OpenAI software engineer interviews cover algorithms, data structures, system design, and coding problems drawn from real interview rounds.
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system design
Hard
Verified Question
#1
1. Top 5 Open AI System Design Questions
Category: Trie-based system design problem
System Design Questions - OpenAI A collection of commonly asked system design questions from OpenAI interviews.
Input: Given inputOutput: Computed result
system design
Hard
data structure
#1
1. [OA] MedianFinder — Implement a data structure for keeping OpenAI's result metrics
OpenAI often needs to compute the median of predicted metrics in real-time. This requires an efficient data structure to retrieve the median quickly while processing continuous streams of data.
Class
Input:
Output:
Explanation: The median of [1,2] is 1.5
Input:
Output:
Explanation: The median of [1,2,3] is 2.
Class Definition
Class
MedianFinder should implement the following methods:addNum(num: int) -> None:Adds a number to the data structure.findMedian() -> float:Returns the median of all added numbers.
Example 1:
Input:
medianFinder = MedianFinder(); medianFinder.addNum(1); medianFinder.addNum(2); medianFinder.findMedian()Output:
1.5Explanation: The median of [1,2] is 1.5
Example 2:
Input:
medianFinder.addNum(3); medianFinder.findMedian()Output:
2Explanation: The median of [1,2,3] is 2.
Constraints:
-10^5 <= num <= 10^5- There will be at most
1,000,000calls toaddNumandfindMedian.
system design
Hard
caching
#2
2. [OA] LRU Cache — Design an LRU Cache for OpenAI's model predictions
OpenAI needs an efficient caching system to store previously computed model predictions, optimizing resource usage and response time.
Class
Input:
Output:
Explanation: Returns 1
Input:
Output:
Explanation: Returns -1 since the cache reached its capacity; LRU (key = 2) was evicted.
Class Definition
Class
LRUCache should implement the following methods:get(key: int) -> int:Returns the value of the key if the key exists, otherwise returns -1.put(key: int, value: int) -> None:Updates the value of the key if the key exists existing and moves the key to the front of the cache. Otherwise, it adds the key/value pair to the cache.
Example 1:
Input:
cache = LRUCache(2); cache.put(1, 1); cache.put(2, 2); cache.get(1)Output:
1Explanation: Returns 1
Example 2:
Input:
cache.put(3, 3); cache.get(2)Output:
-1Explanation: Returns -1 since the cache reached its capacity; LRU (key = 2) was evicted.
Constraints:
- The capacity of the cache will be between
1and1000. - All keys will be unique.
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