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DSA Pythonprogramming~10 mins

Delete Node at Specific Position in DSA Python - Interactive Practice

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

Complete the code to delete the head node of a singly linked list.

DSA Python
def delete_head(head):
    if head is None:
        return None
    new_head = head[1]
    return new_head
Drag options to blanks, or click blank then click option'
A.data
B.prev
C.next
D.head
Attempts:
3 left
💡 Hint
Common Mistakes
Using .prev which is for doubly linked lists.
Trying to access .data instead of moving the head pointer.
2fill in blank
medium

Complete the code to delete the node at position 2 (0-based index) in a singly linked list.

DSA Python
def delete_at_pos(head, pos):
    if pos == 0:
        return head.next
    current = head
    for _ in range(pos - 1):
        current = current[1]
    current.next = current.next.next
    return head
Drag options to blanks, or click blank then click option'
A.prev
B.next
C.data
D.head
Attempts:
3 left
💡 Hint
Common Mistakes
Using .prev which does not exist in singly linked lists.
Trying to access .data instead of moving the pointer.
3fill in blank
hard

Fix the error in the code to delete a node at a given position in a singly linked list.

DSA Python
def delete_node(head, pos):
    if pos == 0:
        return head.next
    current = head
    for i in range(pos - 1):
        if current is None or current.next is None:
            return head
        current = current[1]
    current.next = current.next.next
    return head
Drag options to blanks, or click blank then click option'
A.next
B.head
C.data
D.prev
Attempts:
3 left
💡 Hint
Common Mistakes
Using .prev which is not defined in singly linked lists.
Trying to access .data instead of moving the pointer.
4fill in blank
hard

Fill both blanks to delete the node at position pos in a singly linked list safely.

DSA Python
def delete_node_safe(head, pos):
    if pos == 0:
        return head[1]
    current = head
    for _ in range(pos - 1):
        if current is None or current.next is None:
            return head
        current = current[2]
    current.next = current.next.next
    return head
Drag options to blanks, or click blank then click option'
A.next
B.prev
C.data
D.head
Attempts:
3 left
💡 Hint
Common Mistakes
Using .prev which is not available in singly linked lists.
Trying to access .data instead of moving the pointer.
5fill in blank
hard

Fill all three blanks to delete the node at position pos with boundary checks in a singly linked list.

DSA Python
def delete_node_full(head, pos):
    if pos == 0:
        return head[1]
    current = head
    for _ in range(pos - 1):
        if current is None or current.next is None:
            return head
        current = current[2]
    if current.next is None:
        return head
    current.next = current.next[3]
    return head
Drag options to blanks, or click blank then click option'
A.next
B.prev
D.data
Attempts:
3 left
💡 Hint
Common Mistakes
Using .prev which is not defined in singly linked lists.
Trying to access .data instead of moving the pointer.

Practice

(1/5)
1. What happens when you delete a node at position 0 in a singly linked list?
easy
A. The last node is removed
B. The list remains unchanged
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 0 in linked list

    Position 0 refers to the head node of the list.
  2. Step 2: Effect of deleting head node

    Deleting 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 0 removes head [OK]
Hint: Deleting position 0 always removes the head node [OK]
Common Mistakes:
  • Thinking last node is deleted at position 0
  • 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 Python?
easy
A. head = head.next if pos == 1 else head
B. if pos == 0: head = head.next
C. head.next = head if pos == 0 else head.next
D. head = head.prev if pos == 0 else head.next

Solution

  1. Step 1: Check deletion at position 0

    Deleting at position 0 means moving head to next node.
  2. Step 2: Verify syntax correctness

    if pos == 0: head = head.next correctly assigns head to head.next when pos is 0.
  3. Final Answer:

    if pos == 0: head = head.next -> Option B
  4. Quick Check:

    Correct syntax for deleting head node [OK]
Hint: Deleting head means head = head.next [OK]
Common Mistakes:
  • Using head.prev which doesn't exist in singly linked list
  • Wrong condition for position check
  • Assigning head incorrectly
3. What will be the linked list after deleting the node at position 2 from the list: 10 -> 20 -> 30 -> 40 -> null?
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def delete_node(head, pos):
    if pos == 0:
        return head.next
    current = head
    for _ in range(pos - 1):
        current = current.next
    current.next = current.next.next
    return head

head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head.next.next.next = Node(40)
head = delete_node(head, 2)

# Print list
current = head
result = []
while current:
    result.append(current.data)
    current = current.next
print(result)
medium
A. [10, 20, 40]
B. [10, 30, 40]
C. [20, 30, 40]
D. [10, 20, 30, 40]

Solution

  1. Step 1: Identify node at position 2

    Position 2 node has value 30 in the list 10 -> 20 -> 30 -> 40.
  2. Step 2: Delete node at position 2

    Node 20's next pointer is updated to node 40, skipping node 30.
  3. Final Answer:

    [10, 20, 40] -> Option A
  4. Quick Check:

    Deleted node 30 at position 2 [OK]
Hint: Skip node at position by linking previous to next [OK]
Common Mistakes:
  • Deleting wrong node due to off-by-one error
  • Not updating next pointer correctly
  • Returning wrong head
4. Find the error in the following code snippet that deletes a node at position pos in a singly linked list:
def delete_node(head, pos):
    if pos == 0:
        head = head.next
    current = head
    for _ in range(pos - 1):
        current = current.next
    current.next = current.next.next
    return head
medium
A. No error, code is correct
B. Loop runs one time too many
C. current.next.next may cause AttributeError if current.next is None
D. Missing return after deleting head node

Solution

  1. Step 1: Check deletion at position 0

    When pos is 0, head is updated but not returned immediately.
  2. Step 2: Effect of missing return

    Code continues and tries to access current.next causing errors or wrong behavior.
  3. Final Answer:

    Missing return after deleting head node -> Option D
  4. Quick Check:

    Return immediately after deleting head [OK]
Hint: Return immediately after deleting head node [OK]
Common Mistakes:
  • Not returning head after deleting first node
  • Ignoring edge cases when list is empty
  • Assuming current.next always exists
5. Given a singly linked list 5 -> 10 -> 15 -> 20 -> 25 -> null, write the resulting list after deleting nodes at positions 0, 2, and then 1 in that order. What is the final list?
hard
A. [10, 25]
B. [10, 15, 25]
C. [10, 20]
D. [5, 15, 25]

Solution

  1. Step 1: Delete node at position 0

    Deleting head (5), list becomes 10 -> 15 -> 20 -> 25.
  2. Step 2: Delete node at position 2

    Node at position 2 is 20, deleting it results in 10 -> 15 -> 25.
  3. Step 3: Delete node at position 1

    Node at position 1 is 15, deleting it results in 10 -> 25.
  4. Final Answer:

    [10, 25] -> Option A
  5. Quick Check:

    Sequential deletions update list correctly [OK]
Hint: Delete nodes stepwise, update positions after each deletion [OK]
Common Mistakes:
  • Not updating list after each deletion
  • Confusing positions after deletions
  • Deleting wrong nodes due to static indexing
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10 questions~5 mineasy hard