Remove Nth Node from End of List in DSA Python - Time & Space Complexity

Time Complexity: Remove Nth Node from End of List

O(n)

Understanding Time Complexity

We want to understand how the time needed to remove the nth node from the end of a linked list changes as the list grows.

Specifically, how many steps does the program take when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def removeNthFromEnd(head: ListNode, n: int) -> ListNode:
    fast = slow = head
    for _ in range(n):
        fast = fast.next
    if fast is None:
        return head.next
    while fast.next:
        fast = fast.next
        slow = slow.next
    slow.next = slow.next.next
    return head
    

This code removes the nth node from the end of a singly linked list by using two pointers moving through the list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving pointers through the linked list nodes.
• How many times: The first loop moves the fast pointer n times; the second loop moves both pointers until the fast pointer reaches the end, which can be up to (length - n) times.

How Execution Grows With Input

As the list size grows, the number of steps to move pointers grows roughly in proportion to the list length.

Input Size (length)	Approx. Operations
10	About 10 steps to move pointers
100	About 100 steps to move pointers
1000	About 1000 steps to move pointers

Pattern observation: The steps increase linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to remove the nth node from the end grows linearly with the list size.

Common Mistake

[X] Wrong: "Since we only move the fast pointer n times first, the time complexity is O(n) where n is the given number, not the list size."

[OK] Correct: The second loop moves pointers through the rest of the list, which depends on the total list length, not just the given n. So the total steps depend on the list size.

Interview Connect

Understanding this helps you explain how to efficiently find and remove nodes from linked lists, a common task in coding challenges and real projects.

Self-Check

"What if we used a single pass with a stack instead of two pointers? How would the time complexity change?"

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