero, remove node at position, link neighbors [OK]

Common Mistakes:

• Deleting wrong node by counting from 1
• Not linking previous node to next
• Removing head instead of position 2

4. Given the following code snippet to delete a node at position pos, what is the error?

Node current = head;
for (int i = 0; i < pos - 1; i++) {
    current = current.next;
}
current.next = current.next.next;

medium

A. It does not handle deletion at position 0

B. It causes a syntax error due to missing semicolons

C. It deletes the node after the given position

D. It causes infinite loop

Solution

1. Step 1: Analyze loop for position 0

   For pos=0, loop condition i < pos - 1 (i < -1) is false, current remains head, then current.next = current.next.next; deletes SECOND node (pos 1) instead of head.

2. Step 2: Check deletion logic for other positions

   For pos >=1, loop moves current to node before target and skips target correctly.

3. Final Answer:

   It does not handle deletion at position 0 -> Option A

4. Quick Check:

   Code misses head deletion case [OK]

Hint: Check if code handles deleting head (pos 0) separately [OK]

Common Mistakes:

• Ignoring head deletion case
• Assuming loop runs for pos=0
• Confusing syntax errors with logic errors

5. You have a singly linked list: 10 -> 20 -> 30 -> 40 -> 50 -> null. You want to delete the node at position 4. After deletion, what will be the list? Also, what happens if you try to delete at position 5?

hard

A. 10 -> 20 -> 30 -> 40 -> 50 -> null; Deleting at position 5 deletes head

B. 10 -> 20 -> 30 -> 40 -> null; Deleting at position 5 causes no change (position out of range)

C. 10 -> 20 -> 30 -> 40 -> 50 -> null; Deleting at position 5 deletes last node

D. 10 -> 20 -> 30 -> 40 -> null; Deleting at position 5 causes runtime error (null reference)

Solution

1. Step 1: Delete node at position 4

   Positions: 0->10,1->20,2->30,3->40,4->50. Deleting node 50 removes last node, list becomes 10->20->30->40->null.

2. Step 2: Attempt deletion at position 5

   Position 5 does not exist. Trying to access node at pos 5 causes null reference error (runtime error).

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> null; Deleting at position 5 causes runtime error (null reference) -> Option D

4. Quick Check:

   Delete last node works; out-of-range causes error [OK]

Hint: Deleting beyond list length causes runtime error [OK]

Common Mistakes:

• Assuming out-of-range deletion does nothing
• Deleting wrong node at position 5
• Confusing last node position with list length