Queue Implementation Using Linked List in DSA C++ - Time & Space Complexity

Time Complexity: Queue Implementation Using Linked List

O(1)

Understanding Time Complexity

We want to understand how the time taken by queue operations changes as the number of items grows.

How does adding or removing items affect the speed of the queue?

Scenario Under Consideration

Analyze the time complexity of the enqueue and dequeue operations in this linked list queue.

struct Node {
    int data;
    Node* next;
};

class Queue {
    Node* front;
    Node* rear;
public:
    Queue() : front(nullptr), rear(nullptr) {}

    void enqueue(int x) {
        Node* temp = new Node{x, nullptr};
        if (!rear) {
            front = rear = temp;
            return;
        }
        rear->next = temp;
        rear = temp;
    }

    void dequeue() {
        if (!front) return;
        Node* temp = front;
        front = front->next;
        if (!front) rear = nullptr;
        delete temp;
    }
};
    

This code adds items to the end and removes items from the front of the queue using a linked list.

Identify Repeating Operations

Look for loops or repeated steps inside enqueue and dequeue.

• Primary operation: Adding or removing a single node pointer.
• How many times: Each enqueue or dequeue does this once per call, no loops inside.

How Execution Grows With Input

Each operation works by changing a few pointers, no matter how big the queue is.

Input Size (n)	Approx. Operations
10	1 pointer change
100	1 pointer change
1000	1 pointer change

Pattern observation: The work stays the same even if the queue grows larger.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing an item takes the same short time no matter how many items are in the queue.

Common Mistake

[X] Wrong: "Enqueue or dequeue must take longer as the queue gets bigger because there are more items."

[OK] Correct: The linked list keeps track of front and rear, so operations only change a few pointers, not all items.

Interview Connect

Knowing that queue operations are constant time shows you understand efficient data structure design, a key skill in coding interviews.

Self-Check

"What if we used a singly linked list without a rear pointer? How would the time complexity of enqueue change?"

Practice

1. What is the main advantage of implementing a queue using a linked list instead of an array?

easy

A. Dynamic size without fixed capacity

B. Faster access to middle elements

C. Uses less memory always

D. Simpler to implement than arrays

Solution

1. Step 1: Understand queue size limitation in arrays

   Arrays have fixed size, so queues implemented with arrays have a maximum capacity.

2. Step 2: Linked list dynamic size

   Linked lists can grow or shrink dynamically, so queues using linked lists do not have fixed capacity limits.

3. Final Answer:

   Dynamic size without fixed capacity -> Option A

4. Quick Check:

   Linked list queues grow dynamically = D [OK]

Hint: Linked list queues grow dynamically, arrays don't [OK]

Common Mistakes:

• Thinking linked lists use less memory always
• Confusing access speed advantages
• Assuming linked lists are simpler to implement

2. Which of the following is the correct way to declare a node structure for a queue using a linked list in C++?

easy

A. struct Node { int data; Node* next; };

B. struct Node { int data; Node next; };

C. struct Node { int data; Node* prev; };

D. struct Node { int data; int next; };

Solution

1. Step 1: Understand node structure for linked list

   A node must contain data and a pointer to the next node.

2. Step 2: Analyze options

   struct Node { int data; Node* next; }; correctly uses a pointer to next node. struct Node { int data; Node next; }; uses next as object, which is invalid. struct Node { int data; Node* prev; }; uses prev pointer, which is for doubly linked list. struct Node { int data; int next; }; uses int for next, which is incorrect.

3. Final Answer:

   struct Node { int data; Node* next; }; -> Option A

4. Quick Check:

   Node with data and next pointer = A [OK]

Hint: Node must have pointer to next node, not object or int [OK]

Common Mistakes:

• Using object instead of pointer for next
• Confusing singly and doubly linked list nodes
• Using int instead of pointer for next

3. What will be the output after performing these operations on an empty queue implemented using linked list?

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

Print the queue elements from front to rear.

medium

A. 20 -> 30 -> null

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

C. 30 -> null

D. 10 -> 30 -> null

Solution

1. Step 1: Perform enqueue(10) and enqueue(20)

   Queue becomes: 10 -> 20 -> null

2. Step 2: Perform dequeue()

   Removes 10, queue becomes: 20 -> null

3. Step 3: Perform enqueue(30)

   Queue becomes: 20 -> 30 -> null

4. Step 4: Perform dequeue()

   Removes 20, queue becomes: 30 -> null

5. Final Answer:

   30 -> null -> Option C

6. Quick Check:

   After operations queue front is 30 [OK]

Hint: Track enqueue/dequeue stepwise to find final queue [OK]

Common Mistakes:

• Forgetting to remove front on dequeue
• Appending dequeue element instead of removing
• Mixing order of operations

4. Identify the error in this dequeue function for a queue implemented using linked list:

int dequeue() {
  if (front == nullptr) return -1;
  Node* temp = front;
  front = front->next;
  int val = temp->data;
  delete temp;
  return val;
}

medium

A. No error, function is correct

B. Does not update rear when queue becomes empty

C. Deletes front node before accessing data

D. Returns wrong value after deletion

Solution

1. Step 1: Analyze dequeue steps

   Function removes front node and returns its data correctly.

2. Step 2: Check rear pointer update

   If after dequeue front becomes nullptr (empty queue), rear should also be set to nullptr but this is missing.

3. Final Answer:

   Does not update rear when queue becomes empty -> Option B

4. Quick Check:

   Rear must be nullptr if queue empty [OK]

Hint: Update rear to nullptr when queue becomes empty [OK]

Common Mistakes:

• Ignoring rear pointer update on empty queue
• Deleting node before reading data
• Returning wrong value after deletion

5. You want to implement a function reverseQueue() that reverses the elements of a queue implemented using a linked list without using extra arrays or STL containers. Which approach is best?

hard

A. Use recursion to dequeue elements and enqueue them back in reverse order

B. Traverse linked list and swap data of front and rear nodes repeatedly

C. Create a new linked list and insert elements at front while dequeuing

D. Use two pointers to reverse links between nodes iteratively

Solution

1. Step 1: Understand reversing linked list nodes

   Reversing links between nodes iteratively changes the queue order without extra space.

2. Step 2: Analyze options

   Use recursion to dequeue elements and enqueue them back in reverse order uses recursion but may use call stack (extra space). Traverse linked list and swap data of front and rear nodes repeatedly swaps data but is inefficient and complex. Create a new linked list and insert elements at front while dequeuing creates new list (extra space). Use two pointers to reverse links between nodes iteratively reverses links in place efficiently.

3. Final Answer:

   Use two pointers to reverse links between nodes iteratively -> Option D

4. Quick Check:

   In-place link reversal is best for linked list queue reverse [OK]

Hint: Reverse links in place with two pointers for best solution [OK]

Common Mistakes:

• Using extra arrays or containers
• Swapping data instead of links
• Using recursion causing extra space