Remove Nth Node from End of List in DSA Go - 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?

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

beginner

In Go, how do you define a singly linked list node?

type ListNode struct { Val int Next *ListNode }

intermediate

What happens if N equals the length of the list in the Remove Nth Node from End of List problem?

The node to remove is the head. Using a dummy node and two pointers helps remove it without special code.

intermediate

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 only once.

What is the purpose of the two pointers in the Remove Nth Node from End of List problem?

A To find the node before the one to remove by keeping a gap of N nodes

B To reverse the linked list

C To count the total number of nodes

D To sort the linked list

Two pointers spaced N nodes apart help locate the node before the target node efficiently.

Why is a dummy node used before the head in this problem?

A To store the length of the list

B To add a new node at the end

C To reverse the list

D To simplify removal of the head node

A dummy node provides a stable starting point, making it easier to remove the head node without extra checks.

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

A The 2nd node from the end

B The 2nd node from the start

C The last node

D The head node

The problem removes the Nth node counting from the end, so the 2nd node from the end is removed.

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

A O(1)

B O(N)

C O(N^2)

D O(log N)

The list is traversed once, so the time complexity is linear, O(N).

What should you do if after moving the fast pointer N steps, it reaches null?

A Remove the middle node

B Remove the last node

C Remove the head node

D Return the list as is

If fast pointer reaches null after N steps, the node to remove is the head.

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

Think about spacing between pointers and dummy node usage.

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

Dummy node simplifies edge cases.

Practice

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

easy

A. Reverse the linked list

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

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

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

Solution

1. Step 1: Understand the problem statement

   The problem requires removing the node that is Nth from the end, not from the start or adding nodes.

2. Step 2: Clarify the operation on the linked list

   We must delete the node counting from the end, which is different from reversing or finding a value.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Confusing Nth from end with Nth from start
• Thinking the problem asks to add nodes
• Assuming the list should be reversed

2. Which of the following Go code snippets correctly declares a singly linked list node struct for this problem?

easy

A. type ListNode struct { Val int; Next *ListNode }

B. type ListNode struct { Value int; NextNode *ListNode }

C. struct ListNode { int Val; ListNode* Next; }

D. class ListNode { int Val; ListNode Next; }

Solution

1. Step 1: Identify Go struct syntax

   In Go, structs use 'type Name struct { fields }' and pointers use '*Type'.

2. Step 2: Match correct field names and types

   type ListNode struct { Val int; Next *ListNode } uses 'Val int' and 'Next *ListNode' which is the standard for singly linked list nodes in Go.

3. Final Answer:

   type ListNode struct { Val int; Next *ListNode } -> Option A

4. Quick Check:

   Go struct with pointer field = type ListNode struct { Val int; Next *ListNode } [OK]

Hint: Go structs use 'type' and pointers with '*' [OK]

Common Mistakes:

• Using C/C++ or Java syntax in Go code
• Missing pointer '*' for Next field
• Incorrect field names or types

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

medium

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

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

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

D. 2 -> 3 -> 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 with value 4 and link previous node to next

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

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Removing the wrong node counting from start
• Breaking links incorrectly causing list corruption
• Off-by-one errors in counting nodes

4. Given this Go function snippet to remove the Nth node from end, what is the bug?

func removeNthFromEnd(head *ListNode, n int) *ListNode {
    dummy := &ListNode{0, head}
    first, second := dummy, dummy
    for i := 0; i < n; i++ {
        first = first.Next
    }
    for first.Next != nil {
        first = first.Next
        second = second.Next
    }
    second.Next = second.Next.Next
    return head
}

medium

A. Missing check for empty list before loops

B. The loop runs n times instead of n+1 times

C. second pointer is not initialized correctly

D. Returns head instead of dummy.Next, so removal of first node fails

Solution

1. Step 1: Understand dummy node usage

   Dummy node helps handle edge cases like removing the first node. The result list starts at dummy.Next.

2. Step 2: Check return statement

   Returning 'head' ignores changes if first node was removed. Should return dummy.Next instead.

3. Final Answer:

   Returns head instead of dummy.Next, so removal of first node fails -> Option D

4. Quick Check:

   Return dummy.Next to handle first node removal [OK]

Hint: Always return dummy.Next after removal [OK]

Common Mistakes:

• Returning original head ignoring dummy node
• Not using dummy node for edge cases
• Incorrect loop bounds causing panic

5. You have a linked list: 10 -> 20 -> 30 -> 40 -> 50 -> null. You want to remove the 5th node from the end. What will be the resulting list?

hard

A. 20 -> 30 -> 40 -> 50 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 10 -> 20 -> 30 -> 40 -> 50 -> null

D. 30 -> 40 -> 50 -> null

Solution

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

   The list length is 5, so the 5th from end is the first node with value 10.

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

   Removing node 10 means the new head is node 20, so list becomes 20 -> 30 -> 40 -> 50 -> null.

3. Final Answer:

   20 -> 30 -> 40 -> 50 -> null -> Option A

4. Quick Check:

   Remove 5th from end (first node) = 20->30->40->50 [OK]

Hint: Nth from end equals list length means remove head [OK]

Common Mistakes:

• Removing wrong node when N equals list length
• Not updating head pointer after removal
• Confusing node positions counting from start vs end