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Rippling QA Engineer System Design Questions

41 practice questions for Rippling QA Engineer interviews

Rippling QA engineer interviews test automation frameworks, test strategy, CI integration, performance testing, and debugging complex multi-service systems.

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System Design
system design Hard Verified Question #1

1. Top 5 Rippling System Design Questions


Category: Sliding window system design problem
These are commonly asked system design questions from Rippling interviews
Input: Given input
Output: Computed result
system design Senior ci cd #1

1. [OA] Flaky Test Detector — design a system to identify and manage flaky tests in Rippling’s CI/CD pipeline.

Flaky tests can undermine the reliability of Rippling's software releases. Implement a class that collects results, tracks test history, and flags tests that show consistency in flakiness over time.
- class FlakyTestDetector:
- def __init__(self): Initializes the detector object.
- def add_result(self, test_id: str, result: bool) -> None: Records the result of a test.
- def is_flaky(self, test_id: str) -> bool: Determines if a specific test is flaky based on its results.
Example 1:
Input: detector = FlakyTestDetector(); detector.add_result('testA', True); detector.add_result('testA', False); detector.is_flaky('testA')
Output: True
Explanation: The test has failed at least once in its last few invocations, necessitating review.
Constraints:
- 1 <= test_id <= 10000
- Up to 1000 results can be recorded for each test.
system design Senior caching #2

2. [OA] LRU Cache — implement the caching layer Rippling uses for frequent API responses.

Rippling's services rely on efficient caching solutions to improve response times and reduce database load. You need to implement an LRU Cache that manages API responses effectively.
- class LRUCache:
- def __init__(self, capacity: int): Initializes the LRU Cache with a specified capacity.
- def get(self, key: int) -> int: Returns the value for the specified key, or -1 if not found.
- def put(self, key: int, value: int) -> None: Updates or inserts the value for the specified key.
Example 1:
Input: cache = LRUCache(2); cache.put(1, 1); cache.put(2, 2); cache.get(1)
Output: 1
Explanation: Returns 1, as key 1 is present in the cache.
Constraints:
- 1 <= capacity <= 3000
- 0 <= key, value <= 10^4.

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