622. Design Circular Queue - Explanation

Description

Design and implement 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.

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1. Brute Force

class MyCircularQueue:

    def __init__(self, k: int):
        self.q = []
        self.k = k

    def enQueue(self, value: int) -> bool:
        if len(self.q) == self.k:
            return False
        self.q.append(value)
        return True

    def deQueue(self) -> bool:
        if not self.q:
            return False
        self.q.pop(0)
        return True

    def Front(self) -> int:
        if self.q:
            return self.q[0]
        return -1

    def Rear(self) -> int:
        if self.q:
            return self.q[-1]
        return -1

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

    def isFull(self) -> bool:
        return len(self.q) == self.k

Time & Space Complexity

Time complexity:
- $O(1)$ time for initialization.
- $O(1)$ time for each $enQueue()$ , $Front()$ , $Rear()$ , $isEmpty()$ and $isFull()$ function calls.
- $O(n)$ time for each $deQueue()$ function call.
Space complexity: $O(n)$

Where $n$ is the size of the queue.

2. Array

class MyCircularQueue:

    def __init__(self, k: int):
        self.q = [0] * k
        self.k = k
        self.front = 0
        self.rear = -1
        self.size = 0

    def enQueue(self, value: int) -> bool:
        if self.isFull():
            return False
        self.rear = (self.rear + 1) % self.k
        self.q[self.rear] = value
        self.size += 1
        return True

    def deQueue(self) -> bool:
        if self.isEmpty():
            return False
        self.front = (self.front + 1) % self.k
        self.size -= 1
        return True

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

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

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

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

Time & Space Complexity

Time complexity:
- $O(n)$ time for initialization.
- $O(1)$ time for each $enQueue()$ , $deQueue()$ , $Front()$ , $Rear()$ , $isEmpty()$ and $isFull()$ function calls.
Space complexity: $O(n)$

Where $n$ is the size of the queue.

3. Doubly Linked List

class ListNode:

    def __init__(self, val, nxt, prev):
        self.val, self.next, self.prev = val, nxt, prev

class MyCircularQueue:

    def __init__(self, k: int):
        self.space = k
        self.left = ListNode(0, None, None)
        self.right = ListNode(0, None, self.left)
        self.left.next = self.right

    def enQueue(self, value: int) -> bool:
        if self.isFull(): return False
        cur = ListNode(value, self.right, self.right.prev)
        self.right.prev.next = cur
        self.right.prev = cur
        self.space -= 1
        return True

    def deQueue(self) -> bool:
        if self.isEmpty(): return False
        self.left.next = self.left.next.next
        self.left.next.prev = self.left
        self.space += 1
        return True

    def Front(self) -> int:
        if self.isEmpty(): return -1
        return self.left.next.val

    def Rear(self) -> int:
        if self.isEmpty(): return -1
        return self.right.prev.val

    def isEmpty(self) -> bool:
        return self.left.next == self.right

    def isFull(self) -> bool:
        return self.space == 0

Time & Space Complexity

Time complexity:
- $O(n)$ time for initialization.
- $O(1)$ time for each $enQueue()$ , $deQueue()$ , $Front()$ , $Rear()$ , $isEmpty()$ and $isFull()$ function calls.
Space complexity: $O(n)$

Where $n$ is the size of the queue.

4. Singly Linked List

class ListNode:
    def __init__(self, val, nxt=None):
        self.val = val
        self.next = nxt

class MyCircularQueue:
    def __init__(self, k: int):
        self.space = k
        self.left = ListNode(0)
        self.right = self.left

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

        cur = ListNode(value)
        if self.isEmpty():
            self.left.next = cur
            self.right = cur
        else:
            self.right.next = cur
            self.right = cur

        self.space -= 1
        return True

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

        self.left.next = self.left.next.next
        if self.left.next is None:
            self.right = self.left

        self.space += 1
        return True

    def Front(self) -> int:
        if self.isEmpty(): return -1
        return self.left.next.val

    def Rear(self) -> int:
        if self.isEmpty(): return -1
        return self.right.val

    def isEmpty(self) -> bool:
        return self.left.next is None

    def isFull(self) -> bool:
        return self.space == 0

Time & Space Complexity

Time complexity:
- $O(n)$ time for initialization.
- $O(1)$ time for each $enQueue()$ , $deQueue()$ , $Front()$ , $Rear()$ , $isEmpty()$ and $isFull()$ function calls.
Space complexity: $O(n)$

Where $n$ is the size of the queue.