2424. Longest Uploaded Prefix

Problem Link

Description

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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.

Solution

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Python3

class LUPrefix:
 
    def __init__(self, n: int):
        self.N = n
        self.dp = [False] * (n + 1)
        self.dp[0] = True
        self.curr = 0
 
    def upload(self, video: int) -> None:
        self.dp[video] = True
        while self.curr + 1 <= self.N and self.dp[self.curr + 1]:
            self.curr += 1
 
    def longest(self) -> int:
        return self.curr
 
 
# Your LUPrefix object will be instantiated and called as such:
# obj = LUPrefix(n)
# obj.upload(video)
# param_2 = obj.longest()