53 practice questions for OpenAI Mobile Engineer interviews
OpenAI mobile engineer interviews focus on iOS or Android platform knowledge, memory management, offline-first architecture, and mobile-specific system design.
Category: Trie-based system design problem# System Design Questions - OpenAI A collection of commonly asked system design questions from OpenAI interviews.Input: Given input Output: Computed result
system designSeniorapi design#1
1. Design an Offline-first Sync Engine for OpenAI Mobile
Building robust mobile applications requires careful consideration of user experience and seamless access to AI models from OpenAI, even when users are offline. This is essential for crafting resilient experiences. Your task is to design an Offline-first Sync Engine that manages data synchronization between offline mobile applications and the cloud. Key methods to implement: - sync(data: List<Data>): void: Syncs local data to the server when online. - saveLocally(data: Data): void: Saves data locally when offline. - fetchLocalData(): List<Data>: Fetches data stored on the device. - handleConflict(remoteData: Data, localData: Data): Data: Resolves conflicts between local and remote data.Example 1: Input: sync([{id: 1, value: 'A'}]) Output: [] (Assuming the device is online) Example 2: Input: saveLocally({id: 2, value: 'B'}) Output: nullConstraints: - data.length <= 1000 - 1 <= id <= 10000
system designHardcaching#2
2. [OA] Caching — Manage API response caching for OpenAI's mobile SDK
OpenAI's mobile applications frequently make requests to API endpoints, necessitating a robust caching mechanism to enhance performance and minimize unnecessary network calls. Design and implement an LRU Cache class that stores a limited number of API responses and evicts the least-recently-used responses as new responses are added.Class definition: - LRUCache(int capacity): Initializes the LRU cache with a given capacity. - String get(String key): Returns the value if the key exists in the cache, otherwise returns null. - void put(String key, String value): Updates or adds the value if the key is not already present. When the cache reaches its capacity, it should invalidate the least-recently-used item before inserting a new item.Example 1: Input: cache = new LRUCache(2) cache.put(1, 'A') cache.put(2, 'B') Output: cache.get(1) → 'A' Example 2: Input: cache.put(3, 'C') Output: cache.get(2) → null Constraints: - 1 <= capacity <= 3000 - 0 <= key <= 10^4 - 0 <= value <= 10^4