Leetcode #2424: Longest Uploaded Prefix
In this guide, we solve Leetcode #2424 Longest Uploaded Prefix in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Union Find, Design, Binary Indexed Tree, Segment Tree, Hash Table, Binary Search, Ordered Set, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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()
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.