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

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode fast = head;
    ListNode slow = head;
    for (int i = 0; i < n; i++) {
        fast = fast.next;
    }
    if (fast == null) return head.next;
    while (fast.next != null) {
        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 by using two pointers moving through the list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving pointers through the linked list nodes.
• How many times: The first loop moves the fast pointer n times; the second loop moves both pointers until fast reaches the end, which can be up to (length - n) times.

How Execution Grows With Input

As the list size grows, the number of steps to move pointers grows roughly in proportion to the list length.

Input Size (length)	Approx. Operations
10	About 10 steps to move pointers
100	About 100 steps to move pointers
1000	About 1000 steps to move pointers

Pattern observation: The steps increase linearly as the list gets longer.

Final Time Complexity

Time Complexity: O(n)

This means the time to remove the nth node from the end grows in a straight line with the size of the list.

Common Mistake

[X] Wrong: "Since we only move the fast pointer n times first, the time is O(n) where n is the position from the end, so it is always small."

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

Interview Connect

Understanding this helps you explain how to efficiently handle linked lists with two pointers, a common pattern in coding problems.

Self-Check

"What if we used a single pass 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. Reverse the entire linked list

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

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

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

Solution

1. Step 1: Understand the problem statement

   The problem requires removing the node that is Nth from the end of the linked list, not from the start.

2. Step 2: Clarify the operation on the list

   We do not add or reverse nodes; we specifically remove one node counting from the end.

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 'Nth from end' means counting backwards [OK]

Common Mistakes:

• Confusing Nth from end with Nth from start
• Thinking the problem asks to add or reverse nodes
• Ignoring the direction of counting nodes

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

easy

A. public ListNode removeNthFromEnd(ListNode head, int n)

B. public void removeNthFromEnd(int n, ListNode head)

C. public ListNode removeNode(ListNode head, int index)

D. public ListNode deleteNthNode(ListNode node, int n)

Solution

1. Step 1: Check method parameters and return type

   The method should take the head of the list and an integer n, and return the new head after removal.

2. Step 2: Match signature with problem requirements

   public ListNode removeNthFromEnd(ListNode head, int n) matches the expected signature: it takes (ListNode head, int n) and returns ListNode.

3. Final Answer:

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

4. Quick Check:

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

Hint: Method must return updated head and accept head, n as parameters [OK]

Common Mistakes:

• Wrong parameter order
• Void return type instead of ListNode
• Using incorrect parameter types

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

C. 1 -> 3 -> 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

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

3. Final Answer:

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

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
• Off-by-one errors in counting
• Not updating links correctly after removal

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

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode fast = head;
    ListNode slow = head;
    for (int i = 0; i <= n; i++) {
        fast = fast.next;
    }
    while (fast != null) {
        fast = fast.next;
        slow = slow.next;
    }
    slow.next = slow.next.next;
    return head;
}

medium

A. The for loop should run from i = 0 to i < n, not i <= n

B. The slow pointer should start at null, not head

C. The method should return slow, not head

D. The while loop condition should be fast.next != null

Solution

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

   The loop runs i from 0 to n inclusive (i <= n), moving fast n+1 steps, which is one step too many and causes NullPointerException.

2. Step 2: Correct the loop to run i < n to move fast exactly n steps ahead

   This ensures fast is n nodes ahead of slow, allowing correct removal of the target node.

3. Final Answer:

   The for loop should run from i = 0 to i < n, not i <= n -> Option A

4. Quick Check:

   Loop count off by one = The for loop should run from i = 0 to i < n, not i <= n [OK]

Hint: Check loop bounds carefully for pointer steps [OK]

Common Mistakes:

• Using i <= n instead of i < n in loop
• Starting slow pointer incorrectly
• Returning wrong node after removal

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 after removal?

hard

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

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

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

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

Solution

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

   Since the list has 5 nodes, the 5th from the end is the first node with value 10.

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

   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 C

4. Quick Check:

   Remove first node = 20 -> 30 -> 40 -> 50 -> null [OK]

Hint: If N equals list length, remove head node [OK]

Common Mistakes:

• Not updating head when first node is removed
• Removing wrong node counting from end
• Confusing list length with N