Janestreet Interview Questions (3+ Questions)
Last Updated: June 23, 2026 • 3 Questions • Real Company Interviews
Prepare for your Janestreet interview with our comprehensive collection of 3+ real interview questions and detailed answers. These questions have been curated from actual Janestreet technical interviews across various roles including DevOps Engineer, Data Engineer, QA Engineer, and more.
Table of Contents
- Handle Disk Leak in Docker (medium) 🔒
- Find Highest Salary (easy) 🔒
- Car Fleet (medium)
Our Janestreet interview questions cover a wide range of technical topics and difficulty levels, from entry-level positions to senior roles. Each question includes detailed explanations and answers to help you understand the concepts and prepare effectively for your interview.
💡 Pro Tips for Janestreet Interviews
- Practice each question and understand the underlying concepts
- Review Janestreet's specific technologies and methodologies
- Prepare follow-up questions and edge cases
- Practice explaining your solutions clearly and concisely
Interview Questions & Answers
1. Handle Disk Leak in Docker
We have a container reporting 87% disk usage, but when we look at the actual files, they only account for a small fraction. The rest is phantom usage caused by deleted files that are still held open by a running process. The files are removed from the directory, but the disk blocks aren't freed beca...
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Detailed explanation and solution available for premium members.
2. Find Highest Salary
Objective
Write an SQL query to retrieve the names and salaries of the highest-paid employee(s) from the employees table.
Additional Information
- The
employeestable consists of the following columns:id(integer): The unique identifier for each employee.name(string): The...
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Detailed explanation and solution available for premium members.
3. Car Fleet
def car_fleet(target: int, position: list[int], speed: list[int]) -> int:
pair = [[p, s] for p, s in zip(position, speed)]
stack = []
# Sort by position (reverse) to process closest to target first
for p, s in sorted(pair)[::-1]:
stack.append((target - p) / s)
if len(stack) >= 2 and stack[-1] <= stack[-2]:
stack.pop()
return len(stack)
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