Remove Nth Node from End of List in DSA Python - Execution Trace

Concept Flow - Remove Nth Node from End of List

Create dummy node pointing to head

↓

Set two pointers: first and second at dummy

↓

Move first pointer n+1 steps ahead

↓

Move first and second pointers together until first reaches end

↓

Second pointer now before target node

↓

Remove target node by changing second.next

↓

Return dummy.next as new head

We use two pointers spaced n+1 apart to find the node before the target, then remove the target node.

Execution Sample

DSA Python

def removeNthFromEnd(head, n):
    dummy = ListNode(0, head)
    first = dummy
    second = dummy
    for _ in range(n+1):
        first = first.next
    while first:
        first = first.next
        second = second.next
    second.next = second.next.next
    return dummy.next

Removes the nth node from the end of a singly linked list using two pointers.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Create dummy node pointing to head	0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null	dummy -> 0, dummy.next -> 1	dummy(0) -> 1 -> 2 -> 3 -> 4 -> 5 -> null
2	Set first and second pointers at dummy	0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null	first = dummy(0), second = dummy(0)	dummy(0) -> 1 -> 2 -> 3 -> 4 -> 5 -> null
3	Move first pointer n+1=3 steps ahead	0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null	first moves: 0->1, 1->2, 2->3	dummy(0) -> 1 -> 2 -> 3 -> 4 -> 5 -> null
4	Move first and second pointers together until first reaches end	0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null	first: 3->4, 4->5, 5->null; second: 0->1, 1->2, 2->3	dummy(0) -> 1 -> 2 -> 3 -> 4 -> 5 -> null
5	Second pointer now before target node (3)	0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null	second at node 3	dummy(0) -> 1 -> 2 -> 3 -> 4 -> 5 -> null
6	Remove target node by changing second.next	0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null	second.next changes from 4 to 5 (skips 4)	dummy(0) -> 1 -> 2 -> 3 -> 5 -> null
7	Return dummy.next as new head	1 -> 2 -> 3 -> 5 -> null	head updated to node 1	1 -> 2 -> 3 -> 5 -> null

💡 first pointer reached end (null), loop ends, target node removed

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 6	Final
dummy	node(0) -> 1	node(0) -> 1	node(0) -> 1	node(0) -> 1	node(0) -> 1
first	node(0)	node(3)	null	null	null
second	node(0)	node(0)	node(3)	node(3)	node(3)
head	node(1)	node(1)	node(1)	node(1)	node(1)
List Length	6	6	6	5	5

Key Moments - 3 Insights

Why do we create a dummy node before the head?

The dummy node helps handle cases where the head itself is removed. It simplifies pointer changes by always having a node before the target. See Step 1 and Step 7 in execution_table.

Why move the first pointer n+1 steps ahead, not just n?

Moving first pointer n+1 steps ahead keeps the gap so that when first reaches the end, second is just before the node to remove. This is shown in Step 3 and Step 4.

What happens if n equals the length of the list?

Then the node to remove is the head. The dummy node allows second pointer to be before head, so removal works the same. This is why dummy is important (Step 1 and Step 7).

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, where is the first pointer after moving n+1 steps?

A At node 2

B At node 3

C At node 4

D At dummy node

💡 Hint

Check the 'Pointer Changes' column in Step 3 of execution_table.

At which step does the second pointer move to the node before the target node?

A Step 5

B Step 4

C Step 2

D Step 6

💡 Hint

Look for when 'second at node 3' appears in execution_table.

If we remove the 1st node from the end (n=1), how would Step 6's pointer change differ?

A first pointer would not move

B second.next would skip the head node

C second.next would skip the last node

D dummy node would be removed

💡 Hint

Think about which node is removed when n=1 and check Step 6 pointer change.

Concept Snapshot

Remove Nth Node from End of List:
- Use dummy node before head
- Set two pointers at dummy
- Move first pointer n+1 steps ahead
- Move both pointers until first reaches end
- second.next skips target node
- Return dummy.next as new head

Full Transcript

This concept removes the nth node from the end of a singly linked list. We start by creating a dummy node pointing to the head to handle edge cases. Two pointers, first and second, start at the dummy. We move the first pointer n+1 steps ahead to maintain a gap. Then we move both pointers forward until the first reaches the end. At this point, the second pointer is just before the node to remove. We remove the target node by changing second.next to skip it. Finally, we return dummy.next as the new head of the list. This method works even if the head node is removed.

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