Leetcode #622: Design Circular Queue
In this guide, we solve Leetcode #622 Design Circular Queue in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Queue, Array, Linked List
Intuition
We need level-order exploration or shortest-step expansion, which maps directly to a queue.
BFS guarantees the first time you reach a node is the shortest in unweighted graphs.
Approach
Initialize the queue with starting nodes and expand outward layer by layer.
Track visited nodes to avoid cycles and redundant work.
Steps:
- Initialize a queue with start nodes.
- Pop, process, and enqueue neighbors.
- Track visited nodes.
Example
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
Python Solution
class MyCircularQueue:
def __init__(self, k: int):
self.q = [0] * k
self.size = 0
self.capacity = k
self.front = 0
def enQueue(self, value: int) -> bool:
if self.isFull():
return False
self.q[(self.front + self.size) % self.capacity] = value
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:
return -1 if self.isEmpty() else self.q[self.front]
def Rear(self) -> int:
if self.isEmpty():
return -1
return self.q[(self.front + self.size - 1) % self.capacity]
def isEmpty(self) -> bool:
return self.size == 0
def isFull(self) -> bool:
return self.size == self.capacity
# Your MyCircularQueue object will be instantiated and called as such:
# obj = MyCircularQueue(k)
# param_1 = obj.enQueue(value)
# param_2 = obj.deQueue()
# param_3 = obj.Front()
# param_4 = obj.Rear()
# param_5 = obj.isEmpty()
# param_6 = obj.isFull()
Complexity
The time complexity is O(V+E). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.