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1172. Dinner Plate Stacks

Description

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.

Solutions

Solution 1: Stack Array + Ordered Set

We define the following data structures or variables:

  • capacity: The capacity of each stack;
  • stacks: Stack array, used to store all stacks, each with a maximum capacity of capacity;
  • not_full: Ordered set, used to store the indices of all non-full stacks in the stack array.

For the push(val) operation:

  • We first check if not_full is empty. If it is, it means there are no non-full stacks, so we need to create a new stack and push val into it. At this point, we check if the capacity capacity is greater than $1$. If it is, we add the index of this stack to not_full.
  • If not_full is not empty, it means there are non-full stacks. We take out the smallest index index from not_full, and push val into stacks[index]. At this point, if the capacity of stacks[index] equals capacity, we remove index from not_full.

For the popAtStack(index) operation:

  • We first check if index is within the index range of stacks. If it is not, we directly return $-1$. If stacks[index] is empty, we also directly return $-1$.
  • If stacks[index] is not empty, we pop the top element val from stacks[index]. If index equals the length of stacks minus $1$, it means stacks[index] is the last stack. If it is empty, we loop to remove the index of the last stack from not_full, and remove the last stack from the stack array stacks, until the last stack is not empty, or the stack array stacks is empty. Otherwise, if stacks[index] is not the last stack, we add index to not_full.
  • Finally, return val.

For the pop() operation:

  • We directly call popAtStack(stacks.length - 1).

The time complexity is $(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the number of operations.

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from sortedcontainers import SortedSet


class DinnerPlates:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.stacks = []
        self.not_full = SortedSet()

    def push(self, val: int) -> None:
        if not self.not_full:
            self.stacks.append([val])
            if self.capacity > 1:
                self.not_full.add(len(self.stacks) - 1)
        else:
            index = self.not_full[0]
            self.stacks[index].append(val)
            if len(self.stacks[index]) == self.capacity:
                self.not_full.discard(index)

    def pop(self) -> int:
        return self.popAtStack(len(self.stacks) - 1)

    def popAtStack(self, index: int) -> int:
        if index < 0 or index >= len(self.stacks) or not self.stacks[index]:
            return -1
        val = self.stacks[index].pop()
        if index == len(self.stacks) - 1 and not self.stacks[-1]:
            while self.stacks and not self.stacks[-1]:
                self.not_full.discard(len(self.stacks) - 1)
                self.stacks.pop()
        else:
            self.not_full.add(index)
        return val


# Your DinnerPlates object will be instantiated and called as such:
# obj = DinnerPlates(capacity)
# obj.push(val)
# param_2 = obj.pop()
# param_3 = obj.popAtStack(index)

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class DinnerPlates {
    private int capacity;
    private List<Deque<Integer>> stacks = new ArrayList<>();
    private TreeSet<Integer> notFull = new TreeSet<>();

    public DinnerPlates(int capacity) {
        this.capacity = capacity;
    }

    public void push(int val) {
        if (notFull.isEmpty()) {
            stacks.add(new ArrayDeque<>());
            stacks.get(stacks.size() - 1).push(val);
            if (capacity > 1) {
                notFull.add(stacks.size() - 1);
            }
        } else {
            int index = notFull.first();
            stacks.get(index).push(val);
            if (stacks.get(index).size() == capacity) {
                n