Leetcode #641: Design Circular Deque
In this guide, we solve Leetcode #641 Design Circular Deque 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 double-ended queue (deque). Implement the MyCircularDeque class: MyCircularDeque(int k) Initializes the deque with a maximum size of k.
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
["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
Python Solution
class MyCircularDeque:
def __init__(self, k: int):
self.k = k + 1
self.data = [0] * self.k
self.front = 0
self.rear = 0
def insertFront(self, value: int) -> bool:
if self.isFull():
return False
self.front = (self.front - 1) % self.k
self.data[self.front] = value
return True
def insertLast(self, value: int) -> bool:
if self.isFull():
return False
self.data[self.rear] = value
self.rear = (self.rear + 1) % self.k
return True
def deleteFront(self) -> bool:
if self.isEmpty():
return False
self.front = (self.front + 1) % self.k
return True
def deleteLast(self) -> bool:
if self.isEmpty():
return False
self.rear = (self.rear - 1) % self.k
return True
def getFront(self) -> int:
return -1 if self.isEmpty() else self.data[self.front]
def getRear(self) -> int:
return -1 if self.isEmpty() else self.data[(self.rear - 1) % self.k]
def isEmpty(self) -> bool:
return self.front == self.rear
def isFull(self) -> bool:
return (self.rear + 1) % self.k == self.front
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.