g2401_2500.s2424_longest_uploaded_prefix.LUPrefix Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g2401_2500.s2424_longest_uploaded_prefix;

// #Medium #Binary_Search #Design #Heap_Priority_Queue #Union_Find #Ordered_Set #Segment_Tree
// #Binary_Indexed_Tree #2022_11_19_Time_35_ms_(99.68%)_Space_108.8_MB_(92.53%)

/**
 * 2424 - Longest Uploaded Prefix\.
 *
 * 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`
 * *   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`.
**/
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();
 */