Delete Node at End in DSA Java - Time & Space Complexity

Time Complexity: Delete Node at End

O(n)

Understanding Time Complexity

We want to understand how long it takes to remove the last node from a linked list.

How does the time needed change as the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void deleteAtEnd() {
    if (head == null) return;
    if (head.next == null) {
        head = null;
        return;
    }
    Node current = head;
    while (current.next.next != null) {
        current = current.next;
    }
    current.next = null;
}
    

This code removes the last node from a singly linked list by walking to the second last node and cutting off the last.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves through nodes to find the second last node.
• How many times: It runs once for each node except the last two, so roughly n-2 times for a list of size n.

How Execution Grows With Input

As the list grows, the loop runs more times because it must reach near the end.

Input Size (n)	Approx. Operations
10	About 8 times through the loop
100	About 98 times through the loop
1000	About 998 times through the loop

Pattern observation: The number of steps grows roughly the same as the list size.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete the last node grows linearly with the list size.

Common Mistake

[X] Wrong: "Deleting the last node takes constant time because we just remove it."

[OK] Correct: We must first find the second last node by walking through the list, which takes time proportional to the list length.

Interview Connect

Understanding how linked list operations scale helps you explain your code clearly and shows you know how data structures behave in real situations.

Self-Check

"What if we kept a pointer to the last node? How would the time complexity of deleting the last node change?"

Practice

1. What does the operation Delete Node at End do in a singly linked list?

easy

A. Removes the last node from the list

B. Removes the first node from the list

C. Adds a new node at the end of the list

D. Adds a new node at the beginning of the list

Solution

1. Step 1: Understand the operation meaning

   Deleting a node at the end means removing the last node in the linked list.

2. Step 2: Identify the correct description

   Removing the last node matches Removes the last node from the list, while others describe adding or removing nodes at different positions.

3. Final Answer:

   Removes the last node from the list -> Option A

4. Quick Check:

   Delete Node at End = Removes last node [OK]

Hint: Deleting at end always removes the last node [OK]

Common Mistakes:

• Confusing delete at end with delete at beginning
• Thinking it adds a node instead of removing
• Mixing up first and last node operations

2. Which of the following is the correct Java syntax to delete the last node in a singly linked list named list assuming you have a method deleteAtEnd()?

easy

A. list.deleteAtEnd();

B. deleteAtEnd.list();

C. list.deleteAtEnd;

D. list.deleteAtEnd[];

Solution

1. Step 1: Identify method call syntax in Java

   Calling a method on an object uses dot notation with parentheses: object.methodName();

2. Step 2: Match correct syntax

   list.deleteAtEnd(); uses correct syntax: list.deleteAtEnd(); others have syntax errors or wrong usage.

3. Final Answer:

   list.deleteAtEnd(); -> Option A

4. Quick Check:

   Method call syntax = object.method() [OK]

Hint: Method calls need parentheses after name [OK]

Common Mistakes:

• Omitting parentheses when calling methods
• Using brackets [] instead of parentheses ()
• Reversing object and method names

3. Consider the following Java code for a singly linked list:

class Node {
  int data;
  Node next;
  Node(int d) { data = d; next = null; }
}

class LinkedList {
  Node head;

  void deleteAtEnd() {
    if (head == null) return;
    if (head.next == null) {
      head = null;
      return;
    }
    Node temp = head;
    while (temp.next.next != null) {
      temp = temp.next;
    }
    temp.next = null;
  }

  void printList() {
    Node temp = head;
    while (temp != null) {
      System.out.print(temp.data + " -> ");
      temp = temp.next;
    }
    System.out.println("null");
  }
}

public class Main {
  public static void main(String[] args) {
    LinkedList list = new LinkedList();
    list.head = new Node(1);
    list.head.next = new Node(2);
    list.head.next.next = new Node(3);
    list.deleteAtEnd();
    list.printList();
  }
}

What is the output after running this code?

medium

A. 1 -> 2 -> 3 -> null

B. 1 -> 2 -> null

C. 2 -> 3 -> null

D. null

Solution

1. Step 1: Understand initial list and deleteAtEnd()

   The list starts as 1 -> 2 -> 3 -> null. The deleteAtEnd() removes the last node (3).

2. Step 2: Trace deleteAtEnd() method

   It traverses to node with data 2, sets its next to null, removing node 3.

3. Final Answer:

   1 -> 2 -> null -> Option B

4. Quick Check:

   Last node removed, list ends at 2 [OK]

Hint: Deleting end removes last node, list shortens by one [OK]

Common Mistakes:

• Forgetting to update the second last node's next to null
• Assuming head changes when list has multiple nodes
• Printing old list without deletion

4. The following Java method is intended to delete the last node of a singly linked list. What is the error?

void deleteAtEnd() {
  if (head == null) return;
  Node temp = head;
  while (temp.next != null) {
    temp = temp.next;
  }
  temp = null;
}

medium

A. It causes a NullPointerException

B. It does not handle empty list

C. It deletes the head node incorrectly

D. It does not update the second last node's next pointer

Solution

1. Step 1: Analyze the loop and pointer updates

   The loop moves temp to the last node (where temp.next is null). Setting temp = null only changes local pointer, not list links.

2. Step 2: Identify missing update to second last node

   The second last node's next pointer is not set to null, so the last node remains linked.

3. Final Answer:

   It does not update the second last node's next pointer -> Option D

4. Quick Check:

   Must update second last node's next to null [OK]

Hint: Set second last node's next to null, not just temp = null [OK]

Common Mistakes:

• Setting temp = null only changes local variable
• Not handling single-node list case
• Ignoring list head update if only one node

5. You have a singly linked list with nodes containing values: 10 -> 20 -> 30 -> 40 -> null. You want to delete the last node and then add a new node with value 50 at the end. Which sequence of operations correctly achieves this?

hard

A. Call addAtEnd(50) only

B. Call addAtEnd(50), then call deleteAtEnd()

C. Call deleteAtEnd(), then call addAtEnd(50)

D. Call deleteAtEnd() twice, then addAtEnd(50)

Solution

1. Step 1: Delete the last node

   Calling deleteAtEnd() removes node with value 40, list becomes 10 -> 20 -> 30 -> null.

2. Step 2: Add new node with value 50 at end

   Calling addAtEnd(50) appends node 50, list becomes 10 -> 20 -> 30 -> 50 -> null.

3. Final Answer:

   Call deleteAtEnd(), then call addAtEnd(50) -> Option C

4. Quick Check:

   Delete last, then add new node [OK]

Hint: Delete old last node before adding new one [OK]

Common Mistakes:

• Adding new node before deleting old last node
• Deleting twice unnecessarily
• Not deleting at all before adding