Remove Nth Node from End of List in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind removing the nth node from the end of a linked list?

Use two pointers spaced n nodes apart. Move both until the first pointer reaches the end. The second pointer will be just before the node to remove.

beginner

Why do we use a dummy node at the start of the linked list in this problem?

A dummy node helps handle edge cases, like removing the first node, by providing a stable node before the head.

intermediate

What happens if n equals the length of the linked list when removing the nth node from the end?

The node to remove is the head. Using a dummy node allows easy removal by adjusting dummy's next pointer.

intermediate

In the two-pointer approach, how do you move the pointers initially?

Move the first pointer n+1 steps ahead to keep a gap of n nodes between first and second pointers.

beginner

What is the time complexity of removing the nth node from the end of a linked list using the two-pointer method?

O(L), where L is the length of the list, because we traverse the list at most once.

What is the purpose of the dummy node in the Remove Nth Node from End of List problem?

A To simplify removal of the head node

B To store the value of the node to remove

C To count the length of the list

D To point to the last node

The dummy node helps handle cases where the head node needs to be removed by providing a stable node before the head.

How far ahead should the first pointer move before starting to move both pointers together?

A 2n steps

B n steps

C n-1 steps

D n+1 steps

Moving the first pointer n+1 steps ahead keeps a gap of n nodes between the two pointers.

If the linked list has 5 nodes and n=2, which node will be removed?

A 4th node

B 2nd node

C 3rd node

D 5th node

The 2nd node from the end is the 4th node from the start.

What is the time complexity of the two-pointer approach for this problem?

A O(1)

B O(n)

C O(n^2)

D O(log n)

The algorithm traverses the list at most twice, so time complexity is linear, O(n).

What should you do if after moving the first pointer n+1 steps, it points to null?

A Return the list as is

B Remove the last node

C Remove the head node

D Move the second pointer ahead

If first pointer is null after n+1 steps, it means the node to remove is the head.

Explain step-by-step how to remove the nth node from the end of a linked list using two pointers.

Think about how the gap between pointers helps find the node to remove.

Describe how edge cases like removing the head node are handled in the Remove Nth Node from End of List problem.

Dummy node is key to handling head removal smoothly.

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