Circular Queue Implementation Using Array in DSA C# - Build from Scratch

Circular Queue Implementation Using Array

📖 Scenario: Imagine a small bakery that uses a circular queue to manage orders. The queue holds a fixed number of orders, and when the queue is full, new orders wait until space is free. This helps the bakery serve customers in the order they arrived without losing any orders.

🎯 Goal: You will build a simple circular queue using an array in C#. You will create the queue, set its size, add orders, and then display the queue's current state.

📋 What You'll Learn

Create an array to hold the queue elements

Use variables to track the front and rear positions

Implement enqueue operation to add elements circularly

Print the queue elements in order

💡 Why This Matters

🌍 Real World

Circular queues are used in real-world systems like task scheduling, buffering data streams, and managing resources where fixed-size queues are needed.

💼 Career

Understanding circular queues helps in roles involving system design, embedded programming, and performance optimization where efficient data handling is critical.

1

Create the Circular Queue Array and Variables

Create an integer array called queue with size 5. Also create two integer variables called front and rear, both initialized to -1.

DSA C#

int[] queue = new int[5];
int front = -1;
int rear = -1;
// Your code here

Hint

Use int[] queue = new int[5]; to create the array. Initialize front and rear to -1 to show the queue is empty.

2

Add a Variable for Queue Size

Create an integer variable called size and set it to 5 to store the maximum size of the queue.

DSA C#

int[] queue = new int[5];
int front = -1;
int rear = -1;
int size = 5;
// Your code here

Hint

Use int size = 5; to keep track of the queue's capacity.

3

Enqueue Elements in Circular Manner

Write code to add three elements 10, 20, and 30 to the queue using circular enqueue logic. Update front and rear accordingly. Use the variables queue, front, rear, and size.

DSA C#

int[] queue = new int[5];
int front = -1;
int rear = -1;
int size = 5;

// Add 10, 20, 30 to the queue using circular enqueue logic
// Your code here

Hint

When adding an element, increase rear by 1 modulo size. If front is -1, set it to 0 because the queue was empty.

4

Print the Current Queue Elements

Write code to print all elements in the queue from front to rear in order, handling the circular wrap. Use a while loop and print elements separated by spaces.

DSA C#

int[] queue = new int[5];
int front = -1;
int rear = -1;
int size = 5;

// Enqueue 10
if (front == -1) front = 0;
rear = (rear + 1) % size;
queue[rear] = 10;

// Enqueue 20
rear = (rear + 1) % size;
queue[rear] = 20;

// Enqueue 30
rear = (rear + 1) % size;
queue[rear] = 30;

// Print queue elements
// Your code here

Hint

Start from front and print elements until you reach rear. Use modulo size to wrap around.

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