Insert at End Tail Insert in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does 'Insert at End' mean in a linked list?

It means adding a new node at the very last position of the linked list, so it becomes the new tail.

beginner

Why do we need to traverse the list when inserting at the end in a singly linked list?

Because we only have a reference to the head, we must go through each node until we find the last one to add the new node after it.

intermediate

What is the time complexity of inserting at the end in a singly linked list without a tail pointer?

O(n), where n is the number of nodes, because we must traverse the entire list to find the last node.

beginner

In Java, what is the key step to link the new node at the end of the list?

Set the 'next' pointer of the current last node to the new node, and set the new node's 'next' to null.

beginner

What happens if the linked list is empty when inserting at the end?

The new node becomes the head (and tail) of the list because there are no existing nodes.

What is the first step when inserting a node at the end of a singly linked list?

A Create a new node and set it as head

B Swap the first and last nodes

C Delete the last node

D Traverse the list to find the last node

To insert at the end, you first need to find the last node by traversing the list.

If the linked list is empty, what should the new node become?

A Both head and tail

B Head only

C Tail only

D It should not be inserted

When the list is empty, the new node is both the head and tail.

What pointer of the last node is updated when inserting at the end?

A next

B prev

C head

D tail

The 'next' pointer of the last node is set to the new node.

What is the time complexity of inserting at the end if you keep a tail pointer?

A O(n)

B O(1)

C O(log n)

D O(n^2)

With a tail pointer, insertion at the end is done in constant time O(1).

Which of these is NOT a step in inserting at the end of a singly linked list?

A Create a new node

B Set new node's next to null

C Set new node's next to head

D Update last node's next to new node

The new node's next should be null, not head.

Explain step-by-step how to insert a new node at the end of a singly linked list.

Think about what happens when the list is empty and when it is not.

Describe the difference in time complexity when inserting at the end with and without a tail pointer.

Consider how many nodes you must visit before inserting.

Practice

1. What does the "Insert at End" operation do in a singly linked list?

easy

A. Adds a new node at the end (tail) of the list

B. Deletes the first node of the list

C. Inserts a node at the beginning (head) of the list

D. Searches for a node with a specific value

Solution

1. Step 1: Understand the operation

   Insert at End means adding a new node after the last node of the list.

2. Step 2: Identify the effect on the list

   The new node becomes the new tail, extending the list by one node.

3. Final Answer:

   Adds a new node at the end (tail) of the list -> Option A

4. Quick Check:

   Insert at End = Add node at tail [OK]

Hint: Insert at End always adds after the last node [OK]

Common Mistakes:

• Confusing insert at end with insert at beginning
• Thinking it deletes nodes
• Assuming it searches instead of inserts

2. Which of the following is the correct Java syntax to create a new node with value 10 for a singly linked list?

easy

A. Node newNode = new Node(10);

B. Node newNode = Node(10);

C. newNode = new Node(10);

D. Node newNode = new Node(); newNode.value = 10;

Solution

1. Step 1: Recall Java object creation syntax

   In Java, new objects are created using the 'new' keyword followed by the constructor.

2. Step 2: Match syntax for node creation

   Correct syntax is: Node newNode = new Node(10); assuming Node class has a constructor taking value.

3. Final Answer:

   Node newNode = new Node(10); -> Option A

4. Quick Check:

   Use 'new' keyword and constructor for object creation [OK]

Hint: Use 'new' keyword with constructor to create objects [OK]

Common Mistakes:

• Omitting 'new' keyword
• Calling constructor without 'new'
• Not assigning to a variable

3. What will be the output linked list after executing the following code?

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

public class Test {
  public static void main(String[] args) {
    Node head = new Node(1);
    head.next = new Node(2);
    Node tail = head.next;
    Node newNode = new Node(3);
    tail.next = newNode;
    tail = newNode;
    printList(head);
  }
  static void printList(Node node) {
    while(node != null) {
      System.out.print(node.data + " -> ");
      node = node.next;
    }
    System.out.print("null");
  }
}

medium

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

B. 1 -> 3 -> 2 -> null

C. 1 -> 2 -> 3 -> null

D. 1 -> 2 -> null

Solution

1. Step 1: Trace initial list creation

   head points to node 1, head.next points to node 2, tail points to node 2.

2. Step 2: Insert new node at end and update tail

   New node with data 3 is created, tail.next points to newNode, tail updated to newNode.

3. Step 3: Print list

   List nodes in order: 1 -> 2 -> 3 -> null.

4. Final Answer:

   1 -> 2 -> 3 -> null -> Option C

5. Quick Check:

   Tail insert adds node 3 at end [OK]

Hint: Tail always points to last node; new node added after tail [OK]

Common Mistakes:

• Not updating tail after insertion
• Confusing node order
• Printing incomplete list

4. Identify the error in the following code snippet for inserting a node at the end of a singly linked list:

public void insertAtEnd(int data) {
  Node newNode = new Node(data);
  if (head == null) {
    head = newNode;
  } else {
    Node temp = head;
    while (temp.next != null) {
      temp = temp.next;
    }
    temp = newNode;
  }
}

medium

A. The while loop condition should be 'temp != null'

B. The line 'temp = newNode;' does not link the new node to the list

C. The head is not updated when list is empty

D. The newNode is not created properly

Solution

1. Step 1: Analyze the while loop and insertion

   The loop finds the last node (temp.next == null). After loop, temp points to last node.

2. Step 2: Check how new node is linked

   Assigning 'temp = newNode;' only changes temp variable, not the last node's next pointer.

3. Step 3: Correct linking

   Should assign 'temp.next = newNode;' to link new node at end.

4. Final Answer:

   The line 'temp = newNode;' does not link the new node to the list -> Option B

5. Quick Check:

   Link new node by setting lastNode.next = newNode [OK]

Hint: Assign newNode to last node's next, not to temp variable [OK]

Common Mistakes:

• Assigning newNode to temp instead of temp.next
• Not updating head when list is empty
• Wrong loop condition

5. You have a singly linked list with nodes: 5 -> 10 -> 15 -> null. You want to insert a new node with value 20 at the end. Which sequence of steps correctly updates the list?

hard

A. Create new node with 20, set new node's next to null, but do not link it to list

B. Create new node with 20, set head to new node, set new node's next to old head

C. Create new node with 20, set new node's next to head, set head to new node

D. Create new node with 20, traverse to last node, set last node's next to new node

Solution

1. Step 1: Create new node with value 20

   Use constructor to create new node with data 20 and next as null.

2. Step 2: Traverse to last node

   Start from head, move through nodes until node.next is null (node with 15).

3. Step 3: Link new node at end

   Set last node's next pointer to new node, extending list to 5 -> 10 -> 15 -> 20 -> null.

4. Final Answer:

   Create new node with 20, traverse to last node, set last node's next to new node -> Option D

5. Quick Check:

   Insert at end = link new node after last node [OK]

Hint: Traverse to last node, then link new node's next as null [OK]

Common Mistakes:

• Setting head to new node incorrectly
• Not linking new node to list
• Inserting at beginning instead of end