Circular Queue Implementation Using Array in DSA C# - 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 in a circular queue using an array?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class CircularQueue {
    private int[] data;
    private int front, rear, size, capacity;

    public CircularQueue(int capacity) {
        this.capacity = capacity;
        data = new int[capacity];
        front = 0;
        rear = -1;
        size = 0;
    }

    public void Enqueue(int item) {
        if (size == capacity) throw new InvalidOperationException("Queue full");
        rear = (rear + 1) % capacity;
        data[rear] = item;
        size++;
    }

    public int Dequeue() {
        if (size == 0) throw new InvalidOperationException("Queue empty");
        int item = data[front];
        front = (front + 1) % capacity;
        size--;
        return item;
    }
}
    

This code implements a circular queue using an array with fixed capacity.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Each Enqueue or Dequeue moves a pointer and updates one element.
• How many times: Each operation runs 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
10	1 per operation
100	1 per operation
1000	1 per operation

Pattern observation: The time per operation stays the same no matter how big the queue is.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing an item takes the same short time no matter the queue size.

Common Mistake

[X] Wrong: "Because the queue uses an array, Enqueue or Dequeue must scan or shift many elements."

[OK] Correct: The circular queue uses pointers to track front and rear, so it never moves all elements. Each operation updates only one position.

Interview Connect

Understanding that circular queues provide constant time operations shows you know how to efficiently manage fixed-size buffers, a common real-world need.

Self-Check

"What if we changed the array to a linked list? 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 efficiently uses the array space by reusing freed slots.

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

C. It automatically sorts the elements in the queue.

D. It increases the maximum size of the queue beyond the array length.

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 are empty slots at the front.

2. Step 2: Circular queue advantage

   A circular queue treats the array as circular, so when rear reaches the end, it wraps around to use empty slots at the front, efficiently using all available space.

3. Final Answer:

   It efficiently uses the array space by reusing freed slots. -> Option A

4. Quick Check:

   Circular queue reuses space = Advantage A [OK]

Hint: Circular queue reuses space freed at front [OK]

Common Mistakes:

• Thinking circular queue increases array size
• Believing it sorts elements automatically
• Assuming it stores different data types

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

easy

A. rear == n

B. rear == front

C. (rear + 1) % n == front

D. front == 0 && rear == n - 1

Solution

1. Step 1: Understand circular queue full condition

   In a circular queue, the queue is full when the next position of rear (circularly) is front.

2. Step 2: Express condition mathematically

   This means (rear + 1) modulo n equals front, ensuring no overlap and full queue detection.

3. Final Answer:

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

4. Quick Check:

   Full if next rear equals front = D [OK]

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

Common Mistakes:

• Using rear == front as full (actually empty)
• Checking rear == n ignoring wrap-around
• Using front == 0 && rear == n - 1 only (linear 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 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 operation

   Enqueue increments rear circularly: rear = (3 + 1) % 5 = 4.

2. Step 2: Perform Dequeue operation

   Dequeue increments front circularly: front = (0 + 1) % 5 = 1.

3. Final Answer:

   front = 1, rear = 4 -> Option D

4. Quick Check:

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

Hint: Increment indices modulo size after enqueue/dequeue [OK]

Common Mistakes:

• Not wrapping indices with modulo
• Updating front on enqueue or rear on dequeue
• Confusing initial and final index values

4. Given this snippet of circular queue code, what is the error?

int n = 5;
int front = 0, rear = 0;
int[] queue = new int[n];

// Enqueue operation
rear = (rear + 1) % n;
queue[rear] = 10;

// Dequeue operation
front = (front + 1) % n;
int x = queue[front];

// Check if queue is empty
if (front == rear)
    Console.WriteLine("Queue is empty");

medium

A. The enqueue operation increments rear before inserting, causing off-by-one error.

B. The initial values of front and rear cause the first enqueue to overwrite data.

C. The empty check condition is incorrect; it should be (rear + 1) % n == front.

D. The dequeue operation should decrement front instead of incrementing.

Solution

1. Step 1: Analyze enqueue operation

   Incrementing rear before inserting means the first element is placed at index 1, skipping index 0, which wastes space and causes off-by-one errors.

2. Step 2: Understand correct enqueue logic

   Enqueue should insert at current rear, then increment rear after insertion or handle indices carefully to avoid skipping slots.

3. Final Answer:

   The enqueue operation increments rear before inserting, causing off-by-one error. -> Option A

4. Quick Check:

   Increment rear before insert = off-by-one error B [OK]

Hint: Insert before incrementing rear to avoid skipping slots [OK]

Common Mistakes:

• Confusing empty and full conditions
• Incrementing front on enqueue or rear on dequeue
• Ignoring initial index positions

5. You have a circular queue of size 6. Initially, front = 4 and rear = 2. The queue currently holds 4 elements. Which of the following sequences of operations will cause the queue to become full?

hard

A. Enqueue three times, then dequeue once, then enqueue once.

B. Dequeue once, then enqueue three times.

C. Enqueue three times without any dequeue.

D. Dequeue twice, then enqueue five times.

Solution

1. Step 1: Calculate current queue state

   Size is 6, front=4, rear=2, holding 4 elements means the queue is partially filled with wrap-around.

2. Step 2: Analyze each operation sequence

   A: Dequeue once (remove 1), enqueue three times (add 3) -> net +2 elements, total 6 (full). B: Enqueue three times (add 3), dequeue once (remove 1), enqueue once (add 1) -> net +3 elements, total 7 (overflow). C: Enqueue three times without dequeue would overflow (since only 2 slots free). D: Dequeue twice (remove 2), enqueue five times (add 5) -> net +3 elements, total 7 (overflow).

3. Step 3: Identify which sequence is valid without overflow

   Only Dequeue once, then enqueue three times. correctly balances operations to reach full without overflow errors.

4. Final Answer:

   Dequeue once, then enqueue three times. -> Option B

5. Quick Check:

   Dequeue 1 + Enqueue 3 fills queue = A [OK]

Hint: Track net enqueue minus dequeue to reach full capacity [OK]

Common Mistakes:

• Ignoring wrap-around effect on indices
• Assuming enqueue always succeeds without checking full
• Miscounting elements after operations