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

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

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

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

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

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

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

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

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.

Practice

1. What does the problem Remove Nth Node from End of List ask you to do?

easy

A. Add a new node at the Nth position from the start

B. Find the value of the Nth node from the start

C. Reverse the linked list up to the Nth node

D. Delete the node that is the Nth from the end in a linked list

Solution

1. Step 1: Understand the problem statement

   The problem requires removing a node counting from the end of the linked list, not from the start.

2. Step 2: Identify the correct action

   Deleting the Nth node from the end means locating it and removing it from the list.

3. Final Answer:

   Delete the node that is the Nth from the end in a linked list -> Option D

4. Quick Check:

   Remove Nth node from end = Delete node Nth from end [OK]

Hint: Focus on 'from end' means counting backwards [OK]

Common Mistakes:

• Confusing Nth from start with Nth from end
• Thinking the problem is about reversing the list
• Trying to add nodes instead of removing

2. Which of the following is the correct Python syntax to create a dummy node pointing to the head of a linked list named head?

easy

A. dummy = head.next

B. dummy = ListNode(head)

C. dummy = ListNode(0); dummy.next = head

D. dummy = ListNode(); dummy = head

Solution

1. Step 1: Understand dummy node creation

   A dummy node is usually created with a value (often 0) and its next pointer set to the head.

2. Step 2: Check each option

   dummy = ListNode(0); dummy.next = head correctly creates a dummy node with value 0 and points its next to head.

3. Final Answer:

   dummy = ListNode(0); dummy.next = head -> Option C

4. Quick Check:

   Dummy node points to head = dummy.next = head [OK]

Hint: Dummy node value 0 and next points to head [OK]

Common Mistakes:

• Passing head as value instead of next pointer
• Overwriting dummy after creation
• Not linking dummy to head

3. What will be the output linked list after removing the 2nd node from the end in the list 1 -> 2 -> 3 -> 4 -> 5 -> null?

medium

A. 1 -> 2 -> 3 -> 5 -> null

B. 2 -> 3 -> 4 -> 5 -> null

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

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

Solution

1. Step 1: Identify the 2nd node from the end

   Counting from the end: 5 (1st), 4 (2nd). So node with value 4 is to be removed.

2. Step 2: Remove node 4 and link previous node to next

   Node 3's next pointer should now point to node 5, skipping node 4.

3. Final Answer:

   1 -> 2 -> 3 -> 5 -> null -> Option A

4. Quick Check:

   Remove 2nd from end (4) = 1->2->3->5 [OK]

Hint: Count nodes from end, remove that node [OK]

Common Mistakes:

• Removing the wrong node counting from start
• Not updating the previous node's next pointer
• Confusing node values

4. Given this code snippet to remove the Nth node from end, what is the error?

def removeNthFromEnd(head, n):
    dummy = ListNode(0)
    dummy.next = head
    first = dummy
    second = dummy
    for i in range(n):
        first = first.next
    while first.next:
        first = first.next
        second = second.next
    second.next = second.next.next
    return dummy.next

medium

A. The for loop should run n+1 times to move first pointer

B. The while loop condition should be while first is not None

C. second.next = second.next.next causes error if second.next is None

D. dummy node is not needed for this problem

Solution

1. Step 1: Analyze the for loop moving first pointer

   To position first pointer n nodes ahead, it must move n+1 times to maintain gap with second pointer.

2. Step 2: Understand the impact of moving first pointer only n times

   If first moves only n times, the gap is off by one, causing wrong node removal.

3. Final Answer:

   The for loop should run n+1 times to move first pointer -> Option A

4. Quick Check:

   Move first pointer n+1 times for correct gap [OK]

Hint: Move first pointer n+1 times to keep gap [OK]

Common Mistakes:

• Moving first pointer only n times
• Incorrect while loop condition
• Ignoring dummy node usage

5. You have a linked list 1 -> 2 -> 3 -> 4 -> 5 -> null. If you remove the 5th node from the end, what will be the resulting list?

hard

A. 1 -> 3 -> 4 -> 5 -> null

B. 2 -> 3 -> 4 -> 5 -> null

C. 1 -> 2 -> 3 -> 4 -> null

D. 5 -> null

Solution

1. Step 1: Identify the 5th node from the end

   The list length is 5, so the 5th node from the end is the first node with value 1.

2. Step 2: Remove the first node and link dummy to second node

   Removing node 1 means the new head is node 2, so list becomes 2 -> 3 -> 4 -> 5 -> null.

3. Final Answer:

   2 -> 3 -> 4 -> 5 -> null -> Option B

4. Quick Check:

   Remove 5th from end (head) = 2->3->4->5 [OK]

Hint: Removing Nth from end where N equals list length removes head [OK]

Common Mistakes:

• Removing last node instead of first
• Not updating head pointer after removal
• Confusing node positions