1. 11 months ago
Anonymous

>sharp algorithm skills
>cannot fizzbuzz
how does this happen?

• 11 months ago
Anonymous

by lying on their resume.

• 11 months ago
Anonymous

More likely not knowing what an algorithm is

• 11 months ago
Anonymous

You have no idea the delusional type of people that the money and glamor of Silicon Valley can attract, 150 years ago they’d be dying of exposure in the Klondike convinced they know how the best way to dig gold

2. 11 months ago
Anonymous

I might have to start interviewing candidates this year and if I get someone this retarded I'm gonna fucking lose it.

3. 11 months ago
Anonymous

does anyone have the one that is a notepad++ screenshot where some guy describes all the fizzbuzz interviews he did

• 11 months ago
Anonymous

gotcha covered

• 11 months ago
Anonymous

that made me start to doubt that I could actually do those

• 11 months ago
Anonymous

this made me feel like i could be a professional coder. Are coders really this bad or is it just people lying on their resumes about experience?

• 11 months ago
Anonymous

Theres a lot of bad people out there.

• 11 months ago
Anonymous

These people were looking for a job for a reason.

• 11 months ago
Anonymous

>i = i%1

• 11 months ago
Anonymous

>Find all odd numbers between 0-100
>for(int i = 1, i >= 100, i++)
> i >= 100
> >=
>8 years experience
kek

• 11 months ago
Anonymous

it's also worth noting that 2 is a prime number (it is only divisible by itself and 1), so the interviewer is wrong in saying that odd numbers are 1,3,5 etc and there needs to be an extra statement stating if i == 2 print i

• 11 months ago
Anonymous

um

I-

yes good point

• 11 months ago
Anonymous

>the interviewer is wrong in saying that odd numbers are 1,3,5 etc and there needs to be an extra statement stating if i == 2 print i
Sorry, but we don't like know-it-alls around here, better luck next time.

• 11 months ago
Anonymous

1 is not prime though

• 11 months ago
Anonymous

I'm actually curious, what's a more efficient way to make a list of all odd numbers between x and y?

• 11 months ago
Anonymous

for a in list(range(100)):
if a%2 == 1:
print(a)

or if you like, you can append 'a' to a list and then print the list after the loop is done

• 11 months ago
Anonymous

>not using
>if a&1:

• 11 months ago
Anonymous

dont prioritize shit that your compiler is gonna do anyway over readability

• 11 months ago
Anonymous

Look at the last bit of the number. If it is 1, then it is odd.

• 11 months ago
Anonymous

butts = x || 1

while butts < y:
print(butts)
# or append to list instead
butts += 2

• 11 months ago
Anonymous

If x is odd, start at x and increment by 2 until x is greater than y
otherwise, start at x+1 and increment by 2 until x is greater than y

• 11 months ago
Anonymous

this cant be real

• 11 months ago
Anonymous

Coding interviews are nothing like this.

Either this was written in 1998, or your company is paying garbage-tier wages, or they're bringing in high school students to interview.

Either way, if your interview candidates are that shitty it's totally on you.

• 11 months ago
Anonymous

This reads like an SCP entry listing interactions

• 11 months ago
Anonymous

These people should be contained for their own safety.

• 11 months ago
Anonymous

I feel like you could get some of these if, every time you hit newline, you weren't allowed to see any of the previous lines you wrote.

4. 11 months ago
Anonymous

I've only ever used small amounts of R or python for data analysis or running motors but wouldn't that just be something simple like (mind the pseudocode)

for number in range(1,16):
if number divisible by 3 and 5:
print("FizzBuzz")
elif number divisible by 3:
print("Fizz")
elif number divisible by 5:
print("Buzz")
else print(number)

I don't know what a timecode string is, I'm guessing something like 12:34:24 where it's hours:min:seconds? Seems like an easy enough division.

5. 11 months ago
Anonymous

This is the result of the bootcamps and the campaigns convincing people to learn how to write code.

6. 11 months ago
Anonymous

Everybody get ready to laugh at the CSfags who think they can do better.

• 11 months ago
Anonymous

def product(a, b):
for i in range(b):
a += a
return a

Is this right? I'm a retard who's never coded.

• 11 months ago
Anonymous

oh I made a retard. Replace that section with
c += a
return c

• 11 months ago
Anonymous

You made me curious so I checked and it still works.

• 11 months ago
Anonymous

I disagree, they are the same level.

• 11 months ago
Anonymous

The first one only works for integers up to 10. No floats at all.
The second one works for any integers and 1 float. That's unarguably an improvement, but yes it needs to be fixed.

• 11 months ago
Anonymous

Well obviously they had to stop somewhere, ten just gets the point across. Code overhead is a lot but the runtime is much faster. So there are trade offs.

• 11 months ago
Anonymous

Wtf are you high.

• 11 months ago
Anonymous

>Could fix that with a check, maybe another 3 lines?

Then what do you do with the fractional part of b?

• 11 months ago
Anonymous

Remove it, I'm a chemist it doesn't matter at our quantities when the inherent equipment uncertainties will far outweigh it :^)

• 11 months ago
Anonymous

I set a and b to both be 1.5:
TypeError: 'float' object cannot be interpreted as an integer

• 11 months ago
Anonymous
• 11 months ago
Anonymous

I solved it, awaiting job at MIT:

def product(a,b):
return a/(1/b)

• 11 months ago
Anonymous

genius

• 11 months ago
Anonymous

That was my first thought but isn't the challenge to avoid using any such operators?

• 11 months ago
Anonymous

obey the letter of the law if not the spirit

• 11 months ago
Anonymous

ah yes i see, 3*0 is indeed "ZeroDivisionError: division by zero"

• 11 months ago
Anonymous

This holy fuck, if you are going to write ANY code at least know basic algorithms like Russian multiplication (repeated doubling). It's the base for thousands of other similar algorithms. As long as you have a monoid you can use it.

• 11 months ago
Anonymous

https://pastebin.com/m4s6v4QB

• 11 months ago
Anonymous

I didn't realize C++ doesn't rollover. Very nice.

7. 11 months ago
Anonymous

Imagine the kind of person that posts here in a STEM career gen thread, asking if they need to get a degree in CS or if their HS experience making a couple scripts qualifies them for 80k-250k developer jobs. Those people pumped up on confidence and dunning-kruger waltz into interviews and waste everybody's time. It is usually either them, or boomers who just mass spam every single job listing and hope someone will hire them without checking into their skill level.

• 11 months ago
Anonymous

Listen son, you don't need a resume when you have a firm handshake. And if your a woman, all the same, except you grab the penis.

8. 11 months ago
Anonymous

How do I into coding?
I'm studying chemistry but I need a backup plan because most jobs you can get with an undergrad degree pay like shit and my grades aren't looking too hot atm.

• 11 months ago
Anonymous

Literally just start writing shit and dicking around. Something like freecodecamp or codecadamy can give you a little structure and a nudge in the right direction.

• 11 months ago
Anonymous

come up with a fun problem and solve it
or
take classes that make you suffer and put up with the suffering

both are equally valid

• 11 months ago
Anonymous

http://101.lv/learn/C++/htm/ch01.htm
Think this is supposed to be a decent resource. I had to do some coding for embedded systems classes and the course coordinator recommended it.

9. 11 months ago
Anonymous

Check this one out.

10. 11 months ago
Anonymous

[...]

5 divided by 2 is : 2 and 1 halves
3 divided by 2 is : 1 and 1 halves
1 divided by 2 is : 0 and 1 halves

The % is counting the halves

11. 11 months ago
Anonymous

[...]

Euclidean division, retard. 1 = 2*0 + 1.

12. 11 months ago
Anonymous
• 11 months ago
Anonymous

I don't see the humor though. Cranking the valve open literally will ruin the entire experiment because you will get the wrong amount and over titrate it. And if you already know the amount required beforehand, then fast titrate to within 5 mL of it, then slow titrate. Literally takes less than 2 minutes to go through a 50 mL burette doing this.

13. 11 months ago
Anonymous
14. 11 months ago
Anonymous

This fizz buzz do you have to use (n % 3 == 0 AND n % 5 == 0) print fizzbuzz elseif n % 3 == 0 print fizz elsif n % 5 == 0 print buzz else print n
n++
or is there a compsci way of doing it

• 11 months ago
Anonymous

You can do n % 15 instead of n % 3 and n % 5
Or you could parameterize it like
Fizz : 15n, 15n + 3, 15n + 6, 15n + 9, 15n + 12
Buzz: 15n, 15n + 5, 15n + 10
n : 15n + 1, 15n + 2, 15n + 4, 15n + 6, 15n + 7, 15n + 8, 15n + 11, 15n + 13, 15n + 14

• 11 months ago
Anonymous

for (i = 1, i <=100, i++) {
if (i%15 == 0) {
print "FizzBuzz" }
else if (i%5 == 0) {
print "Buzz" }
else if (i%3 == 0) {
print "Fizz" }
else
print i }

• 11 months ago
Anonymous

in python:

def fizzbuzz(x):
string = ''
if x % 3 == 0:
string = string + 'fizz'
if x % 5 == 0:
string = string + 'buzz'
return string

for x in range(n):
print(fizzbuzz(x))

• 11 months ago
Anonymous

it has to print out the number itself in cases where the string is empty

15. 11 months ago
Anonymous

A primary school boy could do this in 20 minutes if I taught him about the basics of C. How do these people exist?

16. 11 months ago
Anonymous

>SCIENTISTdock the BRILLIANT
How the FUCK does he do it bros?

17. 11 months ago
Anonymous

You laugh and yet my medfag job interview included(aside from med shit) basic arithmetic/algebra tasks along the lines of
>how would you turn 10ml of 40% solution into 200ml of 5% solution(1% in medicine means it contains 10 mg per ml)
We are surrounded by 90iq monkeys in our daily lives, and its best not to think about it

• 11 months ago
Anonymous

You realize you can't turn 10 mL of 40% into 200 mL of 5%, right? Yet you claim everyone else is the 90iq monkey.

• 11 months ago
Anonymous

His problem is asking how much of the 10ml 40% you combine with fluid in the 200ml container to make the 200ml container 5%. Why is that impossible?

• 11 months ago
Anonymous

1% = 10 mg/mL
40% = 400 mg/mL
10 mL of 40% = 4,000 mg
Want 200 mL of 5%, which is 5 x 10 x 200 = 10,000 mg.
Impossible to get 10k from 4k