Remove Nth Node from End of List in DSA C++ - 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 second pointer reaches the end. The first pointer will then be just before the node to remove.

beginner

Why do we use a dummy node before the head in the Remove Nth Node from End of List problem?

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

intermediate

In the two-pointer approach, what happens if the fast pointer reaches the end before moving N steps?

It means N is equal to the length of the list, so the node to remove is the head node.

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.

intermediate

Show the step-by-step process to remove the 2nd node from the end in the list 1 -> 2 -> 3 -> 4 -> 5 -> null.

1. Set dummy before head: dummy -> 1 -> 2 -> 3 -> 4 -> 5 2. Move fast pointer 2 steps ahead: fast at node 3 3. Move slow and fast together until fast reaches end: slow at dummy, fast at 3 slow at 1, fast at 4 slow at 2, fast at 5 fast moves to null, stop 4. Remove node after slow (node 4): 1 -> 2 -> 3 -> 5 -> null

What does the slow pointer point to when the fast pointer reaches the end in the two-pointer method?

A The node before the one to remove

B The node to remove

C The head node

D The last node

The slow pointer stops just before the node that needs to be removed.

Why is a dummy node used in the Remove Nth Node from End of List problem?

A To simplify removal of the head node

B To store the length of the list

C To point to the last node

D To reverse the list

The dummy node helps handle cases where the head node is removed.

If N equals the length of the list, which node is removed?

A The last node

B The head node

C The middle node

D No node is removed

Removing the Nth node from the end where N equals list length means removing the head.

What is the time complexity of the two-pointer approach to remove the Nth node from the end?

A O(1)

B O(N^2)

C O(N)

D O(log N)

The list is traversed at most twice, so time complexity is linear O(N).

Which pointer moves first in the two-pointer approach?

A Slow pointer moves first

B Both move together from start

C Neither moves

D Fast pointer moves N steps first

The fast pointer moves N steps ahead before both pointers move together.

Explain how the two-pointer technique helps remove the Nth node from the end of a linked list.

Think about how moving two pointers at different speeds helps find the target node.

Describe the edge cases to consider when removing the Nth node from the end of a linked list.

Consider what happens when the node to remove is the first node.

Practice

1. What does the problem "Remove Nth Node from End of List" ask you to do with a linked list?

easy

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

B. Remove the node that is Nth from the start of the list

C. Reverse the linked list

D. Remove the node that is Nth from the end of the list

Solution

1. Step 1: Understand the problem statement

   The problem asks to remove the node that is Nth from the end of the linked list, not from the start or to reverse or add nodes.

2. Step 2: Clarify the operation on the list

   Removing the Nth node from the end means counting nodes from the tail backwards and deleting that node.

3. Final Answer:

   Remove the node that is Nth from the end of the list -> Option D

4. Quick Check:

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

Hint: Focus on counting nodes from the list's end, not start [OK]

Common Mistakes:

• Confusing Nth from end with Nth from start
• Thinking the problem asks to reverse the list
• Trying to add nodes instead of removing

2. Which of the following C++ function signatures correctly represents a function to remove the Nth node from the end of a singly linked list?

easy

A. ListNode* removeNthFromEnd(ListNode* head, int n);

B. void removeNthFromEnd(int n, ListNode* head);

C. ListNode removeNthFromEnd(ListNode head, int n);

D. ListNode* removeNthFromEnd(int n);

Solution

1. Step 1: Check parameter order and types

   The function should take the head pointer of the list and the integer n as parameters, with head first and n second.

2. Step 2: Check return type

   Since the head might change (if first node is removed), the function returns a pointer to the new head.

3. Final Answer:

   ListNode* removeNthFromEnd(ListNode* head, int n); -> Option A

4. Quick Check:

   Correct signature = ListNode* removeNthFromEnd(ListNode* head, int n); [OK]

Hint: Function returns new head pointer and takes head, n as parameters [OK]

Common Mistakes:

• Wrong parameter order
• Returning ListNode instead of pointer
• Missing head parameter

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 now points to node 5, skipping node 4.

3. Final Answer:

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

4. Quick Check:

   Remove 2nd from end = Remove node 4 [OK]

Hint: Count nodes from end to find which to remove [OK]

Common Mistakes:

• Removing wrong node counting from start
• Not updating links correctly
• Confusing 2nd from end with 2nd from start

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

ListNode* removeNthFromEnd(ListNode* head, int n) {
    ListNode* fast = head;
    ListNode* slow = head;
    for (int i = 0; i < n; i++) {
        fast = fast->next;
    }
    while (fast->next != nullptr) {
        fast = fast->next;
        slow = slow->next;
    }
    slow->next = slow->next->next;
    return head;
}

medium

A. Uses wrong loop condition in for loop

B. Does not handle case when n equals list length (removing head)

C. Returns wrong pointer type

D. Does not move slow pointer at all

Solution

1. Step 1: Analyze fast pointer movement

   Fast pointer moves n steps ahead. If n equals list length, fast becomes nullptr.

2. Step 2: Check while loop condition

   While loop uses fast->next, but if fast is nullptr, accessing fast->next causes error. Also, removing head node is not handled.

3. Final Answer:

   Does not handle case when n equals list length (removing head) -> Option B

4. Quick Check:

   Edge case for head removal missing [OK]

Hint: Check edge cases when fast pointer reaches null early [OK]

Common Mistakes:

• Ignoring removal of head node
• Accessing fast->next without null check
• Not updating head pointer when needed

5. You want to remove the 3rd node from the end of a linked list with nodes: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> null. Which of the following steps correctly describes the process using two pointers?

hard

A. Move slow pointer 3 steps ahead, then move fast pointer until slow reaches end, then remove fast's next node.

B. Move fast pointer 3 steps ahead, then move slow pointer 3 steps ahead, then remove head node.

C. Move fast pointer 3 steps ahead, then move both pointers until fast reaches end, then remove slow's next node.

D. Move both pointers together from start without offset, then remove the node at position 3.

Solution

1. Step 1: Initialize two pointers at head

   Start both fast and slow pointers at the head of the list.

2. Step 2: Move fast pointer n steps ahead

   Move fast pointer 3 steps ahead to create a gap of 3 nodes between fast and slow.

3. Step 3: Move both pointers until fast reaches end

   Move fast and slow pointers one step at a time until fast reaches the last node.

4. Step 4: Remove the node after slow pointer

   Slow pointer now points to node before the target; remove slow->next node.

5. Final Answer:

   Move fast pointer 3 steps ahead, then move both pointers until fast reaches end, then remove slow's next node. -> Option C

6. Quick Check:

   Two-pointer gap method = Move fast pointer 3 steps ahead, then move both pointers until fast reaches end, then remove slow's next node. [OK]

Hint: Use two pointers with gap n to find node before target [OK]

Common Mistakes:

• Moving slow pointer first instead of fast
• Removing wrong node after pointers move
• Not creating correct gap between pointers