Why Delete Node at Specific Position in DSA C++?

The Big Idea

What if you could instantly remove any item from a list without breaking the order?

The Scenario

Imagine you have a paper list of names and you want to erase the 5th name. You have to count each name one by one and then carefully cross out the right one without messing up the order.

The Problem

Doing this by hand is slow and easy to make mistakes. You might cross out the wrong name or lose track of the order. Fixing errors means starting over or rewriting the whole list.

The Solution

Using a linked list and deleting a node at a specific position lets a computer quickly find and remove the exact item without disturbing the rest. It keeps the list neat and saves time.

Before vs After

✗ Before

string names[] = {"A", "B", "C", "D", "E", "F"};
// To delete 5th name, shift all after it left by one
for(int i = 4; i < 5; i++) {
  names[i] = names[i+1];
}

✓ After

struct Node {
  string data;
  Node* next;
};
void deleteAtPosition(Node*& head, int pos) {
  // Remove node at pos in linked list
}

What It Enables

It enables fast, safe removal of any item from a list without rewriting or losing track.

Real Life Example

Deleting a contact from your phone's call history at a specific spot without erasing the whole list.

Key Takeaways

Manual deletion is slow and error-prone.

Deleting a node in a linked list is efficient and safe.

This skill helps manage dynamic lists easily.

Practice

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

easy

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

B. The last node is removed

C. The list remains unchanged

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: Delete 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 A

4. Quick Check:

   Deleting position 1 removes the head [OK]

Hint: Deleting position 1 means removing 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. head = head->prev;

B. if (pos == 1) { head = head->next; }

C. delete head->next->prev;

D. head->next = NULL;

Solution

1. Step 1: Check deletion at position 1

   If position is 1, we update head to head->next to remove the first node.

2. Step 2: Verify syntax correctness

   if (pos == 1) { head = head->next; } correctly updates head pointer; others are invalid or incorrect for singly linked list.

3. Final Answer:

   if (pos == 1) { head = head->next; } -> Option B

4. Quick Check:

   Update head for position 1 deletion [OK]

Hint: For position 1, assign head = head->next [OK]

Common Mistakes:

• Using prev pointer in singly linked list
• Setting next pointer to NULL incorrectly
• Deleting wrong node pointer

3. What is the output linked list after deleting the node at position 3 from the list: 1 -> 2 -> 3 -> 4 -> 5 -> NULL?

medium

A. 2 -> 3 -> 4 -> 5 -> NULL

B. 1 -> 2 -> 3 -> 5 -> NULL

C. 1 -> 3 -> 4 -> 5 -> NULL

D. 1 -> 2 -> 4 -> 5 -> NULL

Solution

1. Step 1: Identify node at position 3

   Node at position 3 is the node with value 3.

2. Step 2: Remove node 3 and link previous to next

   Link node 2's next to node 4, skipping node 3.

3. Final Answer:

   1 -> 2 -> 4 -> 5 -> NULL -> Option D

4. Quick Check:

   Node 3 removed, list links 2 to 4 [OK]

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

Common Mistakes:

• Removing wrong node
• Not updating previous node's next pointer
• Deleting head instead of position 3

4. Given the code snippet below, what is the error when deleting a node at position 4?

void deleteNode(Node*& head, int pos) {
  Node* temp = head;
  for (int i = 1; i < pos - 1; i++) {
    temp = temp->next;
  }
  Node* toDelete = temp->next;
  temp->next = toDelete->next;
  delete toDelete;
}

medium

A. No check if position is valid or if temp is NULL

B. Incorrect loop range causing off-by-one error

C. Deleting head node incorrectly

D. Using delete on a pointer not allocated with new

Solution

1. Step 1: Analyze loop and pointer usage

   The loop moves temp to node before position to delete, but no checks if temp or temp->next is NULL.

2. Step 2: Identify missing validation

   Without checking if position is valid or temp is NULL, code may crash on invalid positions.

3. Final Answer:

   No check if position is valid or if temp is NULL -> Option A

4. Quick Check:

   Missing boundary checks cause runtime errors [OK]

Hint: Always check pointers before accessing next [OK]

Common Mistakes:

• Ignoring invalid position inputs
• Assuming list length is sufficient
• Not checking for NULL pointers

5. You have a singly linked list: 10 -> 20 -> 30 -> 40 -> 50 -> NULL. Write the resulting list after deleting the node at position 5 and then deleting the node at position 1.

hard

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

B. 10 -> 20 -> 30 -> 40 -> NULL

C. 20 -> 30 -> 40 -> NULL

D. 10 -> 20 -> 30 -> NULL

Solution

1. Step 1: Delete node at position 5

   Node at position 5 is 50, removing it results in list: 10 -> 20 -> 30 -> 40 -> NULL.

2. Step 2: Delete node at position 1

   Deleting head node 10 updates head to 20, final list: 20 -> 30 -> 40 -> NULL.

3. Final Answer:

   20 -> 30 -> 40 -> NULL -> Option C

4. Quick Check:

   Delete last then head node sequentially [OK]

Hint: Delete last node first, then head node for correct order [OK]

Common Mistakes:

• Deleting head first then last causing wrong list
• Not updating head after deleting position 1
• Forgetting to remove last node