Medium
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
video
are distinct.2 * 105
calls in total will be made to upload
and longest
.longest
.public class LUPrefix {
private boolean[] res;
private int count = 0;
public LUPrefix(int n) {
res = new boolean[n + 1];
}
public void upload(int video) {
res[video - 1] = true;
}
public int longest() {
while (res[count]) {
count++;
}
return count;
}
}
/*
* Your LUPrefix object will be instantiated and called as such:
* LUPrefix obj = new LUPrefix(n);
* obj.upload(video);
* int param_2 = obj.longest();
*/