Longest Uploaded Prefix - Practice Coding Problems

Longest Uploaded Prefix - Problem

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You're managing a video streaming platform where videos must be uploaded in a specific sequence. Imagine you have n videos numbered from 1 to n, but they arrive at your server in random order due to network conditions.

Your task is to build a smart system that tracks the longest consecutive prefix of videos that have been successfully uploaded starting from video 1. For example, if videos 1, 2, and 3 are uploaded (regardless of order), your longest prefix is 3. But if videos 1, 3, and 5 are uploaded, your longest prefix is only 1 (since video 2 is missing).

Design the LUPrefix class with these operations:

LUPrefix(int n) - Initialize for a stream of n videos
upload(int video) - Mark video as uploaded
longest() - Return length of longest uploaded prefix

This problem tests your ability to efficiently track consecutive sequences and optimize for frequent queries.

Input & Output

example_1.py — Basic Upload Sequence

$ Input: LUPrefix(4), upload(3), longest(), upload(1), longest(), upload(2), longest()

› Output: [null, null, 0, null, 1, null, 3]

💡 Note: After uploading video 3, longest prefix is 0 (video 1 missing). After uploading video 1, prefix becomes 1. After uploading video 2, videos 1-2-3 are consecutive, so prefix becomes 3.

example_2.py — Sequential Uploads

$ Input: LUPrefix(3), upload(1), longest(), upload(2), longest(), upload(3), longest()

› Output: [null, null, 1, null, 2, null, 3]

💡 Note: Videos uploaded in order 1,2,3. Each upload extends the longest prefix by 1, resulting in prefix lengths 1, 2, and 3 respectively.

example_3.py — Reverse Order Uploads

$ Input: LUPrefix(2), upload(2), longest(), upload(1), longest()

› Output: [null, null, 0, null, 2]

💡 Note: Upload video 2 first - prefix remains 0 since video 1 is missing. Upload video 1 - now videos 1 and 2 are both present, so prefix becomes 2.

Constraints

1 ≤ n ≤ 10⁵
1 ≤ video ≤ n
All values of video are distinct
At most 2 × 10⁵ calls will be made to upload and longest
At least one call will be made to longest

Visualization

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Understanding the Visualization

Track 3 arrives

Can't play anything yet - track 1 is missing, so playable prefix is 0

Track 1 arrives

Now we can play track 1, but track 2 is missing, so prefix length is 1

Track 2 arrives

Perfect! Now tracks 1, 2, and 3 are consecutive, so we can play up to track 3

Key Takeaway

🎯 Key Insight: Instead of scanning from the beginning each time, maintain the current longest prefix and only extend it when the next consecutive video becomes available!

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal approach uses a boolean array to track uploaded videos and maintains a longest_prefix pointer. Key insight: instead of scanning from video 1 every time, we only need to extend the current prefix when consecutive videos arrive. This gives us O(1) longest() queries and amortized O(1) upload operations.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Tracking (Optimal)	O(n) worst case per upload, O(1) per longest()	O(n)	Track uploaded videos and maintain current longest prefix pointer
Brute Force (Linear Scan)	O(n) per longest() call	O(n)	For each longest() query, scan from video 1 until finding a gap

Optimized Tracking (Optimal) — Algorithm Steps

Use boolean array to track which videos are uploaded
Maintain 'longest_prefix' variable for current consecutive length
On upload: mark video as uploaded, then extend prefix if possible
For longest(): simply return the maintained prefix length

Visualization

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Step-by-Step Walkthrough

Initialize

Boolean array tracks uploads, prefix starts at 0

Upload video 3

Mark video 3 as uploaded, but prefix remains 0 (video 1 not uploaded)

Upload video 1

Mark video 1, extend prefix to 1. Check video 2 - not uploaded, stop

Upload video 2

Mark video 2, extend prefix through videos 2 and 3 to reach 3

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

typedef struct {
    int n;
    bool* uploaded;
    int longestPrefix;
} LUPrefix;

LUPrefix* lUPrefixCreate(int n) {
    LUPrefix* obj = (LUPrefix*)malloc(sizeof(LUPrefix));
    obj->n = n;
    obj->uploaded = (bool*)calloc(n + 1, sizeof(bool));
    obj->longestPrefix = 0;
    return obj;
}

void lUPrefixUpload(LUPrefix* obj, int video) {
    obj->uploaded[video] = true;
    
    // Extend prefix if possible
    while (obj->longestPrefix + 1 <= obj->n && 
           obj->uploaded[obj->longestPrefix + 1]) {
        obj->longestPrefix++;
    }
}

int lUPrefixLongest(LUPrefix* obj) {
    return obj->longestPrefix;
}

void lUPrefixFree(LUPrefix* obj) {
    free(obj->uploaded);
    free(obj);
}

// Example usage
int main() {
    LUPrefix* lup = lUPrefixCreate(4);
    lUPrefixUpload(lup, 3);
    printf("%d\n", lUPrefixLongest(lup)); // 0
    lUPrefixUpload(lup, 1);
    printf("%d\n", lUPrefixLongest(lup)); // 1
    lUPrefixUpload(lup, 2);
    printf("%d\n", lUPrefixLongest(lup)); // 3
    lUPrefixFree(lup);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n) worst case per upload, O(1) per longest()

Upload checks consecutive videos to extend prefix. Longest() just returns stored value.

✓ Linear Growth

Space Complexity

O(n)

Boolean array to track all n videos

⚡ Linearithmic Space

28.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Tracking (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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