# Posts published in “Desgin”

Design your implementation of the circular queue. The circular queue is a linear data structure in which the operations are performed based on FIFO (First In First Out) principle and the last position is connected back to the first position to make a circle. It is also called “Ring Buffer”.

One of the benefits of the circular queue is that we can make use of the spaces in front of the queue. In a normal queue, once the queue becomes full, we cannot insert the next element even if there is a space in front of the queue. But using the circular queue, we can use the space to store new values.

Your implementation should support following operations:

• MyCircularQueue(k): Constructor, set the size of the queue to be k.
• Front: Get the front item from the queue. If the queue is empty, return -1.
• Rear: Get the last item from the queue. If the queue is empty, return -1.
• enQueue(value): Insert an element into the circular queue. Return true if the operation is successful.
• deQueue(): Delete an element from the circular queue. Return true if the operation is successful.
• isEmpty(): Checks whether the circular queue is empty or not.
• isFull(): Checks whether the circular queue is full or not.

Example:

MyCircularQueue circularQueue = new MyCircularQueue(3); // set the size to be 3
circularQueue.enQueue(1);  // return true
circularQueue.enQueue(2);  // return true
circularQueue.enQueue(3);  // return true
circularQueue.enQueue(4);  // return false, the queue is full
circularQueue.Rear();  // return 3
circularQueue.isFull();  // return true
circularQueue.deQueue();  // return true
circularQueue.enQueue(4);  // return true
circularQueue.Rear();  // return 4


Note:

• All values will be in the range of [0, 1000].
• The number of operations will be in the range of [1, 1000].
• Please do not use the built-in Queue library.

## Solution: Simulate with an array

We need a fixed length array, and the head location as well as the size of the current queue.

We can use q[head] to access the front, and q[(head + size – 1) % k] to access the rear.

Time complexity: O(1) for all the operations.
Space complexity: O(k)