2424. Longest Uploaded Prefix

Description

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.

Solutions

Solution 1: Simulation

We use a variable $r$ to record the current longest prefix of uploaded videos, and an array or hash table $s$ to record the videos that have been uploaded.

Each time a video is uploaded, we set s[video] to true, then loop to check whether s[r + 1] is true. If it is, we update $r$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the total number of videos.

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class LUPrefix:
    def __init__(self, n: int):
        self.r = 0
        self.s = set()

    def upload(self, video: int) -> None:
        self.s.add(video)
        while self.r + 1 in self.s:
            self.r += 1

    def longest(self) -> int:
        return self.r


# Your LUPrefix object will be instantiated and called as such:
# obj = LUPrefix(n)
# obj.upload(video)
# param_2 = obj.longest()

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class LUPrefix {
    private int r;
    private Set<Integer> s = new HashSet<>();

    public LUPrefix(int n) {
    }

    public void upload(int video) {
        s.add(video);
        while (s.contains(r + 1)) {
            ++r;
        }
    }

    public int longest() {
        return r;
    }
}

/**
 * Your LUPrefix object will be instantiated and called as such:
 * LUPrefix obj = new LUPrefix(n);
 * obj.upload(video);
 * int param_2 = obj.longest();
 */

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class LUPrefix {
public:
    LUPrefix(int n) {
    }

    void upload(int video) {
        s.insert(video);
        while (s.count(r + 1)) {
            ++r;
        }
    }

    int longest() {
        return r;
    }

private:
    int r = 0;
    unordered_set<int> s;
};

/**
 * Your LUPrefix object will be instantiated and called as such:
 * LUPrefix* obj = new LUPrefix(n);
 * obj->upload(video);
 * int param_2 = obj->longest();
 */

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type LUPrefix struct {
	r int
	s []bool
}

func Constructor(n int) LUPrefix {
	return LUPrefix{0, make([]bool, n+1)}
}

func (this *LUPrefix) Upload(video int) {
	this.s[video] = true
	for this.r+1 < len(this.s) && this.s[this.r+1] {
		this.r++
	}
}

func (this *LUPrefix) Longest() int {
	return this.r
}

/**
 * Your LUPrefix object will be instantiated and called as such:
 * obj := Constructor(n);
 * obj.Upload(video);
 * param_2 := obj.Longest();
 */