Design Circular Queue LeetCode solution in JavaScript

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class MyCircularQueue {
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    int head = -1;
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    int tail = -1;
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    int[] q;
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    int k;
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    public MyCircularQueue(int k) {
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        q = new int[k];
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        this.k = k;
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    }
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    public boolean enQueue(int value) {
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        if(head == -1 && tail == -1) {
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            // enqueue when the queue is empty
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            head = tail = 0;
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            q[tail] = value;    // enQueue done
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            return true;
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        }
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        if((tail == k-1 && head == 0) || tail+1 == head) {
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            // queue is full
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            return false;
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        }
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        tail++;
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        if(tail == k) // condition for circularity
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            tail = 0;
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        q[tail] = value;    // enQueue done
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        return true;
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    }
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    public boolean deQueue() {
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        if(head == -1) {
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            // check if q is already empty
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            return false;
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        } 
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        if(head == tail) {
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            // only 1 element is there and head,tail both points same index
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            head = tail = -1;   // deQueue done
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            return true;
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        }
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        head++;       // deQueue done
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        if(head == k) // condition for circularity
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            head = 0;
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        return true;
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    }
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    public int Front() {
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        if(head == -1) 
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            return -1;
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        return q[head];
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    }
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    public int Rear() {
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        if(tail == -1) 
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            return -1;
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        return q[tail];
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    }
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    public boolean isEmpty() {
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        if(head == -1 && tail == -1) 
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            return true;
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        return false;
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    }
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    public boolean isFull() {
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        if(tail == k-1 || tail+1 == head) 
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            return true;
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        return false;
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    }
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}
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/**
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 * Your MyCircularQueue object will be instantiated and called as such:
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 * MyCircularQueue obj = new MyCircularQueue(k);
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 * boolean param_1 = obj.enQueue(value);
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 * boolean param_2 = obj.deQueue();
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 * int param_3 = obj.Front();
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 * int param_4 = obj.Rear();
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 * boolean param_5 = obj.isEmpty();
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 * boolean param_6 = obj.isFull();
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 */