Dinner Plate Stacks - Problem

You have an infinite number of stacks arranged in a row and numbered (left to right) from 0. Each stack 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 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 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 or returns -1 if the stack is empty.

Input & Output

Example 1 — Basic Operations
$ Input: Operations: ["DinnerPlates","push","push","push","push","pop","pop","pop","push","pop"], Values: [[2],[1],[2],[3],[4],[],[],[],[5],[]]
› Output: [null,null,null,null,null,4,3,2,null,5]
💡 Note: Initialize with capacity 2, push 1,2,3,4 (creates stacks [[1,2],[3,4]]), pop returns 4,3,2, push 5 goes to first available spot in stack 0, final pop returns 5
Example 2 — PopAtStack Usage
$ Input: Operations: ["DinnerPlates","push","push","popAtStack","push","pop"], Values: [[1],[1],[2],[0],[3],[]]
› Output: [null,null,null,1,null,2]
💡 Note: Capacity 1, push 1,2 creates [[1],[2]], popAtStack(0) removes 1, push 3 goes to stack 0, pop returns 2 from rightmost stack
Example 3 — Empty Stack Handling
$ Input: Operations: ["DinnerPlates","pop","push","pop","pop"], Values: [[2],[],[1],[],[]]
› Output: [null,-1,null,1,-1]
💡 Note: Pop from empty returns -1, push 1, pop returns 1, pop from empty again returns -1

Constraints

  • 1 ≤ capacity ≤ 2 × 104
  • 1 ≤ val ≤ 2 × 104
  • 0 ≤ index ≤ 105
  • At most 2 × 105 calls to push, pop, and popAtStack

Visualization

Tap to expand
Dinner Plate Stacks INPUT Stacks (capacity=2) 1 2 Stack 0 3 4 Stack 1 empty Stack 2 Operations: push(1), push(2), push(3), push(4), pop(), pop(), pop(), push(5), pop() Values: [[2],[1],[2],[3],[4],[],[],[],[5],[]] Min-Heap tracks available slots rightIndex tracks rightmost ALGORITHM STEPS 1 Initialize Structures Min-heap for leftmost slot Track rightmost non-empty 2 Push Operation Pop from min-heap to get leftmost available stack 3 Pop Operation Pop from rightmost stack Update rightIndex if empty 4 PopAtStack Operation Pop from specific index Add slot back to heap Min-Heap State: 0 1 2 FINAL RESULT Operation Trace: DinnerPlates(2) --> null push(1,2,3,4) --> null x4 pop() --> 4 pop() --> 3 pop() --> 2 push(5) --> null pop() --> 5 Output Array: [null,null,null,null,null, 4, 3, 2, null, 5] Final Stack State: 1 Stack 0 empty Stack 1 empty Stack 2 Key Insight: Use a Min-Heap to efficiently track the leftmost stack with available capacity for O(log n) push. Maintain rightIndex pointer for O(1) access to rightmost non-empty stack for pop operations. When popping creates empty stacks, update rightIndex by scanning left. Heap handles slot recycling. TutorialsPoint - Dinner Plate Stacks | Heap-Based Optimization Approach

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