Design a queue using circular array

Question

Design a queue using circular array Given the skeleton of a Queue class, complete this class by implementing all the queue operations below. > - **Queue(int capacity)** - Initializes the Queue object with the given capacity. > - **size()** - Returns the current size of the queue. > - **empty()** - Returns `true` if the queue is empty and `false` if not. > - **front()** - Returns the element at the front of the queue. If the queue is empty, it returns `-1`. > - **back()** - Returns the element at the back of the queue. If the queue is empty, it returns `-1`. > - **enqueue(int val)** - Adds the given value to the queue and returns `true` if the operation is successful. Returns `false` if the queue is full. > - **dequeue()** - Removes the front element from the queue and returns its value. If the queue is empty, it returns `-1`. You must abide by the following constraints > The input should adhere to the following rules: > 1. The input should contain two arrays of the same size. > 2. The first array should contain the list of operations, while the second should contain the corresponding operands for those operations. > 3. The first index in the first array should contain **Queue**, and the first index in the second array should contain a single positive integer representing the capacity of the queue. This value is used to initialise the queue. > 4. For each index in the first array that contains the **enqueue** operation, the corresponding index in the second array should contain the value that needs to be pushed. > 5. For each index in the first array that contains **size**, **empty**, **front**, **back**, or **dequeue** operations, the corresponding index in the second array should contain an empty array. > > **Example:** > - **Input:** [Queue, enqueue, back, enqueue, front, empty, dequeue, front, enqueue, enqueue, empty] [[2], [2], [], [3], [], [], [], [], [8], [9], []] > > - **Output:** [null, true, 2, true, 2, false, 2, 3, true, false, false] > > **Explanation:** > > **Operation:** Queue queue = new Queue(2) > **Result:** Initializes an empty `Queue` with a capacity of 2 > > **Operation:** queue.enqueue(2) > **Result:** `queue = [2]`, returns `true` > > **Operation:** queue.back() > **Result:** `queue = [2]`, returns `2` > > **Operation:** queue.enqueue(3) > **Result:** `queue = [2, 3]`, returns `true` > > **Operation:** queue.front() > **Result:** `queue = [2, 3]`, returns `2` > > **Operation:** queue.empty() > **Result:** `queue = [2, 3]`, returns `false` > > **Operation:** queue.dequeue() > **Result:** `queue = [3]`, returns `2` > > **Operation:** queue.front() > **Result:** `queue = [3]`, returns `3` > > **Operation:** queue.enqueue(8) > **Result:** `queue = [3, 8]`, returns `true` > > **Operation:** queue.enqueue(9) > **Result:** `queue = [3, 8]`, queue is full, returns `false` > > **Operation:** queue.empty() > **Result:** `queue = [3, 8]`, returns `false`

Accepted Answer

Optimal Solution: Time Complexity All operations in O(1), Space Complexity O(1). Codeintuition provides an interactive coding environment for this problem.