How do you solve Design a Number Container System on LeetCode?

Design a Number Container System (LeetCode #2349) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient lookups and updates.

What is the brute force solution for Design a Number Container System?

The brute force approach for Design a Number Container System is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we can use a simple list to store the numbers at each index. To find the smallest index for a given number, we will iterate through the entire list, checking each index. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Design a Number Container System?

Design a Number Container System uses the following concepts: Hash Table, Design, Heap (Priority Queue), Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Design a Number Container System?

Explain your thought process clearly while designing the data structures. Consider edge cases, such as changing a number that doesn't exist. Discuss the trade-offs of different data structures you could use.

What are common mistakes when solving Design a Number Container System?

Not updating the data structures correctly when replacing numbers. Failing to handle cases where a number has no associated indices.

#2349

Design a Number Container System

Medium

Hash TableDesignHeap (Priority Queue)Ordered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(1)	O(n)

💡

Intuition

Time O(log n)Space O(n)

In the optimal solution, we use two hash maps: one to map indices to numbers and another to map numbers to a sorted set of indices. This allows us to efficiently update and retrieve the smallest index for any number.

⚙️

Algorithm

3 steps

1Step 1: Initialize two hash maps: one for index-to-number and another for number-to-index set.
2Step 2: For 'change', update the index-to-number map and also update the number-to-index set accordingly.
3Step 3: For 'find', directly retrieve the smallest index from the number-to-index set.

solution.py20 lines

1from sortedcontainers import SortedSet
2
3class NumberContainers:
4    def __init__(self):
5        self.index_to_number = {}
6        self.number_to_indices = {}
7
8    def change(self, index: int, number: int) -> None:
9        if index in self.index_to_number:
10            old_number = self.index_to_number[index]
11            self.number_to_indices[old_number].remove(index)
12            if not self.number_to_indices[old_number]:
13                del self.number_to_indices[old_number]
14        self.index_to_number[index] = number
15        if number not in self.number_to_indices:
16            self.number_to_indices[number] = SortedSet()
17        self.number_to_indices[number].add(index)
18
19    def find(self, number: int) -> int:
20        return self.number_to_indices[number][0] if number in self.number_to_indices else -1

ℹ

Complexity note: The time complexity is O(log n) for both 'change' and 'find' operations due to the use of a sorted set (or tree set), which allows for efficient insertions, deletions, and retrievals of the smallest index.

1Using a hash map allows for efficient lookups and updates.
2Maintaining a sorted set for indices enables quick retrieval of the smallest index.

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