Remove Nth Node from End of List in DSA C - 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.

How does the number of steps grow when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    struct ListNode *fast = head, *slow = head;
    for (int i = 0; i < n; i++)
        fast = fast->next;
    if (!fast) 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 using two pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two loops: one moves the fast pointer n steps, the other moves both pointers until fast reaches the end.
• How many times: The first loop runs n times; the second loop runs (length - n) times, where length is the total nodes in the list.

How Execution Grows With Input

As the list length grows, the total steps grow roughly in a straight line with the list size.

Input Size (length)	Approx. Operations
10	About 10 steps to move fast + up to 9 steps to move slow and fast = ~19 steps
100	About 100 steps total
1000	About 1000 steps total

Pattern observation: The number of steps grows roughly in direct proportion to the list size.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Because we only move the fast pointer n steps first, the time complexity is O(n) where n is the position from the end, not the list length."

[OK] Correct: The second loop moves through the rest of the list, which depends on the total list length, so the overall time depends on the full list size, not just n.

Interview Connect

Understanding this linear time approach shows you can efficiently handle linked list problems without extra space, a skill often valued in coding challenges.

Self-Check

"What if we used a single pass approach without 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 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