kpagcha For the learners: you should know that doing something like the setup for this challenge inclines you to do is a bad practice. Never do the following:
def f(): if condition: return True else: return False
This is just dumb. You are returning a boolean, so why even use if blocks in the first place? The correct what of doing this would be:
def f(): return condition
Because this already evaluates as a boolean.
So in this challenge, forget about ifs and elses, and that leap variable, and just do the following:
def is_leap(year): return year % 4 == 0 and (year % 400 == 0 or year % 100 != 0)
Don't be redundant, be DRY.
dujincong_dufe 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
kpagcha because all expressions used in that return statement evaluate to booleans
IndianRockPython thanx
_Kalpesh_ thanks brother
Krishnasaivamsi sir,can you give a clear description about return statements
omj1206 That exactly is not the case. Try to run following code in interpreter:
def is_leap(year):
return (not(year%4) and ((year%100) or not(year%400)))Invoke it using
is_leap(1464)
It returns 64 instead of True. Thats because the expression (year%100) returns 64
omj1206 to mitigate that run following code and that returns either True or False:
def is_leap(year): return (not(year%4) and (bool(year%100) or not(year%400)))
apatil_cs How does that contradict what that guy said? You wrote different code than what he wrote. You're exploiting "truthy/falsey" values; the other guy's just using regular booleans
srious Check out python truthiness
jasar002 [deleted]
ajayi some of us do that for clarity sake for those who may have a hard time processing what is going on.
darshanjadav97 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
arpita0116 the first if statement if year%4==0: return True this will give true for 1900 year also but that it is not true
snath402 you have to give the constraint like
if year in range(1900,pow(10,5)+1): because they clearly mentioned this constraintr
dansiman I thought the constraints meant that you could write your code assuming the constraints to be true - that no test cases would include inputs that don't satisfy the constraints.
ujjwalsrivastav that is true
a_kamal_89 You can just reverse the order and it will be fine.
zrolpin Reversing the order almost works, but you have to add an
elseblock returningFalseat the end to capture cases where none of theifblocks execute (like for the year 1992) and also change allelifstatements toifstatements. I.e: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
This passes all the test cases.
sjayanthkumar191 i will mark your words
bmp17c not passes year %100 == 0 test
danhoffman7 Why wouldn't 1992 return true if % 4 == 0 is the first statement? it divides cleanly by 4 with no remainder, yet it gives false in the original scenario.
anusha7m true
dishapandey77 why we have to change elif statements to if statements sir??
bhatt_siddharth2 i did the same thing but it's not working for year 2100 2100 is a leap year,but according to the compiler we should get False.
navi7299 I reversed the order and it passed all the tests. But i dont know why it did. what did the reversing the order do??
ryoutausami_pyd1 Took a few tries before I understood the conditions but I got it!
Here's my solution:
if year % 4 == 0: if year % 100 != 0: leap = True elif year % 400 == 0: leap = True return leaphhelzer That wound up failing two test cases.
Krishnasaivamsi how can you return heap without any function
Dvcks This quote is the function. The post doesn't include the declaration because it's not relevant.
sjayanthkumar191 fail in two cases
dhagist What about trying something like this:
if year % 4 == 0: leap = True if year % 400 == 0: leap = True elif year % 100 == 0: leap = Falsepratibhakumar225 this works for all test cases
hgahlot50 It fails for the year 2100
snath402 It is giving runtime error
pratibhakumar225 def is_leap(year): leap = False if year % 4 == 0: leap = True if year % 100 == 0: leap = False if year % 400 == 0: leap = True return leap
It woked for me in all cases... Use a proper identation
[deleted] make the last loop as elif year%100!=0: return True
this will work for all the cases
zycarlton09 Good Work ！ I think I ignored the "unliess" in the problem sheet. I learned the if and elif, but never used them. This is an interesting question.
skoonlol This is what I did.
# Write your logic here if (year % 4) == 0: if (year % 100) == 0: if (year % 400) == 0: return True else: return False else: return True else: return False return leapyear = int(input())
yogeshverma28698 def is_leap(year): leap = False if year%4==0: leap=True elif year%100==0 and year%400==0: leap=True else: leap=False return leap
year = int(input()) print(is_leap(year))
yogeshverma28698 why it is giving error on input 2100
tehreemmustafa71 I have the Same Error..what to do?
mateo_villagomez leap = False if year % 4 == 0: leap = True if year % 100 == 0: leap = False if year % 400 == 0: leap = True return leap
avantikas55023 ran your code, 3 out of 6 test cases failed :(
sjayanthkumar191 ohh ... halo
bjohnston370 You can write it without parentheses like this:
def is_leap(year): return year % 4 == 0 and year % 100 != 0 or year % 400 == 0
amitrajitbose9 Adding parenthesis increases readability. I might be wrong though!
def is_leap(y): return ((y%4==0)and(y%100!=0)or(y%400==0))
karthik_bharadh1 what is the logic of this.....man!!!!
agarwal_rishika7 here, the return statement returns in boolean. where the given conditions in the problem are put.a year is a leap year if it is divisible by 4 and 400 but not divisible by 100.
amitrajitbose9 You may find the logic here.
avy0010 Thanks for this , this page cleared everything about Leap years !
yogeshverma28698 def is_leap(year): leap = False if year%4==0: leap=True elif year%100==0 and year%400==0: leap=True else: leap=False return leap
year = int(input()) print(is_leap(year))
yogeshverma28698 why it is giving error on input 2100
ingoglia But then how does python know which one you mean: 1.y%4 AND y%100, or y%400
2.y%4, and y%100 OR y%400?
That's like saying these sentences state the same thing: 1.I will eat pie and roaches, or I will eat pasta.
2.I will eat pie, and either I will eat roaches or pasta.
Am I missing something?
jrmioduszewski Just like there is order of operations for mathematical operators, there is precedence for Python operators. AND has higher precedence than OR, so it is evaluated first.
rfstahl Order of precedence: NOT > AND > OR
ujezidonye Thanks for this. I was having issues with If/ else statements.
valanukonda1999 plss explain this a bit clear how return works their
agarwal_rishika7 here solving the given condition, return turns true
Krishnasaivamsi can you please explain the code.
amitrajitbose9 I think this is a good read, to understand the logic of a leap year. The code is straight forward.
ZackSydney Beautiful I just add parenthesis for me like that (year % 4 == 0) and (year % 100 != 0) or (year % 400 == 0)
sri_sam229 good one man.you seem to have pretty good understanding of python PEP.
zusammen why parenthesis are required and we get error without them??
codeharrier Associativity.
'a and b or c' reads like '(a and b) or c' to the interpreter.
'a and b' is only true every 400 years.
c is true every year except those divisible by 100, so you're marking only 3 years out of every 400 as non-leap years.
hamzabouabid7 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
agarwal_rishika7 def is_leap(year): return year%4==0 and (year%400==0 or year%100!=0)
by using this , you will get ur answer . you dont have to use an extra variable 'leap'
chadwalt this is nice. it avoids repetition. Thanks alot.
intenseCtraining 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
scotmatson 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.
dhinastallion [deleted]
sam_fold Your answer isn't really very clean code though. So while it works, it isn't very readable, which IMO is incredibly important.
abhishekreddy041 IMO means..?
hhelzer IMO = "in my opinion"
also there's
IMHO = "in my humble opinion" (which is often not very humble at all)
abhishekreddy041 thank u
Muttnstuff Thanks that explained it so clearly
Zawaldo Makes perfect sense, thank you very much for your explanation!
Adivareyhanptr 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
wanglei_okay I like your func very much
himanshu_chauhan That's way shorter than my if and else loop's. Thanks.
ftakelait 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
gaurav_sonkusle very effective buddy.... thnk u
beckwithdylan 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
Anaximandrake 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?
sachinsavita95 it is not passing all the test cases?
Veerubhotla thanks :D
zpf7879 Might be clearer:
def is_leap(year): return year % 400 == 0 or year % 4 == 0 and year % 100 != 0
guardadopatrici1 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
Jay_Rockss thanks for advice, keep commenting :)
mayank_sept92 Hi, thanks for the shorter approach but, what about the condition-- 1900<= year <=10**5
[deleted] Thank you sir for the advice.
Zvezdobroeca 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.
chefyunfei 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.ctrlshftenter 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.
mai_tran1992 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
deepika121 how it will come can u explain
sjayanthkumar191 its a mistake he may copied and pasted the code..
godspell but for 1900,its divisible by 4 but also by 100..so how is that code possible
letitslide87 Wow, amazing comment, thank you!
dillingerjames 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/
jadhav_rohan100 failed 2 cases
zrolpin 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
ENIGMA_96 Thanks for the advice. I am new to python and this something that i would have never thought of.
mrknowledge38 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.
aabbron 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..
gmatere why to add OR for % 100 condition? Not able to understand the logic. Please help!!
snhgvh [deleted]megh_scorpio94 Thank you, this helped
laurent_mundell All my respectz.
Neoflo Thank you for the huge help, I kept trying with if's and got nowhere.
edgar_ocampo_b Why:
is_leap(2020) True.
is_leap(2021) False.
... that's not true
dewanaaryan Wow. I did it in a much length way with lots of ifs and elses.
Thank you for sharing this.
friedeggs44 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()
m_kroeger_hello Thanks for your input! I'm new to programming so this is very helpful.
sjayanthkumar191 ok thats good...
shubhendu_verma1 This is a much better approach. Thanks
am110 The calendar builtin module has a function isleap() that returns True if the year is leap, otherwise it returns False
import calendar def is_leap(year): return calendar.isleap(year)
michaelhuynhquoc What I just learnt from you: don't re-invent the wheel :D
super_geek true af
ajayi haha, you are the boss.
cssd1983 Amazing
chefyunfei That kind of defeats the point of this exercise doesn't it?
flmmartins The problem's explanation is not clear. I am not an english speaker and the bullet points were not clear enough for me.
Can that be improved?
Suggested explanation is bellow:
In the Gregorian calendar three criteria must be taken into account to identify leap years
The year can be evenly divided by 4; If the year can be evenly divided by 100, then it is not a leap year The year is also evenly divisible by 400 then it is a leap year.veganXedge i agree. maybe this has something to do with my math skil level but, i couldn't understand how to properly figure out that(code below) was an acceptable way to write it.
year % 4 == 0 and (year % 400 == 0 or year % 100 != 0)
i would like to understand or have some sort of guide on how to better interpret these questions.
iamphanisai yes. i too faced the same problem and had to google what a gregorian calendar was and found a page long explanation about how julias caesar invented it... etc etc before actually understanding what it means!
scotmatson Agree.
I am an English speaker and struggled with this definition. It was poorly written.
xbuildx_matrix I agree
kasunr My answers are ...
def is_leap(year): leap = False if year %4 == 0: if year %400 == 0: leap = True elif year %100 == 0: leap = False else: leap = True else: leap = False return leap
or
return (year %4 == 0 and year %100 != 0 or year %400 == 0)
chefyunfei Pretty much what I did, but I don't htink I nested the arguments under if year%4 so it ended up being wrong.
JonathanV Working
if year%4 == 0 and (year%100 != 0 or year%400 == 0): leap = True
Vangelic_surgeon Thanks this stupid problem took me more time to solve and even the official solution is s**t
KavinRanawella if u are refering to the solution under the editorial tab, I would say, it is more elaborating. I think either you have misread the question, or you haven't noticed that there are no 'elseif' logic in the given solution. Only 'if' statements are there. It suits the requirements perfectly. Read the solution again carefully. U'll get it. :)
tricerabottoms This one is a little longer but I think you'll find it easier to follow:
class is_leap: pass is_leap.__init__ = lambda s,y: setattr(s,'y',y) is_leap.__repr__ = lambda s: str((not s.y%4)^(not s.y%100)^(not s.y%400))
hamichok11 lol :D much easier!
cssd1983 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 LOL
saumitra13325 [deleted]saumitra13325 Ahhh...got the end
bartyl :DDDD good joke
coder_aky 100 % working ,Simple and easy :-)
Here's in Python 3 one liner.....
def is_leap(y):return y%4==0 and (y%400==0 or y%100!=0)
nishant_bhumca17 I think this is pretty starightforward :
def is_leap(year): return year % 4 == 0 if year % 100 != 0 else year % 400 == 0
n_aswath def is_leap(year): if year in range(1900,100000): if (year % 4) == 0 and (year % 100) == 0 and (year % 400) == 0: leap = True elif (year % 4) != 0 and (year % 100) == 0: leap = False elif (year % 4) == 0 and (year % 100) != 0: leap = True elif (year % 4) !=0: leap = False elif (year % 100) !=0: leap = False elif (year % 400) !=0: leap = False return leap
cnhnyu If a year is a leap year: 1. it can be evenly divided by 4 and NOT evenly divided by 100; OR 2. it can be evenly divided by 400
leap = False if any([year % 4 == 0 and year % 4 != 0, year % 400 == 0]): leap = True
alexandar_naray1 I like the use of any here, I didn't think of that. You could set
leap = any([year % 4 == 0 and year % 4 != 0, year % 400 == 0])
since any returns a boolean
cnhnyu Thanks, you're right. It's much simpler.
leap = any([year % 4 == 0 and year % 100 != 0, year % 400 == 0])
Correction: year % 4 == 0 and year % 100 != 0
marcon_zm What is the comma doing here exactly?
year % 100 != 0, year % 400 == 0
Is it testing the output of "%100 != 0", and if it returns false it tests for the "% 400 == 0" conditional?
alexandar_naray1 the any function takes an iterable so it is basically saying: if any of the items in this iterable is true return true. so the comma makes them elements of a list between the brackets.
cnhnyu You can think it as an OR operation in this case.
alexandar_naray1 expains it very well.
mohamedfardinkh1 How to use "any" function? What are it's cases ?
bad_coder_lol for some reason, the Hacker Rank team wasn't able to clarify that clearly
spencer_rhodes The explanation of the rules actually nests the if statements the way you should in your code for this type of problem.
Notice how they indent further each of the three bullet-pointed rules?
You don't want these to be separate checks, but rather a cascading condition.
info_krrajusoni if(year%400==0) or (year%4==0 and year%100!=0): leap=True;
Sort 461 Discussions, By:
Please Login in order to post a comment