Leetcode #2349: Design a Number Container System
In this guide, we solve Leetcode #2349 Design a Number Container System 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
Design a number container system that can do the following: Insert or Replace a number at the given index in the system. Return the smallest index for the given number in the system.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Hash Table, Ordered Set, 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
["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"]
[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]
Output
[null, -1, null, null, null, null, 1, null, 2]
Explanation
NumberContainers nc = new NumberContainers();
nc.find(10); // There is no index that is filled with number 10. Therefore, we return -1.
nc.change(2, 10); // Your container at index 2 will be filled with number 10.
nc.change(1, 10); // Your container at index 1 will be filled with number 10.
nc.change(3, 10); // Your container at index 3 will be filled with number 10.
nc.change(5, 10); // Your container at index 5 will be filled with number 10.
nc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1.
nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20.
nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.
Python Solution
class NumberContainers:
def __init__(self):
self.d = {}
self.g = defaultdict(SortedSet)
def change(self, index: int, number: int) -> None:
if index in self.d:
old_number = self.d[index]
self.g[old_number].remove(index)
self.d[index] = number
self.g[number].add(index)
def find(self, number: int) -> int:
ids = self.g[number]
return ids[0] if ids else -1
# Your NumberContainers object will be instantiated and called as such:
# obj = NumberContainers()
# obj.change(index,number)
# param_2 = obj.find(number)
Complexity
The time complexity is . The space complexity is , where is the number of numbers.
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.