Introduction

In this blog post, we will explore how to solve the LeetCode problem "2424" 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

You are given a stream of n videos, each represented by a distinct number from 1 to n that you need to "upload" to a server. You need to implement a data structure that calculates the length of the longest uploaded prefix at various points in the upload process. We consider i to be an uploaded prefix if all videos in the range 1 to i (inclusive) have been uploaded to the server. The longest uploaded prefix is the maximum value of i that satisfies this definition. Implement the LUPrefix class: LUPrefix(int n) Initializes the object for a stream of n videos. void upload(int video) Uploads video to the server. int longest() Returns the length of the longest uploaded prefix defined above. Example 1: Input ["LUPrefix", "upload", "longest", "upload", "longest", "upload", "longest"] [[4], [3], [], [1], [], [2], []] Output [null, null, 0, null, 1, null, 3] Explanation LUPrefix server = new LUPrefix(4); // Initialize a stream of 4 videos. server.upload(3); // Upload video 3. server.longest(); // Since video 1 has not been uploaded yet, there is no prefix. // So, we return 0. server.upload(1); // Upload video 1. server.longest(); // The prefix [1] is the longest uploaded prefix, so we return 1. server.upload(2); // Upload video 2. server.longest(); // The prefix [1,2,3] is the longest uploaded prefix, so we return 3. Constraints: 1 <= n <= 105 1 <= video <= n All values of video are distinct. At most 2 * 105 calls in total will be made to upload and longest. At least one call will be made to longest.

Explanation

Here's a breakdown of the solution approach, complexity analysis, and the Python code:

• Approach:

  • Use a boolean array (or set) to track which videos have been uploaded.
  • Maintain a longest_prefix variable to store the length of the longest uploaded prefix seen so far.
  • In the longest() method, increment longest_prefix until a video is not found in the uploaded set/array.

• Complexity:

  • Runtime: upload() is O(1), longest() is O(n) in the worst case, but on average, longest() will be much faster as it only iterates until a missing video is found.
  • Storage: O(n) due to the boolean array or set used to track uploaded videos.

Code

    class LUPrefix:

    def __init__(self, n: int):
        self.uploaded = [False] * (n + 1)  # 1-indexed array
        self.longest_prefix = 0

    def upload(self, video: int) -> None:
        self.uploaded[video] = True

    def longest(self) -> int:
        while self.longest_prefix + 1 < len(self.uploaded) and self.uploaded[self.longest_prefix + 1]:
            self.longest_prefix += 1
        return self.longest_prefix