Leetcode 2424 Longest Uploaded Prefix Solution in Java | Hindi Coding Community

 

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.

Java Code :

class LUPrefix {

TreeSet<Integer> tree = new TreeSet<>();

int n;

public LUPrefix(int n) {

for (int i = 1; i <= n ; i++) {

tree.add(i);

}

this.n = n;

}

public void upload(int video) {

tree.remove(video);

}

public int longest() {

return tree.isEmpty() ? n : tree.first() -1;

}

}