Leetcode #1172: Dinner Plate Stacks
In this guide, we solve Leetcode #1172 Dinner Plate Stacks in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Design, Hash Table, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.