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:

• <code>1 <= capacity <= 2 * 10⁴</code>

• <code>1 <= val <= 2 * 10⁴</code>

• <code>0 <= index <= 10⁵</code>

• At most <code>2 * 10⁵</code> calls will be made to push, pop, and popAtStack.