Remove Nth Node from End of List in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize a dummy node pointing to head.

DSA Go

dummy := &ListNode{Val: 0, Next: [1]

Drag options to blanks, or click blank then click option'

Anil

Btail

Cdummy

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dummy.Next to nil loses the list.

Pointing dummy.Next to dummy causes infinite loops.

The dummy node's Next should point to the head of the list to simplify edge cases.

2fill in blank

medium

Complete the code to move the fast pointer n steps ahead.

DSA Go

for i := 0; i < n; i++ {
    fast = [1]
}

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Afast.Prev

Bfast

Cfast.Next

Dfast.Next.Next

Attempts:

3 left

💡 Hint

Common Mistakes

Using fast without .Next does not move the pointer.

Using fast.Prev moves backward, which is incorrect.

To move fast pointer one step forward, assign fast = fast.Next inside the loop.

3fill in blank

hard

Fix the error in the loop condition to move both pointers until fast reaches the end.

DSA Go

for fast != [1] {
    fast = fast.Next
    slow = slow.Next
}

Drag options to blanks, or click blank then click option'

Ahead

Bnil

Cdummy

Dslow

Attempts:

3 left

💡 Hint

Common Mistakes

Using dummy or head in the condition causes infinite loops or wrong behavior.

Using slow in the condition is incorrect.

The loop should continue until fast reaches nil, which means the end of the list.

4fill in blank

hard

Fill both blanks to remove the nth node from the end by adjusting pointers.

DSA Go

slow.Next = slow.[1].[2]

Drag options to blanks, or click blank then click option'

ANext

BPrev

CNext.Next

DPrev.Prev

Attempts:

3 left

💡 Hint

Common Mistakes

Using Prev is incorrect because the list is singly linked.

Using Prev.Prev or Next.Prev causes errors.

To remove the node, link slow.Next to slow.Next.Next, skipping the target node.

5fill in blank

hard

Fill both blanks to return the new head of the list after removal.

DSA Go

return [1].[2]

Drag options to blanks, or click blank then click option'

Adummy

Bhead

CNext

DPrev

Attempts:

3 left

💡 Hint

Common Mistakes

Returning head may return a removed node.

Returning dummy.Prev or dummy itself is incorrect.

Return dummy.Next as the new head after removal, since dummy points before the original head.

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