How does the "return year%4==0..." sentence know what you are returning is a boolean? Is that defined by "def is_leap"? Thank you

some of us do that for clarity sake for those who may have a hard time processing what is going on.

This is what i did.

def is_leap(year):    
    if year%4==0:
        return True
    elif year%100==0:
        return False
    elif year%400==0:
        return True

Thanks for explaining

You can write it without parentheses like this:

def is_leap(year):
    return year % 4 == 0 and year % 100 != 0 or year % 400 == 0

good one man.you seem to have pretty good understanding of python PEP.

why parenthesis are required and we get error without them??

if** year % 4 == 0 and (year % 400 == 0 or year % 100 != 0)** is true the how the true boolean is affected to leap variable since we didn't you didn't make leap=something? I hope you get the idea

this is nice. it avoids repetition. Thanks alot.

learners need to code like this imo. I use some of these programs with my classes and they simply dont understand any inline return equation, hell some of them dont even return values properly

I agree, the multiple return statements are a tad heavy and go against DRY coding practices. However, your recommended solution also sacrifices readability, which IMO is more important than trying to massage your code into a single statement.

Be succinct but be readable. Let your bundlers handle the optimizations.

Your answer isn't really very clean code though. So while it works, it isn't very readable, which IMO is incredibly important.

Thanks that explained it so clearly

Makes perfect sense, thank you very much for your explanation!

can you explain why you use the 'or' operator instead of 'and' for the 'year%100 != 0'? i still don't get it. thank you

I like your func very much

That's way shorter than my if and else loop's. Thanks.

Hi @kpagcha,

Thanks for your detail explanation for this problem. Could you please explaine to me the relationship of your answer and the problem context, please? Because I didn't understand well the answer.

I want to unerstand how did you formulated these steps in this form you had programmed

In the Gregorian calendar three criteria must be taken into account to identify leap years:

1. The year can be evenly divided by 4; 2. If the year can be evenly divided by 100, it is NOT a leap year, unless; 3. The year is also evenly divisible by 400. Then it is a leap year

Thanks for advance. Fouzi

very effective buddy.... thnk u

if you want to reduce redundancy, why not also return the bool of the comparisons to avoid explicitly checking for zero equivalency..

def is_leap(year):
    return bool(not year % 4 and (year % 100 or not year % 400))

irregardless.. you deserve an upvote

Your solution is certainly the most elegant, but when the user is first presented with the problem in the coding box, the return command has already been given the object 'year' and we are given a #Write your logic here note above it. So it's at least forgivable for some of us to construct multiple 'if' blocks to solve it, wouldn't you agree?

it is not passing all the test cases?

thanks :D

Might be clearer:

def is_leap(year):
    return year % 400 == 0 or year % 4 == 0 and year % 100 != 0

Hi just a question about parenthesis placement why does placing a parenthesis like below year % 4 == 0 and (year % 400 == 0 or year % 100 != 0) is correct code versus not placing parenthesis at all or writting it like ((y%4==0)and(y%100!=0)or(y%400==0)) results in error

thanks for advice, keep commenting :)

Hi, thanks for the shorter approach but, what about the condition-- 1900<= year <=10**5

Thank you sir for the advice.

Thanks for this. I've been learning for 4 days now and I really want to be a good coder that abides by the zen of python. Because of your advice I've turned the default code:

def is_leap(year):
    leap = False
    if 1900 <= year <= 10**5 and year % 4 == 0 and \
		(year % 100 != 0 or year % 400 == 0):
        leap = True
    
    return leap
year = int(input())
print(is_leap(year))

into this DRY code:

def is_leap():
    return 1900 <= year <= 10**5 and year % 4 == 0 and \
		(year % 100 != 0 or year % 400 == 0)
year = int(input())
print(is_leap())

I'm sure there are ways to improve this further.

That's crazy simple. Wow. I did do if elif and elif. if year%4 == 0: leap = True elif year % 100 == 0: leap = False elif year % 400 == 0: leap = True

            for some reason it was wrong for 2100 the test case too. That makes so much sense.  and makes it so that both have to be true. Truly a masterful stroke.

agree with you on "dry" . succintness is power and a bloated code is such a waste of python zen.

moreover i dislike it that the code is partially provided: it shoeboxes one into a specific pattern (priming).

the instructions werebloatedtootoomucuunnecessaryinformationpresentedunclearlyinordertoappearsiphisticatedlikethislongword.

Well, the problem is set up with variable leap = False and then return leap at the end, which surely guides the beginners to the if-else method.

def is_leap(year):
    leap = False
    
   ###
    
    return leap

how it will come can u explain

but for 1900,its divisible by 4 but also by 100..so how is that code possible

Wow, amazing comment, thank you!

Thanks for the lesson. For other newbies here is a good article worth reading that will clarify the syntax: https://www.codesdope.com/python-boolean/

failed 2 cases

Your code is certainly shorter, but you also make a slight sacrifice in readability. Of course, the 'readability' of your code will depend on the experience level of the reader. But you won't always be working with talented programmers. For many, I think the following code is more readily understandable and its decrease in computational efficiency is marginal:

def is_leap(year):
    if year % 400 == 0:
        return True
    if year % 100 == 0:
        return False
    if year % 4 == 0:
        return True
    else:
        return False

Thanks for the advice. I am new to python and this something that i would have never thought of.

Thank you for the tip. The logic is so clear when I lok at your code. For whatever reason it isn't straightfoward reading from the problem statement.

everytime I think I have an understanding of a nice logical approach to a problem simple or not, I seem to always be blown away by a simpler methodology. I am completely new to python so I know these lightbulb moments are going to keep happening. Thanks for the simplification. ps. I went straight for the if and else statements..

why to add OR for % 100 condition? Not able to understand the logic. Please help!!

Thank you, this helped

All my respectz.

Thank you for the huge help, I kept trying with if's and got nowhere.

Why:

is_leap(2020) True.

is_leap(2021) False.

... that's not true

Wow. I did it in a much length way with lots of ifs and elses.

Thank you for sharing this.

Indeed the way to go, here are some print statement might help understand the logic:

    print('\nFor year',year,':')
    print('\tyear%4 =',year%4)
    print('\tyear%400 =',year%400)
    print('\tyear%100 =',year%100)
    print('\nTherefore:')
    print('\tyear%4 == 0 is',bool(year%4 == 0))
    print('\tyear%400 == 0 is',bool(year%400 == 0))
    print('\tyear%100 != 0 is',bool(year%100 != 0))
    print('\nSo:')
    print('\tyear%4 == 0 is',bool(year%4 == 0))
    print('\tAND')
    print('\t( year%400 == 0 is',bool(year%400 == 0))
    print('\tOR')
    print('\tyear%100 != 0 is',bool(year%100 != 0),')')
    print('\nBecomes:')
    print('\tyear%4 == 0 is',bool(year%4 == 0))
    print('\tAND')
    print('\t(year%400 == 0 or year%100 != 0) is',bool(year%400 == 0 or year%100 != 0))
    print('\nWhich brings us to:')
    print('\t',bool(year%4 == 0),'AND',bool(year%400 == 0 or year%100 != 0),'is',bool(year%4 == 0 and (year%400 == 0 or year%100 != 0)))
    print()

Thanks for your input! I'm new to programming so this is very helpful.

This is a much better approach. Thanks

man can somebody tell me the right answer so i can copy it?