Problem Description

You need to implement a system that manages dinner plates across multiple stacks. Imagine you have an infinite number of stacks arranged in a row, numbered from left to right starting at 0. Each stack can hold a maximum number of plates (the capacity).

The DinnerPlates class should support the following operations:

1. DinnerPlates(int capacity): Initialize the system with a given maximum capacity for each stack.

2. push(int val): Add a plate with value val to the leftmost stack that isn't full. This means you should find the first stack (with the smallest index) that has room for another plate and add the plate there.

3. pop(): Remove and return the value from the top of the rightmost non-empty stack. If all stacks are empty, return -1.

4. popAtStack(int index): Remove and return the value from the top of the stack at the given index. If the stack at that index is empty or doesn't exist, return -1.

The key challenge is efficiently tracking which stacks are not full (for push operations) and managing the removal of empty stacks from the right side (for pop operations). The solution uses a sorted set to maintain indices of non-full stacks, allowing for quick identification of where to push new plates. When popping from the rightmost stack, the implementation ensures that any trailing empty stacks are removed to maintain the correct "rightmost non-empty stack" for future pop operations.

Intuition

The main challenge in this problem is efficiently finding the leftmost non-full stack for push operations and managing the rightmost non-empty stack for pop operations. Let's think through what data structures we need.

First, we need to store the actual stacks. A simple array of stacks would work since we're numbering them sequentially from 0.

For the push operation, we need to quickly find the leftmost stack that isn't full. If we had to scan through all stacks from left to right each time, it would be inefficient. Instead, we can maintain a collection of indices of all non-full stacks. But we also need this collection to be ordered so we can always get the smallest index quickly. This suggests using an ordered set (like SortedSet in Python).

When we push to a stack, if it becomes full, we remove its index from our ordered set. When we pop from a stack, it becomes non-full, so we add its index back to the ordered set.

The tricky part comes with the pop operation. We need to pop from the rightmost non-empty stack. A naive approach would be to always keep track of the rightmost non-empty stack index, but this gets complicated when stacks become empty.

The key insight is that we can simply try to pop from the last stack in our array. If successful, great! But if that stack becomes empty after popping, we need to clean up. We should remove all trailing empty stacks from our array to maintain the invariant that the last stack in our array is the rightmost stack we care about. This cleanup step is crucial - it ensures that for the next pop operation, we can again just look at the last stack in our array.

This approach elegantly handles both the efficient lookup requirements (using the ordered set for push) and the rightmost stack management (by maintaining a clean array without trailing empty stacks).

Learn more about Stack and Heap (Priority Queue) patterns.

Solution Approach

Let's implement the solution using a stack array combined with an ordered set to efficiently manage our dinner plates system.

Data Structures

We maintain three key components:

• capacity: The maximum capacity for each stack
• stacks: An array of stacks to store all plates
• not_full: A SortedSet to track indices of stacks that aren't at full capacity

Implementation Details

Initialization (__init__)

• Store the capacity limit
• Initialize an empty array for stacks
• Create an empty SortedSet for tracking non-full stack indices

Push Operation (push)

When pushing a value:

1. Check if not_full is empty:

   • If empty, no existing stacks have room, so create a new stack with the value
   • Add the new stack to stacks array
   • If capacity > 1, add this new stack's index to not_full (since it can hold more plates)

2. If not_full has indices:

   • Get the smallest index from not_full (the leftmost non-full stack)
   • Push the value to stacks[index]
   • If the stack reaches capacity, remove its index from not_full

PopAtStack Operation (popAtStack)

When popping from a specific stack at index:

1. Validate the index:

   • Return -1 if index is out of bounds or stacks[index] is empty

2. Pop the value from stacks[index]

3. Handle the aftermath:

   • If this was the last stack (index == len(stacks) - 1) and it's now empty:
     • Clean up all trailing empty stacks
     • Remove their indices from not_full
     • Keep removing empty stacks from the end until we hit a non-empty stack or run out of stacks
   • Otherwise, add index to not_full since the stack now has room

4. Return the popped value

Pop Operation (pop)

Simply delegate to popAtStack(len(stacks) - 1) to pop from the rightmost stack.

Key Optimizations

The use of SortedSet for not_full allows:

• O(log n) insertion and removal of indices
• O(1) access to the minimum index (leftmost non-full stack)

The cleanup of trailing empty stacks ensures:

• The last stack in our array is always the rightmost relevant stack
• We don't waste memory on empty stacks at the end
• The pop() operation remains simple

This design achieves efficient time complexity for all operations while maintaining clean state management throughout the lifetime of the data structure.

Example Walkthrough

Let's walk through a concrete example with capacity = 2 to see how the solution works:

Initial State:

• capacity = 2
• stacks = []
• not_full = {}

Operation 1: push(1)

• not_full is empty, so we create a new stack
• stacks = [[1]]
• Since this stack can hold one more plate (capacity is 2), add index 0 to not_full
• not_full = {0}

Operation 2: push(2)

• not_full = {0}, so we push to the leftmost non-full stack at index 0
• stacks = [[1, 2]]
• Stack 0 is now full (has 2 plates), remove it from not_full
• not_full = {}

Operation 3: push(3)

• not_full is empty, create a new stack at index 1
• stacks = [[1, 2], [3]]
• Stack 1 can hold one more plate, add it to not_full
• not_full = {1}

Operation 4: push(4)

• Push to stack at index 1 (the leftmost non-full)
• stacks = [[1, 2], [3, 4]]
• Stack 1 is now full, remove from not_full
• not_full = {}

Operation 5: push(5)

• Create new stack at index 2
• stacks = [[1, 2], [3, 4], [5]]
• not_full = {2}

Operation 6: popAtStack(0)

• Pop from stack 0: returns 2
• stacks = [[1], [3, 4], [5]]
• Stack 0 now has room, add to not_full
• not_full = {0, 2}

Operation 7: pop()

• Calls popAtStack(2) (rightmost stack)
• Pop from stack 2: returns 5
• stacks = [[1], [3, 4], []]
• Stack 2 is empty and it's the last stack, so clean up trailing empty stacks
• Remove stack 2: stacks = [[1], [3, 4]]
• Remove index 2 from not_full: not_full = {0}

Operation 8: push(6)

• Push to stack 0 (leftmost non-full)
• stacks = [[1, 6], [3, 4]]
• Stack 0 is full, remove from not_full
• not_full = {}

This example demonstrates:

• How not_full tracks which stacks have room (enabling efficient push to leftmost)
• How trailing empty stacks are cleaned up during pop operations
• How the rightmost stack is always at len(stacks) - 1 due to cleanup

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n) where n is the number of operations.

• push(val): The operation involves checking and manipulating the not_full SortedSet. Adding to or accessing the first element of a SortedSet takes O(log m) time where m is the size of the set. In the worst case, m can be proportional to the number of stacks, which grows with the number of operations. Therefore, each push operation is O(log n).

• pop(): This calls popAtStack() with the last index, so its complexity is the same as popAtStack().

• popAtStack(index): The main operations involve:

  • Popping from a list: O(1)
  • Adding to or discarding from the SortedSet: O(log m) where m is the size of the set
  • In the worst case when popping from the last stack, we might need to remove multiple empty stacks. Each removal involves a discard operation from the SortedSet (O(log n)) and a list pop (O(1)). While this cleanup can remove multiple stacks, amortized over all operations, each stack is removed at most once.

  The dominant operation is the SortedSet manipulation, giving O(log n) per operation.

Since we perform n operations total, and each operation has O(log n) complexity, the overall time complexity is O(n × log n).

Space Complexity: O(n) where n is the number of operations.

• The stacks list can grow to store all pushed elements, which in the worst case is O(n).
• The not_full SortedSet can contain at most one entry per stack, which is bounded by O(n).
• Therefore, the total space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Cleaning Up Empty Stacks After popAtStack

One of the most critical pitfalls is forgetting to remove trailing empty stacks when popping from the rightmost stack. This can lead to incorrect behavior in subsequent pop() operations.

Problem Example:

# Without proper cleanup:
plates = DinnerPlates(2)
plates.push(1)  # stacks: [[1]]
plates.push(2)  # stacks: [[1, 2]]
plates.push(3)  # stacks: [[1, 2], [3]]
plates.popAtStack(1)  # Returns 3, stacks: [[1, 2], []]
plates.pop()  # Should pop from [1, 2], but might try to pop from empty stack []

Solution: Always check if we're popping from the last stack and clean up any trailing empty stacks:

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()

2. Incorrect Management of the not_full Set

Failing to properly maintain the not_full set can cause the push operation to behave incorrectly.

Common Mistakes:

• Forgetting to add a stack's index to not_full after popping from it
• Not removing indices when deleting trailing empty stacks
• Adding index 0 to not_full when creating the first stack with capacity 1

Solution: Ensure proper synchronization:

# When creating a new stack:
if self.capacity > 1:  # Only add if stack can hold more than 1 plate
    self.not_full.add(len(self.stacks) - 1)

# When popping creates space:
if not (index == len(self.stacks) - 1 and not self.stacks[-1]):
    self.not_full.add(index)

# When removing trailing stacks:
self.not_full.discard(len(self.stacks) - 1)  # Remove before popping the stack

3. Edge Case: Capacity of 1

When capacity is 1, each stack can only hold one plate. A common mistake is not handling this special case properly when managing the not_full set.

Problem:

plates = DinnerPlates(1)
plates.push(1)  # Creates stack [1]
# If we incorrectly add index 0 to not_full, next push might try to add to full stack

Solution: Check capacity before adding to not_full:

if self.capacity > 1:  # Stack has room for more plates
    self.not_full.add(len(self.stacks) - 1)

4. Using Regular Set Instead of SortedSet

Using a regular set instead of SortedSet breaks the requirement to push to the leftmost non-full stack.

Problem:

# With regular set, iteration order is not guaranteed
not_full = {2, 0, 1}  # Regular set
index = min(not_full)  # O(n) operation each time

Solution: Use SortedSet for O(1) access to minimum:

from sortedcontainers import SortedSet
self.not_full = SortedSet()
index = self.not_full[0]  # O(1) access to minimum

5. Not Validating Index Bounds in popAtStack

Failing to check if the index is valid can cause runtime errors.

Problem:

plates.popAtStack(10)  # When only 3 stacks exist

Solution: Always validate before accessing:

if index < 0 or index >= len(self.stacks) or not self.stacks[index]:
    return -1