Delete Node at Specific Position in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move the pointer to the head of the list.

DSA C

struct Node* current = [1];

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

Ahead

BNULL

Ctail

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using NULL instead of head causes no traversal.

Using tail pointer will not start from the beginning.

We start from the head of the linked list to delete a node at a specific position.

2fill in blank

medium

Complete the code to move the pointer to the node before the one to delete.

DSA C

for (int i = 0; i < position - 1; i++) {
    current = current->[1];
}

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

Aparent

Bnext

Cchild

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' in singly linked list causes errors.

Using 'child' or 'parent' are invalid pointers here.

We move to the next node to reach the node before the one to delete.

3fill in blank

hard

Fix the error in updating the link to skip the deleted node.

DSA C

current->[1] = current->next->next;

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

Aprev

Bchild

Cnext

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' causes compilation errors.

Using 'child' or 'parent' are invalid in this context.

We update current's next pointer to skip the node to delete.

4fill in blank

hard

Fill both blanks to correctly free the deleted node and update the list.

DSA C

struct Node* temp = current->[1];
current->[2] = temp->next;
free(temp);

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

Anext

Bprev

Cchild

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' causes errors in singly linked list.

Not freeing the node causes memory leaks.

We store the node to delete in temp, update current's next to skip it, then free temp.

5fill in blank

hard

Fill all three blanks to handle deleting the head node correctly.

DSA C

if (position == 0) {
    struct Node* temp = [1];
    [2] = [3]->next;
    free(temp);
}

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

Ahead

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating head causes list corruption.

Freeing before updating head causes access errors.

To delete the head node, store it in temp, move head to next node, then free temp.

Practice

1. What happens when you delete a node at position 1 in a singly linked list?

easy

A. The list remains unchanged

B. The last node is removed

C. The head node is removed and the next node becomes the new head

D. All nodes are deleted

Solution

1. Step 1: Understand position 1 in linked list

   Position 1 refers to the head node of the list.

2. Step 2: Deleting the head node

   Removing the head means the next node becomes the new head.

3. Final Answer:

   The head node is removed and the next node becomes the new head -> Option C

4. Quick Check:

   Deleting position 1 removes head [OK]

Hint: Deleting position 1 always removes the head node [OK]

Common Mistakes:

• Thinking position 1 deletes the last node
• Assuming list stays same after deletion
• Deleting all nodes instead of one

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

easy

A. void deleteNode(struct Node* head, int pos);

B. void deleteNode(struct Node** head_ref, int pos);

C. void deleteNode(int pos, struct Node* head);

D. void deleteNode(int pos);

Solution

1. Step 1: Understand pointer usage for head modification

   To delete a node, the function must modify the head pointer if position 1 is deleted, so it needs a pointer to pointer.

2. Step 2: Check function parameters

   Only void deleteNode(struct Node** head_ref, int pos); uses struct Node** head_ref allowing head modification.

3. Final Answer:

   void deleteNode(struct Node** head_ref, int pos); -> Option B

4. Quick Check:

   Use pointer to pointer for head changes [OK]

Hint: Use double pointer to head to delete at any position [OK]

Common Mistakes:

• Using single pointer and expecting head to change
• Wrong parameter order
• Missing head pointer in parameters

3. Given the linked list: 10 -> 20 -> 30 -> 40 -> 50 -> NULL, what is the list after deleting the node at position 3?

medium

A. 10 -> 20 -> 40 -> 50 -> NULL

B. 10 -> 20 -> 30 -> 50 -> NULL

C. 10 -> 30 -> 40 -> 50 -> NULL

D. 20 -> 30 -> 40 -> 50 -> NULL

Solution

1. Step 1: Identify node at position 3

   Position 3 node is 30 in the list.

2. Step 2: Remove node 30 and link previous to next

   Link node 20 to node 40, skipping 30.

3. Final Answer:

   10 -> 20 -> 40 -> 50 -> NULL -> Option A

4. Quick Check:

   Deleted 3rd node (30) correctly [OK]

Hint: Skip the node at position by linking previous to next [OK]

Common Mistakes:

• Deleting wrong node position
• Not updating links correctly
• Removing head instead of position 3

4. Consider this code snippet to delete a node at position pos in a linked list:

void deleteNode(struct Node** head_ref, int pos) {
  struct Node* temp = *head_ref;
  if (pos == 1) {
    *head_ref = temp->next;
    free(temp);
    return;
  }
  for (int i = 1; temp != NULL && i < pos - 1; i++) {
    temp = temp->next;
  }
  if (temp == NULL || temp->next == NULL) return;
  struct Node* next = temp->next->next;
  free(temp->next);
  temp->next = next;
}

What is the bug in this code?

medium

A. It does not handle empty list when pos is 1

B. It frees the wrong node

C. It does not update head pointer correctly

D. It does not check if pos is valid

Solution

1. Step 1: Check deletion at position 1

   If list is empty (*head_ref is NULL), temp is NULL, so temp->next causes error.

2. Step 2: Identify missing empty list check

   Code should check if temp is NULL before accessing temp->next when pos == 1.

3. Final Answer:

   It does not handle empty list when pos is 1 -> Option A

4. Quick Check:

   Empty list check missing at head deletion [OK]

Hint: Always check if list is empty before deleting [OK]

Common Mistakes:

• Ignoring empty list case
• Assuming pos is always valid
• Not freeing the deleted node

5. You have a singly linked list: 5 -> 10 -> 15 -> 20 -> 25 -> NULL. You want to delete the node at position 6. What should happen after calling the delete function?

hard

A. The last node (25) is deleted

B. The function crashes due to invalid position

C. The head node (5) is deleted

D. The list remains unchanged because position 6 is invalid

Solution

1. Step 1: Check if position 6 exists in list

   List has 5 nodes, so position 6 is invalid.

2. Step 2: Behavior on invalid position

   Proper delete function should do nothing if position is invalid, leaving list unchanged.

3. Final Answer:

   The list remains unchanged because position 6 is invalid -> Option D

4. Quick Check:

   Invalid position means no deletion [OK]

Hint: Invalid position means no change to list [OK]

Common Mistakes:

• Deleting last node incorrectly
• Crashing due to no position check
• Deleting head by mistake