How do you solve Design Circular Queue on LeetCode?

Design Circular Queue (LeetCode #622) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(n) space complexity. Key insight: Circular queues efficiently utilize space by wrapping around.

What is the brute force solution for Design Circular Queue?

The brute force approach for Design Circular Queue is Brute Force, with O(n²) time complexity and O(1) space complexity. In a brute-force approach, we can use a simple list to represent the queue and manage the front and rear indices. Each time we enqueue or dequeue, we will adjust these indices manually, but this can lead to inefficiencies when the queue wraps around. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Design Circular Queue?

Design Circular Queue uses the following concepts: Array, Linked List, Design, Queue. The recommended patterns to study are: Array, Queue.

What are the interview tips for Design Circular Queue?

Clearly explain your thought process while coding. Test edge cases like enqueueing into a full queue. Be prepared to discuss the trade-offs of different implementations.

What are common mistakes when solving Design Circular Queue?

Not handling the wrap-around condition correctly. Confusing the front and rear pointers during enqueue and dequeue.

#622

Design Circular Queue

Medium

ArrayLinked ListDesignQueueArrayQueue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(n)

💡

Intuition

Time O(1)Space O(n)

The optimal solution uses a fixed-size array and two pointers to efficiently manage the front and rear of the queue. This avoids the need for shifting elements, making operations O(1).

⚙️

Algorithm

5 steps

1Step 1: Initialize an array of size k, and two pointers: front and rear, both set to -1.
2Step 2: For enQueue, check if the queue is full. If not, increment rear and add the value. If the queue was empty, set front to 0.
3Step 3: For deQueue, check if the queue is empty. If not, increment front. If front and rear meet, reset them to -1.
4Step 4: Implement Front and Rear methods to return the respective values from the queue using the pointers.
5Step 5: Implement isEmpty and isFull methods to check the status of the queue based on the pointers.

solution.py36 lines

1class MyCircularQueue:
2    def __init__(self, k: int):
3        self.queue = [0] * k
4        self.size = k
5        self.front = self.rear = -1
6
7    def enQueue(self, value: int) -> bool:
8        if self.isFull():
9            return False
10        if self.isEmpty():
11            self.front = self.rear = 0
12        else:
13            self.rear = (self.rear + 1) % self.size
14        self.queue[self.rear] = value
15        return True
16
17    def deQueue(self) -> bool:
18        if self.isEmpty():
19            return False
20        if self.front == self.rear:
21            self.front = self.rear = -1
22        else:
23            self.front = (self.front + 1) % self.size
24        return True
25
26    def Front(self) -> int:
27        return -1 if self.isEmpty() else self.queue[self.front]
28
29    def Rear(self) -> int:
30        return -1 if self.isEmpty() else self.queue[self.rear]
31
32    def isEmpty(self) -> bool:
33        return self.front == -1
34
35    def isFull(self) -> bool:
36        return (self.rear + 1) % self.size == self.front

ℹ

Complexity note: The time complexity is O(1) for all operations because we are using fixed-size array and pointers to manage the queue without shifting elements. The space complexity is O(n) due to the array storage.

1Circular queues efficiently utilize space by wrapping around.
2Pointers help manage the front and rear without shifting elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.