Delete Node at Specific Position in DSA C - Execution Trace

Concept Flow - Delete Node at Specific Position

Start

↓

Check if list empty

No↓

Check if position is 1

↓

Delete head

↓

Update head

↓

Free node

↓

End

↓

End

The flow checks if the list is empty or if the position is the first node, deletes accordingly, otherwise traverses to the node before the target and deletes the target node.

Execution Sample

DSA C

void deleteAtPosition(Node** head, int pos) {
  if (*head == NULL) return;
  Node* temp = *head;
  if (pos == 1) {
    *head = 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;
  Node* next = temp->next->next;
  free(temp->next);
  temp->next = next;
}

Deletes the node at the given position in a singly linked list.

Execution Table

Step	pos	temp points to	Condition Checked	Action	List State
1	3	Node 1 (value=10)	*head != NULL	Continue	10 -> 20 -> 30 -> 40 -> null
2	3	Node 1 (value=10)	pos == 1	No	10 -> 20 -> 30 -> 40 -> null
3	3	Node 1 (value=10)	i=1 < pos-1=2	temp = temp->next (Node 2)	10 -> 20 -> 30 -> 40 -> null
4	3	Node 2 (value=20)	i=2 < pos-1=2	Loop ends	10 -> 20 -> 30 -> 40 -> null
5	3	Node 2 (value=20)	temp != NULL && temp->next != NULL	Yes	10 -> 20 -> 30 -> 40 -> null
6	3	Node 2 (value=20)	Prepare to delete Node 3 (value=30)	next = Node 4	10 -> 20 -> 30 -> 40 -> null
7	3	Node 2 (value=20)	free(Node 3)	Delete Node 3	10 -> 20 -> 40 -> null
8	3	Node 2 (value=20)	temp->next = next	Link Node 2 to Node 4	10 -> 20 -> 40 -> null
9	3	-	End of function	Return	10 -> 20 -> 40 -> null

💡 Function ends after deleting node at position 3 and updating links.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 6	Final
pos	3	3	3	3	3
temp	Node 1 (10)	Node 2 (20)	Node 2 (20)	Node 2 (20)	Node 2 (20)
next	NULL	NULL	NULL	Node 4 (40)	Node 4 (40)
*head	Node 1 (10)	Node 1 (10)	Node 1 (10)	Node 1 (10)	Node 1 (10)

Key Moments - 3 Insights

Why do we check if temp == NULL or temp->next == NULL before deleting?

Because if temp is NULL or temp->next is NULL, it means the position is beyond the list length, so no deletion happens. See execution_table row 5 where this check prevents errors.

Why do we handle position 1 separately?

Position 1 means deleting the head node, so we update the head pointer directly. This is shown in execution_table row 2 where pos == 1 is checked.

How does the loop find the node before the one to delete?

The loop runs pos-2 times moving temp forward, so temp points to the node before the target. See execution_table rows 3 and 4 for temp moving from Node 1 to Node 2.

Visual Quiz - 3 Questions

Look at the execution_table, what node does temp point to after the loop ends?

A Node 3 (value=30)

B Node 1 (value=10)

C Node 2 (value=20)

D Node 4 (value=40)

💡 Hint

Check execution_table rows 3 and 4 where temp moves and loop ends.

At which step is the node at position 3 deleted?

A Step 5

B Step 7

C Step 6

D Step 8

💡 Hint

Look at execution_table row 7 where free(Node 3) happens.

If pos was 1, what would happen to the head pointer?

A It would point to the second node

B It would point to NULL

C It would remain unchanged

D It would point to the last node

💡 Hint

Refer to key_moments about position 1 deletion and execution_table row 2.

Concept Snapshot

Delete Node at Specific Position in Linked List:
- If list empty, do nothing
- If pos == 1, update head to next node and free old head
- Else, move to (pos-1)th node
- If next node exists, unlink and free node at pos
- Update links to maintain list integrity

Full Transcript

This concept shows how to delete a node at a specific position in a singly linked list. First, it checks if the list is empty. If the position is 1, it deletes the head node by updating the head pointer. Otherwise, it moves to the node before the target position. If the target node exists, it unlinks it from the list and frees its memory. The execution table traces each step, showing how the pointer moves and how the list changes after deletion. Key moments clarify why certain checks are needed to avoid errors. The visual quiz tests understanding of pointer positions and list updates during deletion.

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