Python Coding Interview Preparation - For Beginners

In this video I'll be doing some coding interview preparation by answering an fairly easy python programming problem. This is meant to help you prepare for your coding interviews!

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Tags:
- Tech With Tim
- Python Tutorials
- Python Coding Interview
- Coding Interview Python
- Python Coding Interview Prep
- Interview Prep

#Python #CodingInterviewPrep

[nizaranyonemos]

Thank you my friend for this information

[malmiteria]

I would use generators for this:

def iterFib():
yield 0
yield 1
first, second = 0, 1
while True:
first, second = second, first + second
yield second

def getNthFib(n):
fib_seq = iterFib()
for _ in range(n-1):
next(fib_seq)
return next(fib_seq)

[jacobbringham]

Isn’t there a way to solve in constant time using Binet’s formula?

[gingerboy]

Dude... You're blowing ma mind with that recursion....🤐

[andrewbarzu]

Both solutions are kinda bad, instead of doing recursion and filling the stack and calling a lot of functions which is expensive in time you could just do it iteratively, which is a lot faster, doesn't put anything on the stack, and has constant memory

[kuchigyz]

I did it with a list and a for loop in the most simple way possible but I forgot about the odd case where n is 0, 1 or 2 xd

[paulschmidt]

11:37 I always avoid the "in" statement as a it is very slow because it has to check every instance. It would be much more efficient to use a try except block instead. try: return calculated[n] except: (something)

[mandeepubhi]

1st case without storing values runs in O(2^n), However 2nd case in O(n). But we can achieve O(n) without using hash table, But using tuples, It is followed as:-
Def Good_fibonacci(n) :
If n<=1:
Return (n, 0)
Else:
a, b=Good_fibonacci(n-1)
Return a+b, a
In this case, function has been called recursively for one time for one operation which eliminates O(2^n). Hence, it runs with O(n)

[davidmurphy]

It's contra to pep8 to put a space around the equals when assigning an attribute in a function. Tut, tut.

[jupahe]

why would you use recursion?

if n==2:
return 1;
elif n==1

return 0;

int prev = 0;
int curr = 1;
for i in range(1, n):
int next = curr + prev;
prev = curr;
curr = next;


return curr;

does the job just fine and I assume it's way faster

[marilynbrazzell]

I have never done any coding before but i am really interested in it! Can anyone tell me some websites that are for total nubes?

[SybotLV]

No need to cover fib(3) in the if part, 0+1=1 ;)

[pushpajitbiswas]

Was that a hash table which you used in your 2nd code which executes your code more faster?

[hasantao]

Time complexity of this method is exponential. Avoid recursion when you can solve the problem in an iterative fashion.

[kaushallyadealwis]

or you could do this

def Fibonacci(n):
fibNum = [0, 1]

if n <= 1:
fibNum = [0]

while len(fibNum) != n and n >= 3:
fibNum.append(fibNum[-2] + fibNum[-1])

return fibNum

Though this is longer.

[ashvinpkumar]

Literally me seeing this video after watchin algo expert add

[andrerodriguez]

What's the advantage over premium leetcode + free YouTube problem solving channels ?

[Kirchert]

hey tim, i tried solving the question befor watching your/ the solution. this is what i came up with:


def getNthFib(n):
list = [0, 1]
for number in range(2, n):
fib_number = list[number-1] + list[number-2]
list.append(fib_number)

print(f'Input: {n}')
print(f'Output: {list[n-1]}, {list}')


from my testing it finds the right number and makes the right list. but it failes all the tests :(..

[adithmanu]

Does this work?
nlist=[0, 1]
for i in range(10):

[ehizuelenabhulimen]

Your solution didnt show the result stored in a list, it still spits out single values