Dinner Plate Stacks

You have an infinite number of stacks arranged in a row and numbered (left to right) from 0, each of the stacks has the same maximum capacity.

Implement the DinnerPlates class:

• DinnerPlates(int capacity) Initializes the object with the maximum capacity of the stacks.

• void push(int val) pushes the given positive integer val into the leftmost stack with size less than capacity.

• int pop() returns the value at the top of the rightmost non-empty stack and removes it from that stack, and returns -1 if all stacks are empty.

• int popAtStack(int index) returns the value at the top of the stack with the given index and removes it from that stack, and returns -1 if the stack with that given index is empty.

Example:

Input: 
["DinnerPlates","push","push","push","push","push","popAtStack","push","push","popAtStack","popAtStack","pop","pop","pop","pop","pop"]
[[2],[1],[2],[3],[4],[5],[0],[20],[21],[0],[2],[],[],[],[],[]]
Output: 
[null,null,null,null,null,null,2,null,null,20,21,5,4,3,1,-1]

Explanation: 
DinnerPlates D = DinnerPlates(2);  // Initialize with capacity = 2
D.push(1);
D.push(2);
D.push(3);
D.push(4);
D.push(5);         // The stacks are now:  2  4
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.popAtStack(0);   // Returns 2.  The stacks are now:     4
                                                       1  3  5
                                                       ﹈ ﹈ ﹈
D.push(20);        // The stacks are now: 20  4
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.push(21);        // The stacks are now: 20  4 21
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.popAtStack(0);   // Returns 20.  The stacks are now:     4 21
                                                        1  3  5
                                                        ﹈ ﹈ ﹈
D.popAtStack(2);   // Returns 21.  The stacks are now:     4
                                                        1  3  5
                                                        ﹈ ﹈ ﹈ 
D.pop()            // Returns 5.  The stacks are now:      4
                                                        1  3 
                                                        ﹈ ﹈  
D.pop()            // Returns 4.  The stacks are now:   1  3 
                                                        ﹈ ﹈   
D.pop()            // Returns 3.  The stacks are now:   1 
                                                        ﹈   
D.pop()            // Returns 1.  There are no stacks.
D.pop()            // Returns -1.  There are still no stacks.

Constraints:

• 1 <= capacity <= 20000

• 1 <= val <= 20000

• 0 <= index <= 100000

• At most 200000 calls will be made to push, pop, and popAtStack.

class DinnerPlates {
    int capacity;
    List<Stack<Integer>> stacks;
    TreeSet<Integer> pushable;
    TreeSet<Integer> popable;

    public DinnerPlates(int cap) {
        capacity = cap;
        stacks = new ArrayList<>();
        pushable = new TreeSet<>();
        popable = new TreeSet<>();
    }

    public void push(int val) {
        if (pushable.isEmpty()) {
            pushable.add(stacks.size());
            stacks.add(new Stack<>());
        }
        int index = pushable.first();
        Stack<Integer> stack = stacks.get(index);
        stack.push(val);
        popable.add(index);
        if (stack.size() == capacity) {
            pushable.remove(index);
        }
    }

    public int pop() {
        if (popable.size() == 0) {
            return -1;
        }
        int index = popable.last();
        return popAtStack(index);
    }

    public int popAtStack(int index) {
        if (!popable.contains(index)) {
            return -1;
        }
        Stack<Integer> stack = stacks.get(index);
        int res = stack.pop();
        pushable.add(index);
        if (stack.size() == 0) {
            popable.remove(index);
        }
        return res;
    }
}