Introduction

In this blog post, we will explore how to solve the LeetCode problem "641" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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 double-ended queue (deque). Implement the MyCircularDeque class: MyCircularDeque(int k) Initializes the deque with a maximum size of k. boolean insertFront() Adds an item at the front of Deque. Returns true if the operation is successful, or false otherwise. boolean insertLast() Adds an item at the rear of Deque. Returns true if the operation is successful, or false otherwise. boolean deleteFront() Deletes an item from the front of Deque. Returns true if the operation is successful, or false otherwise. boolean deleteLast() Deletes an item from the rear of Deque. Returns true if the operation is successful, or false otherwise. int getFront() Returns the front item from the Deque. Returns -1 if the deque is empty. int getRear() Returns the last item from Deque. Returns -1 if the deque is empty. boolean isEmpty() Returns true if the deque is empty, or false otherwise. boolean isFull() Returns true if the deque is full, or false otherwise. Example 1: Input ["MyCircularDeque", "insertLast", "insertLast", "insertFront", "insertFront", "getRear", "isFull", "deleteLast", "insertFront", "getFront"] [[3], [1], [2], [3], [4], [], [], [], [4], []] Output [null, true, true, true, false, 2, true, true, true, 4] Explanation MyCircularDeque myCircularDeque = new MyCircularDeque(3); myCircularDeque.insertLast(1); // return True myCircularDeque.insertLast(2); // return True myCircularDeque.insertFront(3); // return True myCircularDeque.insertFront(4); // return False, the queue is full. myCircularDeque.getRear(); // return 2 myCircularDeque.isFull(); // return True myCircularDeque.deleteLast(); // return True myCircularDeque.insertFront(4); // return True myCircularDeque.getFront(); // return 4 Constraints: 1 <= k <= 1000 0 <= value <= 1000 At most 2000 calls will be made to insertFront, insertLast, deleteFront, deleteLast, getFront, getRear, isEmpty, isFull.

Explanation

• Circular Buffer: Implement the deque using a fixed-size array (circular buffer). This allows efficient insertion and deletion at both ends using modulo arithmetic to wrap around the array.
  • Pointers: Use two pointers, head and tail, to track the front and rear of the deque. size variable keeps track of current number of elements.
  • Handle Edge Cases: Carefully handle edge cases like empty and full deque conditions to ensure correct operation and avoid errors.

• Runtime Complexity: O(1) for all operations. Storage Complexity: O(k), where k is the maximum size of the deque.

Code

    class MyCircularDeque:

    def __init__(self, k: int):
        self.deque = [0] * k
        self.head = 0
        self.tail = 0
        self.size = 0
        self.capacity = k

    def insertFront(self, value: int) -> bool:
        if self.isFull():
            return False
        self.head = (self.head - 1) % self.capacity
        self.deque[self.head] = value
        self.size += 1
        return True

    def insertLast(self, value: int) -> bool:
        if self.isFull():
            return False
        self.deque[self.tail] = value
        self.tail = (self.tail + 1) % self.capacity
        self.size += 1
        return True

    def deleteFront(self) -> bool:
        if self.isEmpty():
            return False
        self.head = (self.head + 1) % self.capacity
        self.size -= 1
        return True

    def deleteLast(self) -> bool:
        if self.isEmpty():
            return False
        self.tail = (self.tail - 1) % self.capacity
        self.size -= 1
        return True

    def getFront(self) -> int:
        if self.isEmpty():
            return -1
        return self.deque[self.head]

    def getRear(self) -> int:
        if self.isEmpty():
            return -1
        return self.deque[(self.tail - 1) % self.capacity]

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

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