Remove Nth Node from End of List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move the fast pointer n steps ahead.

DSA Python

for _ in range([1]):
    fast = fast.next

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A0

B1

Cn

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of n causes the gap to be incorrect.

Using 0 means the fast pointer doesn't move ahead at all.

We need to move the fast pointer exactly n steps ahead to maintain the gap.

2fill in blank

medium

Complete the condition to move both pointers until fast reaches the end.

DSA Python

while fast.next [1] None:
    fast = fast.next
    slow = slow.next

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A!=

B>

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' None stops the loop too early.

Using '<' or '>' with None is invalid.

We continue moving while fast.next is not None, meaning fast has not reached the last node.

3fill in blank

hard

Fix the error in removing the target node by correctly updating the next pointer.

DSA Python

slow.next = slow.next[1]next

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A::

B->

C=>

D.

Attempts:

3 left

💡 Hint

Common Mistakes

Using '->' causes syntax errors in Python.

Using '=>' or '::' are invalid in this context.

In Python, the dot operator '.' is used to access attributes like 'next' of an object.

4fill in blank

hard

Fill both blanks to correctly handle the edge case when the head node is removed.

DSA Python

if fast is [1]:
    return [2].next

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ANone

Bhead

Ctail

Dslow

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if fast is head is incorrect here.

Returning head instead of head.next does not remove the node.

If fast is None after moving n steps, it means we remove the head node, so return head.next.

5fill in blank

hard

Fill all three blanks to complete the function that removes the nth node from the end.

DSA Python

def removeNthFromEnd(head, n):
    dummy = ListNode(0)
    dummy.next = head
    fast = dummy
    slow = dummy
    for _ in range([1]):
        fast = fast.next
    while fast.next [2] None:
        fast = fast.next
        slow = slow.next
    slow.next = slow.next[3]next
    return dummy.next

Drag options to blanks, or click blank then click option'

An

B!=

C.

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators in the while loop.

Using wrong attribute access operator.

Not moving fast pointer n steps first.

Move fast pointer n steps, loop while fast.next != None, then remove node by linking slow.next to slow.next.next.