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      Entretien pour Software Engineer

      19 mars 2024
      Employé (anonyme)
      Paris
      Offre acceptée
      Expérience neutre
      Entretien moyen

      Candidature

      J'ai postulé en ligne. Le processus a pris 3 mois. J'ai passé un entretien chez Prophesee (France) (Paris) en sept. 2023

      Entretien

      Introduction with HR manager about the role and why you applied, basic discussion about salary expectations, etc. Introduction with manager of your department about what prophesee does and what they want to do in the future. Also about your background and how its applicable with their line of work. Coding assignment with 2 code tests and some multiple choice questions about C/C++, Git, agile, swe in general. Technical interview with team members, (prepare your 2 coding tests solution, as it will be discussed why you made choice xyz). Final discussion with manager about knowledge of CI/CD.

      Questions d'entretien [4]

      Question 1

      Due to current leaks in the event-based sensor pixels, noisy events are generated from random pixels at a low frequency. A simple strategy to filter this background noise is to consider that an input event ([xi,yi],ti) is due to noise unless at least one event ([x,y],t) was recently received in its spatial neighborhood. More formally, given a spatial radius R>0 and a temporal threshold T>0 an input event ([xi,yi),ti) is considered noisy unless we can find at least one past event ([x,y],t) such that: • The past event is in the spatial neighborhood of the input event: x¡-R <= x <= xi+R and yi-R <= y <= yi+R • The past event was received recently: t₁ - T <= t < 640 and 0<=y<480 for all events • the output vector is initially empty
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      Question 2

      Unlike conventional cameras, which generate a stream of images, event-based cameras generate a stream of timestamped events carrying pixel coordinates where changes of luminosity happened: struct Event { int x, y; //position unsigned long long t; //timestamp }; In practice, the event stream is typically sliced in successive buffers of events for further processing and stored in standard vectors: std::vector buffer; Real processing pipelines often need to apply different algorithms on the same input buffers and merge their results later in the processing chain. The goal of this exercise is to write a function taking as input two event buffers, both assumed to be ordered chronologically, and returning the merged event buffer still ordered chronologically: void chronological_merge(const std::vector& input1, const std::vector& input2, std::vector& output) { // TODO } As an example, if input1 = {([0,0],0), ([1,0],2)} and input2 = {([0,1],1)}, then the expected output is output={([0,0],0), ([0,1],1), ([1,0],2)}. To simplify, you can assume that all input events have different timestamps (i.e. there is no duplicate timestamps in the input events) and that the output vector is initially empty.
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      Question 3

      We define the following notations for ordering the complexities of sorting algorithms A and B A B means "A has lower average complexity than B, ie. A will be faster than B on average when sorting a sufficiently high number of elements A B means "A has higher averoge complexity than B, ie. A will be slower than B an average when sorting a sufficiently high number of elements AB means "A and 8 have similar average complexity Which ordering of the following sorting algorithms is correct? quick sort merge sort bubble sort insertion sort quick sort merge sort bubble sort insertion sort quick sort merge sort bubble sort insertion sort quick sort merge sort bubble sort insertion sort none of the above
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      Question 4

      Which statement about classes provided in the STL is true? Insertion, suppression and lookup in a stat unordered map take a constant time on average, le, the same time no matter the number of elements stored Insertion, suppression and lookup in a std map take a linear time on average, l.e. a time proportional to N where N is the number of elements stored Insertion and suppression in a stad:list take a logarithmic time on overage, le a time proportional to log(N) where N is the number of elements stored
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