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DSA Typescriptprogramming~10 mins

Delete Node at Specific Position in DSA Typescript - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

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

DSA Typescript
if (position < 0 || position >= this.length) {
  return [1];
}
Drag options to blanks, or click blank then click option'
Anull
Bfalse
Cundefined
Dtrue
Attempts:
3 left
💡 Hint
Common Mistakes
Returning true or false instead of null.
Not checking for position less than 0.
2fill in blank
medium

Complete the code to delete the head node when position is 0.

DSA Typescript
if (position === 0) {
  this.head = this.head.[1];
  this.length--;
  return;
}
Drag options to blanks, or click blank then click option'
Aprev
Bnext
Cchild
Dpointer
Attempts:
3 left
💡 Hint
Common Mistakes
Using prev instead of next.
Not decrementing the length.
3fill in blank
hard

Fix the error in updating the previous node's next pointer during deletion.

DSA Typescript
let current = this.head;
let previous = null;
let index = 0;
while (index < position) {
  previous = current;
  current = current.[1];
  index++;
}
previous.next = current.next;
this.length--;
Drag options to blanks, or click blank then click option'
Achild
Bprev
Cnext
Dpointer
Attempts:
3 left
💡 Hint
Common Mistakes
Using prev which is undefined.
Not updating previous.next correctly.
4fill in blank
hard

Fill both blanks to complete the loop that finds the node before the one to delete.

DSA Typescript
let current = this.head;
let index = 0;
while (index < position - 1) {
  current = current.[1];
  index[2];
}
Drag options to blanks, or click blank then click option'
Anext
B++
C--
Dprev
Attempts:
3 left
💡 Hint
Common Mistakes
Using prev which does not exist.
Decrementing index instead of incrementing.
5fill in blank
hard

Fill all three blanks to complete the delete function that removes a node at a given position.

DSA Typescript
deleteAtPosition(position: number): void {
  if (position < 0 || position >= this.length) {
    return null;
  }
  if (position === 0) {
    this.head = this.head.[1];
  } else {
    let current = this.head;
    let index = 0;
    while (index < position - 1) {
      current = current.[2];
      index[3];
    }
    current.next = current.next.next;
  }
  this.length--;
}
Drag options to blanks, or click blank then click option'
Anext
B++
C--
Dprev
Attempts:
3 left
💡 Hint
Common Mistakes
Using prev which does not exist.
Decrementing index instead of incrementing.
Not updating the length after deletion.

Practice

(1/5)
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
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10 questions~5 mineasy hard