Why Delete Node at Specific Position in DSA Typescript?

The Big Idea

What if you could remove any link from a chain without breaking it or guessing where to cut?

The Scenario

Imagine you have a long paper chain made of connected rings. You want to remove the 5th ring from the chain, but you only have scissors and no way to easily find and cut that exact ring without breaking the whole chain.

The Problem

Manually counting and cutting the right ring is slow and risky. You might cut the wrong ring or break the chain accidentally. It's hard to keep track of positions without a clear method.

The Solution

Using a method to delete a node at a specific position in a linked list lets you find and remove the exact ring safely and quickly. It keeps the chain connected and intact without guesswork.

Before vs After

✗ Before

let current = head;
let count = 1;
while (count < position - 1 && current.next) {
  current = current.next;
  count++;
}
current.next = current.next.next;

✓ After

function deleteAtPosition(head: Node | null, position: number): Node | null {
  if (!head) return null;
  if (position === 1) return head.next;
  let current = head;
  for (let i = 1; i < position - 1 && current.next; i++) {
    current = current.next;
  }
  if (current.next) current.next = current.next.next;
  return head;
}

What It Enables

This lets you quickly and safely remove any item from a chain of connected items without breaking the rest.

Real Life Example

Think of a playlist of songs where you want to remove the 3rd song without disturbing the order of others. This method helps you do that smoothly.

Key Takeaways

Manual removal is slow and error-prone.

Deleting at a specific position keeps the structure intact.

It enables precise and safe removal from linked sequences.

Practice

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

easy

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

B. Adding a new node at that position

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

D. Counting the total nodes in the list

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: Apply to specific position

   At a specific position, we find the node before it and change its next pointer to skip the node to delete.

3. Final Answer:

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

4. Quick Check:

   Delete node = remove and link neighbors [OK]

Hint: Delete means remove and link neighbors, not add or swap [OK]

Common Mistakes:

• Thinking deletion adds a node
• Confusing deletion with swapping nodes
• Ignoring the need to link previous and next nodes

2. Which of the following TypeScript code snippets correctly deletes a node at position pos in a singly linked list starting with head?

easy

A. if (pos === 0) head = head.next; else { let temp = head; for (let i = 0; i < pos - 1; i++) temp = temp.next; temp.next = temp.next.next; }

B. if (pos === 0) head = head.next; else { let temp = head; for (let i = 0; i < pos; i++) temp = temp.next; temp.next = temp.next.next; }

C. if (pos === 0) head = head.next; else { let temp = head; for (let i = 0; i <= pos; i++) temp = temp.next; temp.next = temp.next.next; }

D. if (pos === 0) head = head.next; else { let temp = head; for (let i = 1; i <= pos; i++) temp = temp.next; temp.next = temp.next.next; }

Solution

1. Step 1: Handle deletion at head (position 0)

   If position is 0, simply move head to next node to delete the first node.

2. Step 2: Traverse to node before position

   Loop from 0 to pos-1 to reach node before the one to delete, then link its next to next.next.

3. Final Answer:

   if (pos === 0) head = head.next; else { let temp = head; for (let i = 0; i < pos - 1; i++) temp = temp.next; temp.next = temp.next.next; } -> Option A

4. Quick Check:

   Loop to pos-1, then skip node [OK]

Hint: Loop to pos-1 to delete node at pos [OK]

Common Mistakes:

• Looping to pos instead of pos-1
• Not handling deletion at head correctly
• Accessing temp.next.next when temp.next is null

3. Given the linked list: 10 -> 20 -> 30 -> 40 -> 50 -> null and the code below, what will be the list after deleting the node at position 2?

let head = { val: 10, next: { val: 20, next: { val: 30, next: { val: 40, next: { val: 50, next: null } } } } };
let pos = 2;
if (pos === 0) head = head.next;
else {
  let temp = head;
  for (let i = 0; i < pos - 1; i++) temp = temp.next;
  temp.next = temp.next.next;
}

medium

A. 10 -> 30 -> 40 -> 50 -> null

B. 10 -> 20 -> 30 -> 40 -> 50 -> null

C. 10 -> 20 -> 40 -> 50 -> null

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

Solution

1. Step 1: Identify node at position 2

   Position 2 node has value 30 (0-based indexing: 10 at 0, 20 at 1, 30 at 2).

2. Step 2: Delete node at position 2

   Traverse to node at position 1 (value 20), then set its next to node at position 3 (value 40), skipping 30.

3. Final Answer:

   10 -> 20 -> 40 -> 50 -> null -> Option C

4. Quick Check:

   Node 30 removed, neighbors linked [OK]

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

Common Mistakes:

• Deleting wrong node due to off-by-one error
• Not updating links properly
• Confusing position indexing

4. The following TypeScript code attempts to delete a node at position pos in a singly linked list but causes a runtime error. What is the error?

function deleteNode(head: Node | null, pos: number): Node | null {
  if (pos === 0) return head.next;
  let temp = head;
  for (let i = 0; i < pos; i++) {
    temp = temp.next!;
  }
  temp.next = temp.next!.next;
  return head;
}

medium

A. Incorrect loop range causes skipping the wrong node

B. Accessing temp.next when temp is null causes runtime error

C. Missing check for empty list before deletion

D. Returning head.next instead of head after deletion

Solution

1. Step 1: Analyze loop traversal

   The loop runs from 0 to pos, moving temp forward pos times, which can make temp null if pos >= list length.

2. Step 2: Identify null access

   After loop, temp may be null, so accessing temp.next causes runtime error.

3. Final Answer:

   Accessing temp.next when temp is null causes runtime error -> Option B

4. Quick Check:

   Loop overshoots, temp becomes null, then temp.next fails [OK]

Hint: Loop to pos-1, not pos, to avoid null access [OK]

Common Mistakes:

• Looping too far causing null pointer
• Not checking for empty list
• Misunderstanding return value after deletion

5. You have a singly linked list representing a playlist: Song1 -> Song2 -> Song3 -> Song4 -> Song5 -> null. You want to delete the 4th song (position 3). After deletion, you want to insert a new song Song6 at the same position. Which sequence of operations correctly achieves this?

hard

A. Delete node at position 3, then insert new node at head

B. Delete node at position 3, then traverse to position 3 and insert new node after it

C. Delete node at position 2, then traverse to position 3 and insert new node after it

D. Delete node at position 3, then traverse to position 2 and insert new node after it

Solution

1. Step 1: Delete node at position 3

   Remove the node at position 3 (Song4) by linking node at position 2 to node at position 4.

2. Step 2: Insert new node at position 3

   Traverse to position 2, then insert new node after it to place Song6 at position 3.

3. Final Answer:

   Delete node at position 3, then traverse to position 2 and insert new node after it -> Option D

4. Quick Check:

   Delete at pos 3, insert after pos 2 = new pos 3 [OK]

Hint: Insert after pos-1 to place new node at pos [OK]

Common Mistakes:

• Inserting at wrong position after deletion
• Deleting wrong node position
• Inserting at head instead of correct position