Introduction

In this blog post, we will explore how to solve the LeetCode problem "2349" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. Implement the NumberContainers class: NumberContainers() Initializes the number container system. void change(int index, int number) Fills the container at index with the number. If there is already a number at that index, replace it. int find(int number) Returns the smallest index for the given number, or -1 if there is no index that is filled by number in the system. Example 1: 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. Constraints: 1 <= index, number <= 109 At most 105 calls will be made in total to change and find.

Explanation

Here's an efficient implementation of the NumberContainers class in Python, addressing the problem constraints and optimizing for performance.

• Key Approach:

  • Use a dictionary (index_map) to store the number at each index. This allows for O(1) change operations (insert/replace).
  • Use a dictionary (number_map) to store a sorted set of indices for each number. This facilitates finding the smallest index for a given number efficiently. We use a sorted set to automatically keep indices sorted and ensure the smallest index can be retrieved quickly.

• Complexity:

  • Runtime Complexity: change: O(log n) for updating the sorted set, find: O(1) on average.
  • Storage Complexity: O(N) where N is the number of unique indices used.

Code

    from sortedcontainers import SortedSet

class NumberContainers:
    def __init__(self):
        self.index_map = {}  # index: number
        self.number_map = {}  # number: SortedSet of indices

    def change(self, index: int, number: int) -> None:
        if index in self.index_map:
            old_number = self.index_map[index]
            self.number_map[old_number].remove(index)
            if not self.number_map[old_number]:
                del self.number_map[old_number]

        self.index_map[index] = number
        if number not in self.number_map:
            self.number_map[number] = SortedSet()
        self.number_map[number].add(index)

    def find(self, number: int) -> int:
        if number in self.number_map:
            return self.number_map[number][0]
        else:
            return -1