How do you solve Dinner Plate Stacks on LeetCode?

Dinner Plate Stacks (LeetCode #1172) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) for push and pop operations, O(1) for popAtStack time complexity and O(n) space complexity. Key insight: Using a priority queue helps efficiently manage available stacks.

What is the brute force solution for Dinner Plate Stacks?

The brute force approach for Dinner Plate Stacks is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we maintain a list of stacks and simply iterate through them to find the leftmost stack with available capacity for pushing. For popping, we check the rightmost stack for the top value. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(log n) for push and pop operations, O(1) for popAtStack.

What algorithm or data structure is used to solve Dinner Plate Stacks?

Dinner Plate Stacks uses the following concepts: Hash Table, Stack, Design, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Dinner Plate Stacks?

Explain your thought process clearly while coding. Discuss edge cases, such as pushing to an empty stack. Practice writing clean and efficient code.

What are common mistakes when solving Dinner Plate Stacks?

Not managing the available stacks correctly, leading to inefficient pushes. Failing to check if stacks are empty before popping.

#1172

Dinner Plate Stacks

Hard

Hash TableStackDesignHeap (Priority Queue)Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n) for push and pop operations, O(1) for popAtStack
Space	O(n)	O(n)

💡

Intuition

Time O(log n) for push and pop operations, O(1) for popAtStackSpace O(n)

In the optimal solution, we maintain a list of stacks along with a min-heap (priority queue) to efficiently track the leftmost available stack for pushing and the rightmost non-empty stack for popping. This allows us to perform operations in a more efficient manner.

⚙️

Algorithm

4 steps

1Step 1: Initialize a list to hold stacks and a min-heap to track available stacks.
2Step 2: For push, check the min-heap for the leftmost available stack and push the value. If no stack is available, create a new one.
3Step 3: For pop, use the last stack in the list to pop the top value. If it becomes empty, remove it from the list.
4Step 4: For popAtStack, directly access the specified index and pop the top value if it exists.

solution.py33 lines

1import heapq
2
3class DinnerPlates:
4    def __init__(self, capacity: int):
5        self.capacity = capacity
6        self.stacks = []
7        self.available = []
8
9    def push(self, val: int) -> None:
10        if self.available:
11            idx = heapq.heappop(self.available)
12            self.stacks[idx].append(val)
13            if len(self.stacks[idx]) == self.capacity:
14                return
15            heapq.heappush(self.available, idx)
16        else:
17            self.stacks.append([val])
18            heapq.heappush(self.available, len(self.stacks) - 1)
19
20    def pop(self) -> int:
21        while self.stacks and not self.stacks[-1]:
22            self.stacks.pop()
23        if not self.stacks:
24            return -1
25        return self.stacks[-1].pop()
26
27    def popAtStack(self, index: int) -> int:
28        if 0 <= index < len(self.stacks) and self.stacks[index]:
29            plate = self.stacks[index].pop()
30            if len(self.stacks[index]) < self.capacity:
31                heapq.heappush(self.available, index)
32            return plate
33        return -1

ℹ

Complexity note: The time complexity is O(log n) for push and pop operations due to the use of a priority queue to manage available stacks. Space complexity remains O(n) for storing the plates in stacks.

1Using a priority queue helps efficiently manage available stacks.
2Direct access to specific stacks allows for targeted operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.