Circular Queue Implementation Using Array in DSA Typescript - Time & Space Complexity

Time Complexity: Circular Queue Implementation Using Array

O(1)

Understanding Time Complexity

We want to understand how the time taken by circular queue operations changes as the queue size grows.

Specifically, how fast can we add or remove items when using an array in a circular way?

Scenario Under Consideration

Analyze the time complexity of the enqueue and dequeue operations in this circular queue code.

class CircularQueue {
  private arr: number[];
  private front: number = -1;
  private rear: number = -1;
  private size: number;

  constructor(size: number) {
    this.size = size;
    this.arr = new Array(size);
  }

  enqueue(value: number): boolean {
    if ((this.rear + 1) % this.size === this.front) return false; // full
    if (this.front === -1) this.front = 0;
    this.rear = (this.rear + 1) % this.size;
    this.arr[this.rear] = value;
    return true;
  }

  dequeue(): number | null {
    if (this.front === -1) return null; // empty
    const value = this.arr[this.front];
    if (this.front === this.rear) {
      this.front = -1;
      this.rear = -1;
    } else {
      this.front = (this.front + 1) % this.size;
    }
    return value;
  }
}
    

This code adds (enqueue) and removes (dequeue) items in a fixed-size array used as a circular queue.

Identify Repeating Operations

Look for loops or repeated steps inside enqueue and dequeue.

• Primary operation: Simple index calculations and assignments.
• How many times: Each enqueue or dequeue runs these steps once per call, no loops inside.

How Execution Grows With Input

Each enqueue or dequeue does a fixed number of steps regardless of queue size.

Input Size (n)	Approx. Operations per enqueue/dequeue
10	About 5 steps
100	About 5 steps
1000	About 5 steps

Pattern observation: The number of steps stays the same no matter how big the queue is.

Final Time Complexity

Time Complexity: O(1)

This means each enqueue or dequeue operation 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 the array is bigger."

[OK] Correct: The circular queue uses simple math to jump directly to the right spot, so it never scans or moves many items.

Interview Connect

Understanding constant time operations like enqueue and dequeue in circular queues shows you can design efficient data structures that handle growing data smoothly.

Self-Check

"What if we changed the circular queue to a normal queue using a simple array without wrapping? How would the time complexity of enqueue and dequeue change?"

Practice

1. What is the main advantage of using a circular queue over a simple linear queue when implemented using an array?

easy

A. It allows storing different data types in the same queue.

B. It efficiently uses all available space by wrapping around the array.

C. It automatically sorts the elements in the queue.

D. It uses less memory than a linear queue.

Solution

1. Step 1: Understand linear queue limitation

   In a linear queue using an array, once the rear reaches the end, no more elements can be added even if there is free space at the front.

2. Step 2: Circular queue advantage

   A circular queue connects the end of the array back to the front, allowing reuse of empty spaces and efficient use of the array.

3. Final Answer:

   It efficiently uses all available space by wrapping around the array. -> Option B

4. Quick Check:

   Circular queue = efficient space use [OK]

Hint: Circular queue reuses space by wrapping indices [OK]

Common Mistakes:

• Thinking circular queue sorts elements
• Believing it uses less memory than linear queue
• Assuming it stores different data types automatically

2. Which of the following is the correct way to check if a circular queue is full when using an array of size size with front and rear indices?

easy

A. if (rear === front)

B. if (front === -1)

C. if ((rear + 1) % size === front)

D. if (rear === size - 1)

Solution

1. Step 1: Understand full condition in circular queue

   The queue is full when the next position of rear (rear + 1) modulo size equals front, meaning no space to insert.

2. Step 2: Apply modulo for wrap-around

   Using modulo ensures the rear wraps to the start of the array, so the condition is (rear + 1) % size === front.

3. Final Answer:

   if ((rear + 1) % size === front) -> Option C

4. Quick Check:

   Full condition = (rear + 1) % size === front [OK]

Hint: Full when next rear wraps to front [OK]

Common Mistakes:

• Checking rear === front means empty, not full
• Using rear === size - 1 ignores wrap-around
• Checking front === -1 is for empty queue

3. Consider a circular queue of size 5 (indices 0 to 4). Initially, front = 0 and rear = 3. After performing enqueue(10) and dequeue(), what are the new values of front and rear?

medium

A. front = 0, rear = 4

B. front = 1, rear = 3

C. front = 0, rear = 3

D. front = 1, rear = 4

Solution

1. Step 1: Perform enqueue(10)

   rear moves from 3 to (3 + 1) % 5 = 4. So rear = 4, front remains 0.

2. Step 2: Perform dequeue()

   front moves from 0 to (0 + 1) % 5 = 1. So front = 1, rear remains 4.

3. Final Answer:

   front = 1, rear = 4 -> Option D

4. Quick Check:

   Enqueue rear+1=4, Dequeue front+1=1 [OK]

Hint: Update rear on enqueue, front on dequeue with modulo [OK]

Common Mistakes:

• Not applying modulo for wrap-around
• Updating front on enqueue instead of rear
• Confusing initial and final pointer positions

4. Given the following TypeScript snippet for dequeue operation in a circular queue, identify the bug:

function dequeue() {
  if (front === -1) return null;
  const value = queue[front];
  if (front === rear) {
    front = rear = -1;
  } else {
    front = front + 1;
  }
  return value;
}

medium

A. Missing modulo operation when updating front, causing index overflow.

B. Incorrect check for empty queue; should check rear instead of front.

C. Should update rear instead of front after dequeue.

D. Returning value before updating front causes wrong output.

Solution

1. Step 1: Analyze front update in dequeue

   The code updates front as front = front + 1 without modulo, which can exceed array bounds.

2. Step 2: Correct update with modulo

   To wrap around in circular queue, front should be updated as front = (front + 1) % size.

3. Final Answer:

   Missing modulo operation when updating front, causing index overflow. -> Option A

4. Quick Check:

   Front update needs modulo [OK]

Hint: Always use modulo when moving front or rear [OK]

Common Mistakes:

• Forgetting modulo causes index errors
• Confusing front and rear updates
• Wrong empty queue condition check

5. You have a circular queue of size 4. You perform these operations in order: enqueue(5), enqueue(10), enqueue(15), dequeue(), enqueue(20), enqueue(25). What will be the final state of the queue array (index 0 to 3) and the positions of front and rear?

hard

A. Queue: [25, 10, 15, 20], front = 1, rear = 0

B. Queue: [5, 10, 15, 20], front = 1, rear = 3

C. Queue: [20, 25, 15, 10], front = 0, rear = 1

D. Queue: [10, 15, 20, 25], front = 0, rear = 3

Solution

1. Step 1: Perform enqueues and dequeue step-by-step

   Start: front = -1, rear = -1, queue size = 4 - enqueue(5): front=0, rear=0, queue=[5,_,_,_] - enqueue(10): rear=1, queue=[5,10,_,_] - enqueue(15): rear=2, queue=[5,10,15,_] - dequeue(): front=1, queue unchanged - enqueue(20): rear=3, queue=[5,10,15,20] - enqueue(25): rear=(3+1)%4=0, queue[0]=25, queue=[25,10,15,20]

2. Step 2: Final front and rear positions

   front=1 (pointing to 10), rear=0 (pointing to 25). Queue array is [25,10,15,20].

3. Final Answer:

   Queue: [25, 10, 15, 20], front = 1, rear = 0 -> Option A

4. Quick Check:

   Wrap rear with modulo, front moves on dequeue [OK]

Hint: Track front/rear with modulo after each operation [OK]

Common Mistakes:

• Not wrapping rear index on enqueue
• Not updating front on dequeue
• Mixing up front and rear positions