Dinner Plate Stacks

Dinner Plate Stacks

What is this problem about?

The Dinner Plate Stacks coding problem asks you to design a data structure that manages multiple stacks of a fixed capacity arranged in a row. You need to implement:

1. push(val): Adds val to the leftmost stack that is not full.
2. pop(): Removes and returns the value from the rightmost non-empty stack.
3. popAtStack(index): Removes and returns the value from the specific stack at the given index. The challenge is efficiently finding the "leftmost available" and "rightmost non-empty" spots after multiple operations.

Why is this asked in interviews?

Amazon uses this "Hard" level design problem to test a candidate's mastery of complex state management and choice of data structures. It evaluations whether you can coordinate multiple structures (like a list of stacks and a priority queue) to maintain $O(\log N)$ or $O(1)$ efficiency. This mirrors real-world problems in memory management or logistics, where resources are allocated based on specific spatial rules.

Algorithmic pattern used

This is a Design problem using Hash Tables and a Min-Heap (Priority Queue).

• Use a list of stacks to store the plates.
• Use a Min-Heap to keep track of the indices of all stacks that currently have at least one empty slot. This allows push to find the leftmost available stack in $O(\log N)$.
• Keep a pointer or use a map to find the rightmost non-empty stack for pop. When popAtStack is called, a previously full stack might become available, so you add its index back to the Min-Heap.

Example explanation

Capacity = 2.

1. push(1), push(2): Stack 0 is full [1, 2].
2. push(3): Stack 0 is full, push to Stack 1. Stacks: [1, 2], [3].
3. popAtStack(0): Remove 2. Stacks: [1], [3]. Stack 0 is now the leftmost available stack.
4. push(4): Since Stack 0 has space, it goes there. Stacks: [1, 4], [3].

Common mistakes candidates make

• Slow Search: Searching for the first non-full stack using a linear loop ($O(N)$), which is too slow.
• Stale Indices: Forgetting to remove indices from the Priority Queue once a stack becomes full during a push.
• Empty Stacks: Failing to remove empty stacks from the end of the list during a pop, which makes finding the "rightmost non-empty" stack difficult.

Interview preparation tip

When you need to find the "minimum" or "leftmost" index in a dynamically changing set, a Priority Queue is your best tool. In design problems, always consider how one operation (like popAtStack) might affect the state required for another operation (like push).