Enqueue Using Linked List in DSA Java - Time & Space Complexity

Time Complexity: Enqueue Using Linked List

O(1)

Understanding Time Complexity

We want to understand how long it takes to add an item to a queue when using a linked list.

Specifically, how the time changes as the queue grows larger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

class Queue {
    Node front, rear;

    void enqueue(int value) {
        Node newNode = new Node(value);
        if (rear == null) {
            front = rear = newNode;
            return;
        }
        rear.next = newNode;
        rear = newNode;
    }
}
    

This code adds a new item to the end of a queue implemented with a linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding a new node at the end by updating pointers.
• How many times: This happens once per enqueue call, no loops or recursion inside enqueue.

How Execution Grows With Input

Each enqueue adds one node by changing a few pointers, no matter how big the queue is.

Input Size (n)	Approx. Operations
10	About 3 pointer updates
100	About 3 pointer updates
1000	About 3 pointer updates

Pattern observation: The work stays the same no matter how many items are in the queue.

Final Time Complexity

Time Complexity: O(1)

This means adding an item takes the same small amount of time, no matter how big the queue is.

Common Mistake

[X] Wrong: "Enqueue takes longer as the queue grows because we have to find the end each time."

[OK] Correct: We keep a pointer to the end (rear), so we add directly without searching, keeping time constant.

Interview Connect

Knowing that enqueue in a linked list queue is fast and constant time shows you understand efficient data structure design.

Self-Check

"What if we did not keep a rear pointer and had to find the end each time? How would the time complexity change?"

Practice

1. What does the enqueue operation do in a linked list implementation of a queue?

easy

A. Adds an element at the end (rear) of the linked list

B. Removes an element from the front of the linked list

C. Deletes the entire linked list

D. Adds an element at the front (head) of the linked list

Solution

1. Step 1: Understand enqueue operation in queue

   Enqueue means adding an element to the queue, which is done at the rear end.

2. Step 2: Linked list enqueue adds at the rear

   In linked list, adding at the rear means inserting the new node at the end of the list.

3. Final Answer:

   Adds an element at the end (rear) of the linked list -> Option A

4. Quick Check:

   Enqueue = add at rear [OK]

Hint: Enqueue always adds at the rear of the queue [OK]

Common Mistakes:

• Confusing enqueue with dequeue operation
• Thinking enqueue adds at the front
• Mixing linked list head and tail roles

2. Which of the following is the correct way to create a new node for enqueue in Java linked list?

easy

A. Node newNode = new Node(); newNode.data = value; newNode.next = null;

B. Node newNode = Node(value); newNode.next = null;

C. Node newNode = new Node(value); newNode.next = null;

D. Node newNode = new Node; newNode.data = value;

Solution

1. Step 1: Understand Java object creation syntax

   In Java, to create an object with a constructor that takes a value, use new Node(value).

2. Step 2: Check next pointer initialization

   After creating the node, set next to null to mark end of list.

3. Final Answer:

   Node newNode = new Node(value); newNode.next = null; -> Option C

4. Quick Check:

   Java new object with constructor = new Node(value) [OK]

Hint: Use 'new' keyword with constructor for object creation [OK]

Common Mistakes:

• Missing 'new' keyword when creating object
• Incorrect constructor call syntax
• Not setting next pointer to null

3. What will be the state of the queue after the following enqueue operations on an empty linked list queue?

enqueue(10); enqueue(20); enqueue(30);

medium

A. 10 -> 20 -> 30 -> null

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

C. null -> 10 -> 20 -> 30

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

Solution

1. Step 1: Enqueue 10 on empty queue

   Queue becomes: 10 -> null

2. Step 2: Enqueue 20 and 30 sequentially

   After enqueue(20): 10 -> 20 -> null
   After enqueue(30): 10 -> 20 -> 30 -> null

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option A

4. Quick Check:

   Enqueue adds at rear, order preserved [OK]

Hint: Enqueue adds at rear, so order is insertion order [OK]

Common Mistakes:

• Reversing order of elements after enqueue
• Confusing enqueue with push (stack)
• Assuming new elements added at front

4. Identify the error in the following enqueue method for a linked list queue:

void enqueue(int value) {
  Node newNode = new Node(value);
  if (rear == null) {
    front = newNode;
  } else {
    rear.next = newNode;
  }
}

medium

A. Should set front.next to newNode instead of rear.next

B. Missing update of rear pointer to newNode

C. New node should be created with next pointing to front

D. Should check if front is null instead of rear

Solution

1. Step 1: Analyze enqueue logic

   When queue is empty, front is set to newNode but rear is not updated.

2. Step 2: Check rear pointer update

   After adding newNode, rear should point to newNode to keep track of last element.

3. Final Answer:

   Missing update of rear pointer to newNode -> Option B

4. Quick Check:

   Rear must always point to last node [OK]

Hint: Always update rear to new node after enqueue [OK]

Common Mistakes:

• Forgetting to update rear pointer
• Updating front.next instead of rear.next
• Not handling empty queue case properly

5. Given a linked list queue with elements 5 -> 15 -> 25 -> null, what will be the queue state after enqueueing 35 and then dequeueing one element?

hard

A. 35 -> 15 -> 25 -> null

B. 5 -> 15 -> 25 -> 35 -> null

C. 25 -> 35 -> null

D. 15 -> 25 -> 35 -> null

Solution

1. Step 1: Enqueue 35 at rear

   Queue becomes: 5 -> 15 -> 25 -> 35 -> null

2. Step 2: Dequeue removes front element 5

   After dequeue: 15 -> 25 -> 35 -> null

3. Final Answer:

   15 -> 25 -> 35 -> null -> Option D

4. Quick Check:

   Enqueue adds rear, dequeue removes front [OK]

Hint: Enqueue rear, dequeue front; order shifts accordingly [OK]

Common Mistakes:

• Removing wrong element on dequeue
• Adding new element at front instead of rear
• Not updating front pointer after dequeue