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

Time Complexity: Circular Queue Implementation Using Array

O(1)

Understanding Time Complexity

We want to understand how fast operations like adding or removing items work in a circular queue using an array.

How does the time needed change as the queue gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func (q *CircularQueue) Enqueue(value int) bool {
    if (q.rear+1)%q.size == q.front {
        return false // Queue is full
    }
    q.rear = (q.rear + 1) % q.size
    q.array[q.rear] = value
    return true
}

func (q *CircularQueue) Dequeue() (int, bool) {
    if q.front == q.rear {
        return 0, false // Queue is empty
    }
    q.front = (q.front + 1) % q.size
    return q.array[q.front], true
}
    

This code adds (Enqueue) and removes (Dequeue) items in a circular queue using an array with fixed size.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Single step index update and assignment or retrieval in array.
• How many times: Each operation runs once per call, no loops or recursion 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	5 steps
100	5 steps
1000	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 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 takes longer as the queue gets full or empty."

[OK] Correct: The operations only update indexes and access one array spot, so time does not grow with queue size.

Interview Connect

Knowing that circular queue operations run in constant time shows you understand efficient data handling, a key skill in many coding tasks.

Self-Check

"What if we used a linked list instead of an array for the circular queue? How would the time complexity 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 the available space by reusing freed slots.

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 connects the end of the array back to the front, allowing reuse of empty slots freed by dequeued elements.

3. Final Answer:

   It efficiently uses the available space by reusing freed slots. -> Option B

4. Quick Check:

   Space reuse = Advantage [OK]

Hint: Circular queue reuses space freed by dequeues [OK]

Common Mistakes:

• Thinking circular queue increases array size
• Believing it sorts elements automatically
• Confusing data type flexibility with queue type

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

easy

A. if front == -1

B. if rear == front

C. if rear == n - 1

D. if (rear + 1) % n == front

Solution

1. Step 1: Understand full condition in circular queue

   The queue is full when the next position of rear (circularly) equals front.

2. Step 2: Apply modulo for circular increment

   Using modulo n ensures wrap-around, so (rear + 1) % n == front means full.

3. Final Answer:

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

4. Quick Check:

   Full condition uses modulo wrap [OK]

Hint: Full when next rear equals front modulo array size [OK]

Common Mistakes:

• Checking rear == front means empty, not full
• Using rear == n - 1 ignores circular wrap
• Confusing front == -1 as full condition

3. Consider a circular queue of size 5 (indices 0 to 4). Initially, front = -1 and rear = -1. The following operations are performed:
enqueue(10), enqueue(20), enqueue(30), dequeue(), enqueue(40), enqueue(50), enqueue(60)
What will be the values of front and rear after these operations?

medium

A. front = 1, rear = 0

B. front = 0, rear = 4

C. front = 1, rear = 3

D. front = 2, rear = 1

Solution

1. Step 1: Trace enqueue and dequeue operations

   Start: front = -1, rear = -1
   enqueue(10): front=0, rear=0
   enqueue(20): rear=1
   enqueue(30): rear=2
   dequeue(): front=1 (removes 10)
   enqueue(40): rear=3
   enqueue(50): rear=4
   enqueue(60): rear=(4+1)%5=0 (wraps around)

2. Step 2: Final front and rear values

   front = 1, rear = 0 after last enqueue.

3. Final Answer:

   front = 1, rear = 0 -> Option A

4. Quick Check:

   Wrap-around rear = 0, front moved after dequeue [OK]

Hint: Track front/rear changes step-by-step with modulo [OK]

Common Mistakes:

• Not wrapping rear index after reaching end
• Forgetting to update front after dequeue
• Assuming front stays at -1 after first enqueue

4. Given the following Go code snippet for a circular queue, what is the bug that causes incorrect enqueue operation?

func enqueue(q []int, front *int, rear *int, n int, value int) {
    if (*rear+1)%n == *front {
        fmt.Println("Queue is full")
        return
    }
    if *front == -1 {
        *front = 0
    }
    *rear = (*rear + 1) % n
    q[*rear] = value
}

medium

A. The check for full queue is incorrect; it should be rear == front.

B. The queue slice q is passed by value, so changes to q are not saved.

C. The front pointer is not updated when queue is empty.

D. The rear index is incremented before checking if queue is full.

Solution

1. Step 1: Check handling of empty queue

   The code updates front to 0 only if front == -1, handling the initial empty state correctly.

2. Step 2: Issue after full dequeues

   After dequeuing all elements, dequeue updates front = (front + 1) % n, making (*rear + 1) % n == *front true, falsely detecting full. Enqueue fails to update/reset front for subsequent empty states.

3. Final Answer:

   The front pointer is not updated when queue is empty. -> Option C

4. Quick Check:

   Handle non-initial empty by resetting front [OK]

Hint: Reset front when queue empty after dequeues [OK]

Common Mistakes:

• Confusing full condition with rear == front
• Overlooking empty queue handling after dequeues
• Thinking slice passing causes the issue

5. You have a circular queue implemented using an array of size 6. Initially, front = 4 and rear = 2. Which of the following sequences of operations will cause the queue to become full?

hard

A. dequeue, enqueue, enqueue, enqueue

B. enqueue, dequeue, enqueue, dequeue

C. dequeue, dequeue, enqueue, enqueue

D. dequeue, enqueue, dequeue

Solution

1. Step 1: Understand initial state and full condition

   Array size n=6, front=4, rear=2 (5 elements occupied, 1 free). Full: (rear + 1) % 6 == front.

2. Step 2: Simulate each option

   Only D causes final full state:
   dequeue: front=5, rear=2 (4 elem)
   enqueue: rear=3 (5 elem)
   enqueue: rear=4 (6 elem, now full since (4+1)%6=5==front)
   enqueue: full detected, no change. Final: front=5, rear=4 full.
   Others end not full.

3. Final Answer:

   dequeue, enqueue, enqueue, enqueue -> Option A

4. Quick Check:

   Simulate steps to check final full condition [OK]

Hint: Simulate front/rear moves to check full condition [OK]

Common Mistakes:

• Assuming enqueue always fills queue immediately
• Ignoring circular wrap in index calculations
• Not updating front on dequeue