LeetCode : 622. Design Circular Queue

// 53 / 58 test cases passed.
// Ref : https://leetcode.com/submissions/detail/946042071
class MyCircularQueue {
    ArrayList<Integer> list;
    int front;
    int rear = -1;
    int size;

    public MyCircularQueue(int k) {
        list = new ArrayList<>(k);
        for (int i = 0; i < k; i++) {
            list.add(-1);
        }
        size = k;
    }

    public boolean enQueue(int value) {
        int nextPos = (rear + 1) % size;
        if (list.get(nextPos) == -1) {
            list.set(nextPos, value);
            rear = nextPos;
            return true;
        }
        return false;
    }

    public boolean deQueue() {
        if (list.get(front) != -1) {
            list.set(front, -1);
            front = (front + 1) % size;
            return true;
        }
        return false;
    }

    public int Front() {
        return list.get(front);
    }

    public int Rear() {
        return list.get(rear);
    }

     public boolean isEmpty() {
        for (int i : list) {
            if (i != -1) {
                return false;
            }
        }
        return true;
    }

    public boolean isFull() {
        return (rear + 1) % size == front;
    }
}

// Runtime: 5 ms
// Memory Usage: 42.8 MB
// Ref : https://leetcode.com/submissions/detail/946051399
// Ref : https://leetcode.com/problems/design-circular-queue/discuss/2620288/LeetCode-The-Hard-Way-Explained-Line-By-Line
// Time Complexity: O(1)
// Space Complexity: O(N)
class MyCircularQueue {

    public MyCircularQueue(int k) {
        // the queue holding the elements for the circular queue
        q = new int[k];
        // the number of elements in the circular queue
        cnt = 0;
        // queue size
        sz = k;
        // the idx of the head element
        headIdx = 0;
    }
    
    public boolean enQueue(int value) {
         // handle full case
        if (isFull()) return false;
        // set the value 
        // Given an array of size of 4, we can find the position to be inserted using the formula
        // targetIdx = (headIdx + cnt) % sz
        // e.g. [1, 2, 3, _]
        // headIdx = 0, cnt = 3, sz = 4, targetIdx = (0 + 3) % 4 = 3
        // e.g. [_, 2, 3, 4]
        // headIdx = 1, cnt = 3, sz = 4, targetIdx = (1 + 3) % 4 = 0
        q[(headIdx + cnt) % sz] = value;
        // increase the number of elements by 1
        cnt += 1;
        return true;
    }
    
    public boolean deQueue() {
        // handle empty case
        if (isEmpty()) return false;
        // update the head index
        headIdx = (headIdx + 1) % sz;
        // decrease the number of elements by 1
        cnt -= 1;
        return true;
    }
    
    public int Front() {
        // handle empty queue case
        if (isEmpty()) return -1;
        // return the head element
        return q[headIdx];
    }
    
    public int Rear() {
        // handle empty queue case
        if (isEmpty()) return -1;
        // Given an array of size of 4, we can find the tailIdx using the formula
        // tailIdx = (headIdx + cnt - 1) % sz
        // e.g. [0 1 2] 3
        // headIdx = 0, cnt = 3, sz = 4, tailIdx = (0 + 3 - 1) % 4 = 2
        // e.g. 0 [1 2 3]
        // headIdx = 1, cnt = 3, sz = 4, tailIdx = (1 + 3 - 1) % 4 = 3
        // e.g. 0] 1 [2 3
        // headIdx = 2, cnt = 3, sz = 4, tailIdx = (2 + 3 - 1) % 4 = 0
        return q[(headIdx + cnt - 1) % sz];
    }
    
    public boolean isEmpty() {
        // no element in the queue
        return cnt == 0;
    }
    
    public boolean isFull() {
        // return true if the count is equal to the queue size
        // else return false
        return cnt == sz;
    }
    
    private int[] q;
    private int headIdx, cnt, sz;
}