LeetCode 2424. Longest Uploaded Prefix Solution in Java, C++, Python & Go | Explanation + Code

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.

PythonJavaC++Go

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()(code-box)

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(); */(code-box)

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(); */(code-box)

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(); */(code-box)