public class DinnerPlates
extends Object
1172 - Dinner Plate Stacks\.
Hard
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 `capacity`.
* `void push(int val)` Pushes the given integer `val` into the leftmost stack with a 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 the stacks are empty.
* `int popAtStack(int index)` Returns the value at the top of the stack with the given index `index` and removes it from that stack or returns `-1` if the stack with that given index is empty.
**Example 1:**
**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 <= 2 * 104
* 1 <= val <= 2 * 104
* 0 <= index <= 105
* At most 2 * 105 calls will be made to `push`, `pop`, and `popAtStack`.