I applied online. The process took 3 weeks. I interviewed at MathWorks (Boston, MA) in Sept 2018
Interview
Pretty much the same as what people said here except the onsite.
Onsite is a terrible experience. Basically, one guy handling and control the entire interview process. Look at your resume tried to pick a problem from it.
My first hour during the tech interview, the guy is trying to ask me to turn a project on my resume which served as an online scheduling server into a school general service website.
The next hour is working on a question LeetCode 255. When I tried to code or start on something he asked me to stop because my direction is not the same as the solution he has in hand.
2 hour passed.
It was a technical interview and contain 2 questions, mainly medium level. But it was fairly doable and takes time to comprehend. But interviewer can explain the prompt to you.
I applied online. The process took 4 weeks. I interviewed at MathWorks (Cambridge, England) in Jan 2022
Interview
There were 3 stages:
- Technical interview (1 hour) where they tested mathematics and computing knowledge
- Behavioral interview (1 hour) whit questions on your CV
- HR interview (20 mins) concerning logistics matters
Interview questions [1]
Question 1
Compute function for Fourier series using different methods and compare computation time for different solutions
coding interview 2 coding Questions + 15 MCQ's
Coding only in C,C++,Java
Test conducted on Hacker rank. Dynamic programming and basic DSA / Algo were tested. C++ requires knowledge of STL
Interview questions [1]
Question 1
1. Given a sequence of binary numbers (0, 1) of size ‘n’ in which every 0 represent a decrement to the sum and every 1 represents the increment to the sum. You have to find the minimum possible value of ‘k’ such that the sum of the first ‘k’ elements is greater than the sum of remaining elements.
The initial sum for both the half is 0.
Eg.
given sequence – 101011
The minimum possible value of k – 6
given sequence – 1001001
The minimum possible value of k – 0