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 at the start of the linked list in this problem?

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

intermediate

In the two-pointer approach, what happens if the fast pointer moves N steps and reaches NULL immediately?

It means we need to remove the head node because the list length equals N.

beginner

What is the time complexity of removing the Nth node from the end of a singly linked list using the two-pointer method?

O(L), where L is the length of the list, because we traverse the list at most twice.

intermediate

How do you update the pointers to remove the target node once the first pointer is just before it?

Set first->next = first->next->next to skip the target node and free its memory if needed.

What does the fast pointer do initially in the two-pointer approach?

A Moves N steps ahead

B Moves one step

C Stays at the head

D Moves to the end immediately

The fast pointer moves N steps ahead to create a gap of N nodes between fast and slow pointers.

Why is a dummy node useful when removing the Nth node from the end?

A It stores the value of the removed node

B It prevents memory leaks

C It helps handle removal of the head node easily

D It speeds up the traversal

A dummy node simplifies edge cases like removing the head by providing a stable node before the head.

If the list has 5 nodes and N=5, which node is removed?

A First node

B Last node

C Third node

D Second node

Removing the 5th node from the end means removing the first node in a 5-node list.

What is the space complexity of the two-pointer method for this problem?

A O(N^2)

B O(N)

C O(log N)

D O(1)

The method uses only a few pointers, so space complexity is constant O(1).

After positioning the pointers, how do you remove the target node?

A Set target->next = NULL

B Set previous->next = target->next

C Delete the entire list

D Swap values with next node

You skip the target node by linking the previous node directly to the node after the target.

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

Think about how spacing the pointers helps find the node to remove.

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

Consider what happens when N equals the list length.

Practice

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

easy

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

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

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

D. Reverse the linked list up to the Nth node from the end

Solution

1. Step 1: Understand the problem statement

   The problem requires removing the node that is exactly N positions from the end of the linked list.

2. Step 2: Clarify the difference from other options

   Options A, C, and D describe different operations: removing from start, adding nodes, or reversing, which are not asked here.

3. Final Answer:

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

4. Quick Check:

   Remove Nth from end = Remove the node that is Nth from the end of the list [OK]

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

Common Mistakes:

• Confusing Nth from start with Nth from end
• Thinking the problem asks to add nodes
• Assuming the list needs reversing

2. Which of the following is the correct way to declare a pointer to the head of a singly linked list in C?

easy

A. struct Node head*;

B. Node head*;

C. struct Node* head;

D. Node* head;

Solution

1. Step 1: Recall C struct pointer syntax

   In C, to declare a pointer to a struct Node, use 'struct Node* head;'.

2. Step 2: Identify incorrect syntax

   Options A, B, and D misuse the order or omit 'struct' keyword which is required unless typedef is used.

3. Final Answer:

   struct Node* head; -> Option C

4. Quick Check:

   Pointer to struct Node = struct Node* head; [OK]

Hint: Remember 'struct' keyword before struct name in C [OK]

Common Mistakes:

• Omitting 'struct' keyword in declaration
• Placing '*' incorrectly after variable name
• Using typedef name without defining it

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 -> 4 -> 5 -> null

B. 2 -> 3 -> 4 -> 5 -> null

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

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

Solution

1. Step 1: Identify the 2nd node from the end

   The list is 1->2->3->4->5. Counting from 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 to next

   After removal, the list becomes 1->2->3->5->null.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Removing the wrong node counting from start
• Skipping the link update after removal
• Confusing 2nd from end with 2nd from start

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

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

medium

A. The slow pointer is moved too many times causing segmentation fault

B. The loop does not check if fast becomes NULL before accessing fast->next

C. The function does not handle the case when n is zero

D. The function returns the wrong pointer instead of head

Solution

1. Step 1: Analyze the for loop moving fast pointer

   The loop moves fast pointer n times without checking if fast is NULL, which can cause a crash if n is larger than list length.

2. Step 2: Check while loop condition and pointer usage

   The while loop uses fast->next without null check before the loop, which is unsafe if fast is NULL.

3. Final Answer:

   The loop does not check if fast becomes NULL before accessing fast->next -> Option B

4. Quick Check:

   Missing NULL check on fast pointer causes error [OK]

Hint: Always check pointers for NULL before dereferencing [OK]

Common Mistakes:

• Not checking if fast is NULL before fast->next
• Assuming list length is always >= n
• Ignoring edge case when removing head node

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 (no change)

D. 30 -> 40 -> 50 -> 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 10.

2. Step 2: Remove the head node and update head pointer

   Removing the first node results in the list: 20 -> 30 -> 40 -> 50 -> null.

3. Final Answer:

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

4. Quick Check:

   Remove 5th from end (head) = list without first node [OK]

Hint: Removing N equals list length means remove head [OK]

Common Mistakes:

• Not updating head pointer when removing first node
• Confusing Nth from end with Nth from start
• Assuming no change if N equals list length