# Solving Leetcode Interviews in Seconds with AI: Design Circular Queue


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "622" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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. Implement the MyCircularQueue class:  MyCircularQueue(k) Initializes the object with the size of the queue to be k. int Front() Gets the front item from the queue. If the queue is empty, return -1. int Rear() Gets the last item from the queue. If the queue is empty, return -1. boolean enQueue(int value) Inserts an element into the circular queue. Return true if the operation is successful. boolean deQueue() Deletes an element from the circular queue. Return true if the operation is successful. boolean isEmpty() Checks whether the circular queue is empty or not. boolean isFull() Checks whether the circular queue is full or not.  You must solve the problem without using the built-in queue data structure in your programming language.    Example 1:  Input ["MyCircularQueue", "enQueue", "enQueue", "enQueue", "enQueue", "Rear", "isFull", "deQueue", "enQueue", "Rear"] [[3], [1], [2], [3], [4], [], [], [], [4], []] Output [null, true, true, true, false, 3, true, true, true, 4]  Explanation MyCircularQueue myCircularQueue = new MyCircularQueue(3); myCircularQueue.enQueue(1); // return True myCircularQueue.enQueue(2); // return True myCircularQueue.enQueue(3); // return True myCircularQueue.enQueue(4); // return False myCircularQueue.Rear();     // return 3 myCircularQueue.isFull();   // return True myCircularQueue.deQueue();  // return True myCircularQueue.enQueue(4); // return True myCircularQueue.Rear();     // return 4    Constraints:  1 <= k <= 1000 0 <= value <= 1000 At most 3000 calls will be made to enQueue, deQueue, Front, Rear, isEmpty, and isFull.  

	# Explanation
	*   **Circular Array:** Use a fixed-size array to store the queue elements. The `front` and `rear` pointers will move circularly within this array.
*   **Modulo Arithmetic:** Employ the modulo operator (%) to ensure the `front` and `rear` pointers wrap around to the beginning of the array when they reach the end, achieving the circular behavior.
*   **Empty/Full Conditions:** Differentiate between an empty and a full queue using the `front` and `rear` pointers. An empty queue occurs when `front == rear`. A full queue occurs when `(rear + 1) % capacity == front`.

*   **Runtime Complexity:** O(1) for all operations. **Storage Complexity:** O(k), where k is the size of the queue.

	
	# Code
	```python
	class MyCircularQueue:

    def __init__(self, k: int):
        self.queue = [None] * k
        self.capacity = k
        self.front = 0
        self.rear = 0
        self.size = 0

    def enQueue(self, value: int) -> bool:
        if self.isFull():
            return False

        self.queue[self.rear] = value
        self.rear = (self.rear + 1) % self.capacity
        self.size += 1
        return True

    def deQueue(self) -> bool:
        if self.isEmpty():
            return False

        self.front = (self.front + 1) % self.capacity
        self.size -= 1
        return True

    def Front(self) -> int:
        if self.isEmpty():
            return -1
        return self.queue[self.front]

    def Rear(self) -> int:
        if self.isEmpty():
            return -1
        return self.queue[(self.rear - 1) % self.capacity]

    def isEmpty(self) -> bool:
        return self.size == 0

    def isFull(self) -> bool:
        return self.size == self.capacity
	```