Design Circular Deque | Simplest Explanation | 2 Ways | Leetcode 641 | codestorywithMIK
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In this video we will try to solve a good Design DSA Problem : Design Circular Deque | Simplest Explanation | 2 Ways | Leetcode 641 | codestorywithMIK
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Problem Name : Design Circular Deque | Simplest Explanation | 2 Ways | Leetcode 641 | codestorywithMIK
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Summary :
The implementation uses an array (deq) to simulate a circular deque with a fixed size K. The class maintains pointers (front and rear) to track the current front and rear positions in the array, along with currentCount to store the number of elements currently present.
Initialization:
The deque is represented as a vector of size K.
front is initialized to 0 and rear to K - 1.
Insertions:
insertFront(value): Decrements the front index by 1 (using modulo arithmetic to wrap around) and assigns the value to deq[front].
insertLast(value): Increments the rear index by 1 (also using modulo) and assigns the value to deq[rear].
Deletions:
deleteFront(): Increments front to remove the element at the front.
deleteLast(): Decrements rear to remove the element at the rear.
Queries:
getFront(): Returns the element at front or -1 if the deque is empty.
getRear(): Returns the element at rear or -1 if empty.
State Checks:
isEmpty(): Returns true if currentCount is 0.
isFull(): Returns true if currentCount equals K.
The approach ensures constant time operations for insertions, deletions, and access by leveraging a circular array structure.
✨ Timelines✨
00:00 - Introduction
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