# 2 mil

Perguntas de entrevista para o cargo de Machine Learning Engineer compartilhadas pelos candidatos

## Principais perguntas de entrevista

Pergunta feita para Machine Learning Software Engineer...21 de janeiro de 2010

### Suppose you have a matrix of numbers. How can you easily compute the sum of any rectangle (i.e. a range [row_start, row_end, col_start, col_end]) of those numbers? How would you code this?

7 respostas

Compute the sum of the rectangles, for all i,j, bounded by (i,j), (i,m), (n,j), (n,m), where (n,m) is the size of the matrix M. Call that sum s(i,j). You can calculate s(i,j) by dynamic programming: s(i,j) = M(i,j) + s(i+1,j) + s(i,j+1) - s(i+1,j+1). And the sum of any rectangle can be computed from s(i,j). Menos

Awesome!!

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6 respostas

What were the online coding questions like? Could you elaborate?

Object detection. Is that what yours was?

it is same as mine. Could you give me more details about the online coding? what algorithm did they test on object detection part? Menos

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### What are some of the projects that you have done?

4 respostas

Do you mind to share what are the hard leetcode questions they asked during the interview? Menos

I dont think it's fair to share which question they asked. But the exact same question is on leetcode and the difficulty level is hard. Menos

What topic you are being ask from in leetcode? also did they ask you system design and CS fundamentals. Menos

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### Give an image, when we take 2 sub images from it, calculate the ratio similar to AnB/AuB.

4 respostas

Coded in python but wasn't able to finish it

Can you elaborate on the question

Given a matrix and coordinates of 2 rectangles calculate the weighted IoU in linear/constant time. Menos

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### how to sort in O(Logn) time

3 respostas

I don't think you can sort in O(logn) because you will need to go through the whole data at least once, making it O(n). Indeed, you can do it in O(logn) if the data is guarantee with some specific constrain or relationship. I think the best you can sort a completely random data is O(nlogn). Menos

I didn't come up with the answer. it is not difficult, just not prepared

what is the question

### Why does one use MSE as a measure of quality. What is the scientific/mathematical reason for the same?

3 respostas

Mean-Square error is an error metric for measuring image or video quality it is popular video and image quality metric because the analysis and mathematics is easier with this L2-Norm metric. Most video and image quality experts will agree that MSE is not a very good measure of perceptual video and image quality. Menos

The mathematical reasoning behind the MSE is as follows: For any real applications, noise in the readings or the labels is inevitable. We generally assume this noise follows Gaussian distribution and this holds perfectly well for most of the real applications. Considering 'e' follows gaussian distribution in y=f(x) + e and calculating the MLE, we get MSE which is also L2 distance. Note: Assuming some other noise distribution may lead to other MLE estimate which will not be MSE. Menos

MSE is used for understanding the weight of the errors in any model. This helps us understand model accuracy in a way that is helpful when choosing different types of models. Check out more answers on InterviewQuery.com Menos

### Please code up and send me a function that takes two integer arrays and returns their intersection. This answer must take less than n^2 time.

3 respostas

Use a hash table or tree.

modify merge sort

sample outline of O(n log n) algorithm : a.sort(); b.sort(); list c={}; int i1=0,i2=0; while(true) { if(i1==n || i2==n) break; if(a[i1]==b[i2]) { c.insert(a[i1]); i1++; i2++; }else { if(a[i1] &lt; b[i2]) i1++; else i2++; } } return c; Menos

### Définir une fonction qui met à plat des listes de liste.

3 respostas

flatten = lambda l: [item for sublist in l for item in sublist]

the lambda is useless here. just do : f = [x for xs in a for x in xs]

@Paolo You need either a lambda or a def, else you don't define a function that flatten a list of lists. Menos