LeetCode #2424 — MEDIUM

Longest Uploaded Prefix

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

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.

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Binary Search · Union-Find · Design · Segment Tree

Example 1

["LUPrefix","upload","longest","upload","longest","upload","longest"]
[[4],[3],[],[1],[],[2],[]]

Core Insight

What unlocks the optimal approach

Maintain an array keeping track of whether video “i” has been uploaded yet.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2424: Longest Uploaded Prefix
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();
 */

// Accepted solution for LeetCode #2424: Longest Uploaded Prefix
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();
 */

# Accepted solution for LeetCode #2424: Longest Uploaded Prefix
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()

// Accepted solution for LeetCode #2424: Longest Uploaded Prefix
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2424: Longest Uploaded Prefix
// 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();
//  */

// Accepted solution for LeetCode #2424: Longest Uploaded Prefix
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2424: Longest Uploaded Prefix
// 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();
//  */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.