Design Video Sharing Platform solution and explanation
Google
The problem
You have a video sharing platform where users can upload and delete videos. Each video is a string of digits, where the ith digit of the string represents the content of the video at minute i. For example, the first digit represents the content at minute 0 in the video, the second digit represents the content at minute 1 in the video, and so on. Viewers of videos can also like and dislike videos. Internally, the platform keeps track of the number of views, likes, and dislikes on each video.
When a video is uploaded, it is associated with the smallest available integer videoId starting from 0. Once a video is deleted, the videoId associated with that video can be reused for another video.
Implement the VideoSharingPlatform class:
VideoSharingPlatform()Initializes the object.int upload(String video)The user uploads avideo. Return thevideoIdassociated with the video.void remove(int videoId)If there is a video associated withvideoId, remove the video.String watch(int videoId, int startMinute, int endMinute)If there is a video associated withvideoId, increase the number of views on the video by1and return the substring of the video string starting atstartMinuteand ending atmin(endMinute, video.length - 1)(inclusive). Otherwise, return"-1".void like(int videoId)Increases the number of likes on the video associated withvideoIdby1if there is a video associated withvideoId.void dislike(int videoId)Increases the number of dislikes on the video associated withvideoIdby1if there is a video associated withvideoId.int[] getLikesAndDislikes(int videoId)Return a 0-indexed integer arrayvaluesof length2wherevalues[0]is the number of likes andvalues[1]is the number of dislikes on the video associated withvideoId. If there is no video associated withvideoId, return[-1].int getViews(int videoId)Return the number of views on the video associated withvideoId, if there is no video associated withvideoId, return-1.
Example 1:
Input
["VideoSharingPlatform", "upload", "upload", "remove", "remove", "upload", "watch", "watch", "like", "dislike", "dislike", "getLikesAndDislikes", "getViews"]
[[], ["123"], ["456"], [4], [0], ["789"], [1, 0, 5], [1, 0, 1], [1], [1], [1], [1], [1]]
Output
[null, 0, 1, null, null, 0, "456", "45", null, null, null, [1, 2], 2]
Explanation
VideoSharingPlatform videoSharingPlatform = new VideoSharingPlatform();
videoSharingPlatform.upload("123"); // The smallest available videoId is 0, so return 0.
videoSharingPlatform.upload("456"); // The smallest available videoId is 1, so return 1.
videoSharingPlatform.remove(4); // There is no video associated with videoId 4, so do nothing.
videoSharingPlatform.remove(0); // Remove the video associated with videoId 0.
videoSharingPlatform.upload("789"); // Since the video associated with videoId 0 was deleted,
// 0 is the smallest available videoId, so return 0.
videoSharingPlatform.watch(1, 0, 5); // The video associated with videoId 1 is "456".
// The video from minute 0 to min(5, 3 - 1) = 2 is "456", so return "456".
videoSharingPlatform.watch(1, 0, 1); // The video associated with videoId 1 is "456".
// The video from minute 0 to min(1, 3 - 1) = 1 is "45", so return "45".
videoSharingPlatform.like(1); // Increase the number of likes on the video associated with videoId 1.
videoSharingPlatform.dislike(1); // Increase the number of dislikes on the video associated with videoId 1.
videoSharingPlatform.dislike(1); // Increase the number of dislikes on the video associated with videoId 1.
videoSharingPlatform.getLikesAndDislikes(1); // There is 1 like and 2 dislikes on the video associated with videoId 1, so return [1, 2].
videoSharingPlatform.getViews(1); // The video associated with videoId 1 has 2 views, so return 2.
Example 2:
Input ["VideoSharingPlatform", "remove", "watch", "like", "dislike", "getLikesAndDislikes", "getViews"] [[], [0], [0, 0, 1], [0], [0], [0], [0]] Output [null, null, "-1", null, null, [-1], -1] Explanation VideoSharingPlatform videoSharingPlatform = new VideoSharingPlatform(); videoSharingPlatform.remove(0); // There is no video associated with videoId 0, so do nothing. videoSharingPlatform.watch(0, 0, 1); // There is no video associated with videoId 0, so return "-1". videoSharingPlatform.like(0); // There is no video associated with videoId 0, so do nothing. videoSharingPlatform.dislike(0); // There is no video associated with videoId 0, so do nothing. videoSharingPlatform.getLikesAndDislikes(0); // There is no video associated with videoId 0, so return [-1]. videoSharingPlatform.getViews(0); // There is no video associated with videoId 0, so return -1.
Constraints:
1 <= video.length <= 105- The sum of
video.lengthover all calls touploaddoes not exceed105 videoconsists of digits.0 <= videoId <= 1050 <= startMinute < endMinute < 105startMinute < video.length- The sum of
endMinute - startMinuteover all calls towatchdoes not exceed105. - At most
105calls in total will be made to all functions.
Approach overview
Problem Overview: You need to design a simplified video sharing platform that supports uploading videos, deleting them, watching a substring of the video, and tracking likes, dislikes, and views. The tricky part is reusing the smallest available video ID after deletions while keeping all operations efficient.
Approach 1: Sequential ID Scan with Hash Table (O(n) allocation, O(1) average ops)
A straightforward design stores videos in a hash map where the key is the video ID and the value stores the video string plus metadata (views, likes, dislikes). When uploading, scan from 0 upward to find the first unused ID. Operations like watch, like, and dislike become constant-time hash lookups. The downside is the upload step: scanning for the smallest free ID can degrade to O(n) time when many IDs are occupied. This works for small datasets but does not scale well when uploads and deletions are frequent.
Approach 2: Hash Table + Ordered Set / Min-Heap for Reusable IDs (O(log n) per update)
A better design keeps two structures: a hash table mapping video IDs to video objects, and an ordered set (or min-heap) storing IDs freed after deletions. When uploading, check the ordered set first. If it contains IDs, remove the smallest one and reuse it; otherwise assign the next sequential ID. Deleting a video removes it from the hash map and inserts its ID into the ordered set for reuse.
Each stored video keeps the raw video string and counters for views, likes, and dislikes. The watch operation extracts a substring from startMinute to endMinute and increments the view counter. Like/dislike operations update counters directly. Hash lookups give O(1) average time for video access, while ID reuse costs O(log n) due to ordered set operations. This design matches the constraints of most system design style problems.
Some implementations replace the ordered set with a reusable ID stack if the problem does not require the smallest ID specifically. However, this problem requires the smallest available ID, making a min-heap or ordered set the correct choice.
Recommended for interviews: The hash table + ordered set approach. The brute sequential scan shows you understand the ID reuse requirement, but interviewers expect you to optimize allocation using a min-heap or ordered set. It keeps uploads and deletions efficient while maintaining constant-time video operations, which demonstrates strong design thinking.
The written solution for this problem is still being prepared. Try it yourself in the meantime.
Try solving it yourselfComplexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Sequential ID Scan + Hash Map | Upload O(n), other ops O(1) | O(n) | Simple implementation when uploads are small and ID reuse frequency is low |
| Hash Map + Min-Heap / Ordered Set | Upload/Delete O(log n), access O(1) | O(n) | General optimal solution when IDs must be reused in ascending order |
| Hash Map + Stack for Recycled IDs | O(1) | O(n) | When smallest-ID ordering is not required and constant-time reuse is preferred |
Frequently asked questions
Is Design Video Sharing Platform easy or hard?
Design Video Sharing Platform is rated Hard because it requires building a full data structure with multiple operations and efficient ID reuse. The challenge lies in coordinating hash-based storage with ordered ID management while keeping operations efficient.
Design Video Sharing Platform Python/Java solution
The typical Python or Java solution implements a class with internal structures: a dictionary or HashMap for video storage and a PriorityQueue (min-heap) for reusable IDs. Each method updates counters or retrieves substrings while maintaining O(1) access and O(log n) ID management.
How to solve Design Video Sharing Platform in O(log n)?
Maintain a hash map from videoId to video metadata and a min-heap containing IDs freed after deletions. When uploading, reuse the smallest ID from the heap if available; otherwise assign the next sequential ID. This keeps allocation and deletion at O(log n) while all other operations remain O(1).
What is the best approach for Design Video Sharing Platform?
The most effective approach uses a hash map to store video data and a min-heap or ordered set to track reusable video IDs. The hash map enables O(1) average lookup for operations like watch, like, and dislike. The ordered set ensures the smallest available ID is reused after deletions, keeping uploads and removals at O(log n) time.
Is Design Video Sharing Platform asked at Google/Amazon/Meta?
Design-style problems like this frequently appear in interviews at companies such as Amazon, Meta, and Google. They evaluate understanding of data structure design, efficient ID management, and handling multiple operations on shared state.
What data structure is used in Design Video Sharing Platform?
The core structures are a hash table for storing video metadata and an ordered set or min-heap for tracking reusable video IDs. Each video record stores the video string along with counters for views, likes, and dislikes.
What is the time complexity of Design Video Sharing Platform?
Using the optimal design with a hash map and min-heap, uploads and deletions run in O(log n) time due to heap operations. Video access operations such as watch, like, dislike, and retrieving statistics run in O(1) average time. Space complexity is O(n) for storing videos and metadata.