Is it a leap year?

Question

This challenge is quite simple. You will take an input which will be a year from 1801 to 2400, and output if it is a leap year or not.

Your input will have no newlines or trailing spaces:

You will output in any way that you like that clearly tells the user if it is or isn't a leap year (I will accept y or n for yes/no)

You can get a list of leap years here: http://kalender-365.de/leap-years.php I would note that leap years are not ever four years always. 1896 is a leap year, but 1900 is not. The years that follow this "skip" are:

Test cases:

1936 ->  y
1805 ->  n
1900 ->  n
2272 ->  y
2400 ->  y

EDIT: This is based on a standard Gregorian calendar: http://www.epochconverter.com/date-and-time/daynumbers-by-year.php

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You should be more clear: A given year is a leap year if and only if it is (divisible by 4)∧((divisible by 100)→(divisible by 400)). — LegionMammal978, May 26, 2015 at 11:34
Your input will have no newlines or trailing spaces. Dang it, that would have saved me 2 bytes... — Dennis, May 26, 2015 at 17:10
You should extend the accepted input range to AD 1601 thru 2400. This covers two 400-year Gregorian cycles (which proleptically start on Monday). — David R Tribble, May 26, 2015 at 18:17
Does falsy if leap year and truthy if not a leap year count as "clearly tells the user if it is or isn't"? — lirtosiast, May 28, 2015 at 21:27

Community · Accepted Answer · 2020-06-17 09:04:33Z

PowerShell, 31 bytes

I am excited to say that I golfed this shorter than the builtin!

param($a)!($a%(4,16)[!($a%25)])

Outputs true for leap years and false otherwise.

Builtin:

[datetime]::IsLeapYear($args[0])

Though, if I wanted to stretch the statement 'clearly tells the user if it is or isn't a leap year' and do something non-standard, I could save 3 bytes and use:

param($a)$a%(4,16)[!($a%25)]

This outputs 0 for leap years and 1 or higher for non-leap years, which I don't like since I'd prefer to return a more standard truthy value for leap years.

Community · Accepted Answer · 2020-06-17 09:04:33Z

LOLCODE, 228 202 159 bytes

HOW IZ I f YR a
MOD OF a AN 100
O RLY?
YA RLY
MOD OF a AN 4
O RLY?
YA RLY
b R 1
OIC
NO WAI
MOD OF a AN 400
O RLY?
YA RLY
b R 0
NO WAI
b R 1
OIC
OIC
IF U SAY SO

Ungolfed:

HAI 1.3 BTW "HAI" does nothing functionally in current versions and does not throw an error if you omit it.
HOW IZ I leap YR input
    I HAS A output
    DIFFRINT MOD OF input AN 100 AN 0 BTW Thanks @LeakyNun, In LOLCODE any non-empty values, i.e. 0, "", etc. default to WIN.
    O RLY?
        YA RLY
            BOTH SAEM MOD OF a AN 4 AN 0
            O RLY?
                YA RLY
                    output R WIN BTW "WIN" is true, but in the actual program I used 1 as a truthy value because it's shorter.
            OIC
        NO WAI
            DIFFRINT MOD OF a AN 400 AN 0
            O RLY?
                YA RLY
                    output R FAIL BTW "Fail" is false, but in the actual program I used 0 as a falsy value.
                NO WAI
                    output R WIN
            OIC
    OIC
    FOUND YR output BTW This statement is implied in the golfed version.
IF U SAY SO BTW "KTHXBYE", just like "HAI" has no functional purpose and throws no error on omission.
KTHXBYE

In Python ungolfed, because LOLCODE is confusing:

def leap:
    if(year % 100 != 0):
        if(year % 4 == 0):
            output = true
    else:
        if(year % 400 != 0):
            output = false
        else:
            output = true
    return(output)

You have updated the main code to be a function, but not the ungolfed code? — Destructible Lemon, Jul 23, 2016 at 2:25
I thought LOLCODE has automatic type coercion, meaning that any non-zero value is equivalent to WIN.. — Leaky Nun, Jul 23, 2016 at 2:38
It is, but how could I use that? I don't think I am doing any casting. — AAM111, Jul 23, 2016 at 2:41

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Mathematica, 15 bytes

LeapYearQ@*List

Mathematica has a built-in for everything. List is there to make the integer input into a compact DateList object. Returns True or False.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 21, 2016 at 22:03

JungHwan Min

13.7k3 gold badges21 silver badges68 bronze badges

\$\begingroup\$ LeapYearQ@{#}& for 14 \$\endgroup\$
– att
Aug 15, 2019 at 6:39
\$\begingroup\$ Just LeapYearQ works \$\endgroup\$
– hakr14
Nov 11, 2022 at 21:03

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

PHP, 28 bytes

Why calculate when there´s a builtin?

<?=checkdate(2,29,$argv[1]);

prints 1 for leap year, empty string for no leap year.
Add + after <?= for 1/0.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Feb 2, 2017 at 2:34

Titus

14.6k1 gold badge21 silver badges41 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

SmileBASIC, 40 bytes

INPUT Y?0DTREAD STR$(Y)+"/02/29"OUT,,,?1

Prints 0 1 if it's a leap year, otherwise prints 0 and errors.

DTREAD parses a date string in the form of YYYY/MM/DD and gives the year, month, and day. It can also return the day of the week, and trying to get the day of the week of a day that doesn't exist (Feb. 29 of a non-leap year) will cause an error.

This is shorter than the obvious answer using mod, because SB uses MOD instead of %, which takes up 3-5 characters.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Feb 2, 2017 at 14:01

12Me21

6,5611 gold badge20 silver badges40 bronze badges

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aross · Accepted Answer · 2017-03-03 16:43:23Z

2

PHP, 34 30 bytes

echo+!($argn%($argn%25?4:16));

I'm using the + sign to convert false (which would normally be converted to an empty string) to 0, because that seems to comply closer to the "clearness" requirement (maybe not?). With standard truthy/falsy rules, 1 byte can be saved.

Note: I shamelessly implemented the divisibility by 25 trick from @David Hammen. Without his algorithm it would be 37 bytes:

<?=+!(($b=$argv[1])%400^$b%100^$b%4);

Run like this:

echo 1900 | php -nR 'echo+!($argn%($argn%25?4:16));';echo

Tweaks

Saved 4 bytes by using $argn

edited Mar 3, 2017 at 16:43

answered Jul 20, 2016 at 14:00

aross

1,65310 silver badges12 bronze badges

1

\$\begingroup\$ Save one more byte on standard truthy/falsy with <1 instead of !() \$\endgroup\$
– Titus
Feb 2, 2017 at 2:45

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AplusKminus · Accepted Answer · 2019-03-02 11:05:12Z

2

Haskell, 43 41 39 bytes

l x=mod x(if mod x 25<1then 16else 4)<1

If x is divisible by 25, check if it is also divisible by 16, making it divisible by the least common multiple of 25 and 16, which is 400.
If x is not divisible by 25, check if it is divisible by 4.

Inspired by @David Hammen's answer in JavaScript.

edited Mar 2, 2019 at 11:05

answered May 28, 2015 at 20:49

AplusKminus

1712 silver badges7 bronze badges

\$\begingroup\$ I don't know Haskell, but can ==0 be <1? \$\endgroup\$
– lirtosiast
May 28, 2015 at 21:13
\$\begingroup\$ You are right, it can. \$\endgroup\$
– AplusKminus
May 29, 2015 at 9:04
1

\$\begingroup\$ You can drop the spaces before then and else. \$\endgroup\$
– dfeuer
Mar 1, 2019 at 7:38
1

\$\begingroup\$ The conditional can be equivalently expressed as (4+sum[12|mod x 25<1]) for a total of 33 bytes: Try it online! \$\endgroup\$
– Laikoni
Mar 3, 2019 at 9:24

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Community · Accepted Answer · 2020-06-17 09:04:33Z

05AB1E, 9 7 bytes

т‰0Kθ4Ö

Try it online or verify all test cases.

Explanation:

т‰         # Divmod the (implicit) input by 100
           #  i.e. 1900 → [19,00]
           #  i.e. 1936 → [19,36]
           #  i.e. 1991 → [19,91]
           #  i.e. 2000 → [20,00]
  0K       # Remove all 0s
           #  i.e. [19,00] → [19]
           #  i.e. [19,36] → [19,36]
           #  i.e. [19,91] → [19,91]
           #  i.e. [20,00] → [20]
    θ      # Pop and get the last item
           #  i.e. [19] → 19
           #  i.e. [19,36] → 36
           #  i.e. [19,91] → 91
           #  i.e. [20] → 20
     4Ö    # Check if it's divisible by 4 (and output the result implicitly)
           #  i.e. 19 → 0 (falsey)
           #  i.e. 36 → 1 (truthy)
           #  i.e. 91 → 0 (falsey)
           #  i.e. 20 → 1 (truthy)

Community · Accepted Answer · 2020-06-17 09:04:33Z

x86 machine code, 14 bytes

00000000: adfd ad2d 3030 74fa 86c4 d50a a803       ...-00t.......

Unassembled

AD         LODSW           ; SI += 2 to start with last two digits
FD         STD             ; set direction for LODSW to decrement
AD         LODSW           ; load two ASCII digit chars into AX
2D 3030    SUB  AX, '00'   ; convert to decimal, test if year ends in 00?
74 FA      JZ   -6         ; if so, look at the first two digits instead
86 C4      XCHG AL, AH     ; endian convert
D5 0A      AAD             ; base convert from 10 to binary
A8 03      TEST AL, 3      ; mod 4 = 0?

Input string in SI, output is ZF if is a leap year.

Explanation

If the year does not end in 00, it is a leap year if last two digits mod 4 is 0. If it does end in 00, it will be a leap year if the first two digits mod 4 is 0. Example:

Leap year:

1936: 36 mod 4 == 0
2400: 24 mod 4 == 0

NOT leap year:

1805: 05 mod 4 == 1
1900: 19 mod 4 == 3

Test Program Output

Download and test LEAP.COM.

orthoplex · Accepted Answer · 2020-09-18 08:00:49Z

2

Excel, 30 25 bytes

Because of an intentional bug in Excel to maintain compatibility with Lotus 1-2-3 the year 1900 is treated as a leap year. We can however shift the year by 400 before doing the calculation. The leap year bug then occurs in 1500, which we don't care about anyway. For the following formula A1 serves as input.

=DAY(DATE(A1+400,2,29))>1

edited Sep 18, 2020 at 8:00

answered Apr 2, 2019 at 16:07

orthoplex

3892 silver badges6 bronze badges

1

\$\begingroup\$ If you can accept 1 for leap year and and #VALUE! for other years, you can get this down to 19 bytes: =("2/29/"&E1+400)^0. \$\endgroup\$
– Engineer Toast
Jan 10, 2022 at 16:18

Add a comment |

Jitse · Accepted Answer · 2020-11-13 12:56:01Z

2

Python 3, 26 bytes

lambda n:n%4**-~(n%25<1)<1

Try it online!

Using Kevin Cruijssen's Java approach.

Python 3, 28 bytes

lambda n:(n%100or n/100)%4<1

Try it online!

Using Kevin Cruijssen's 05AB1E approach.

edited Nov 13, 2020 at 12:56

answered Nov 13, 2020 at 12:38

Jitse

6,7862 gold badges17 silver badges40 bronze badges

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knugie · Accepted Answer · 2022-01-07 09:22:08Z

2

Ruby, 28 bytes

The lambda takes an Integer y representing the year and returns true or false.

->y{y%4<1&&y%100>0||y%400<1}

Use like this:

f=->y{y%4<1&&y%100>0||y%400<1}
f[2004]
# => true

or this:

f=->y{y%4<1&&y%100>0||y%400<1}
[1936,1805,1900,2272,2400].map(&f)
# => [true, false, false, true, true]

Try it online

Explanation

y%4<1&&y%100>0
- Years that are divisible by 4 but not by 100 are leap years, e.g. 2004
- Years that are divisible by 4 and also by 100 are not leap years, e.g. 1700
y%400<1
- Years that are divisible by 400 are leap years, e.g. 1600

edited Jan 7, 2022 at 9:22

answered Jan 6, 2022 at 17:30

knugie

1212 bronze badges

1

\$\begingroup\$ Welcome to Code Golf, and nice answer! According to site rules, you'll need to use a full program or function instead of assuming the input is in y. \$\endgroup\$
– Rydwolf Programs
Jan 6, 2022 at 18:20
\$\begingroup\$ Welcome to Code Golf, and nice first answer! Unfortunately, this isn't quite valid as submissions here must be full programs or functions. If you made this a function taking y as an argument, that'd be fine.Also, it helps to provide a TIO link or similar so others can test your code themselves. \$\endgroup\$
– emanresu A
Jan 6, 2022 at 18:29
\$\begingroup\$ @RedwolfPrograms thanks for pointing out, provided function (with additional 9 characters). \$\endgroup\$
– knugie
Jan 7, 2022 at 9:09
\$\begingroup\$ I just realized there's a 22 bytes Ruby solution already 🤷 \$\endgroup\$
– knugie
Jan 7, 2022 at 9:21

Add a comment |

chunes · Accepted Answer · 2022-11-11 21:12:47Z

2

Arturo, 5 bytes

leap?

Try it!

answered Nov 11, 2022 at 21:12

chunes

16.9k3 gold badges15 silver badges42 bronze badges

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Brian · Accepted Answer · 2015-05-27 09:26:22Z

1

SpecBAS - 77

Not the shortest (but not the longest either).

1 INPUT y: PRINT y;"->";"ny"((y MOD 4=0 AND (y MOD 100<>0) OR y MOD 400=0)+1)

Uses the standard formula, then prints "n" or "y" based on return value (the final +1 at the end is due to strings being 1-based).

answered May 27, 2015 at 9:26

Brian

1,2378 silver badges9 bronze badges

\$\begingroup\$ I don't know the language, but based the fact that there are no booleans, you may be able to save some characters by using something like y MOD (4+12*(y MOD 100=0))=0. \$\endgroup\$
– lirtosiast
May 28, 2015 at 22:49

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Daniel · Accepted Answer · 2016-07-20 13:03:40Z

1

Python, 50 48 bytes

import calendar as y
def a(b):print(y.isleap(b))

edited Jul 20, 2016 at 13:03

answered Jul 20, 2016 at 5:55

Daniel

1,9081 gold badge13 silver badges24 bronze badges

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aloisdg · Accepted Answer · 2016-07-20 13:53:36Z

1

C#, 26 bytes

y=>DateTime.IsLeapYear(y);

C# lambda (Predicate) where the input is a int and the output is a bool. I use a builtin.

answered Jul 20, 2016 at 13:53

aloisdg

1,96716 silver badges26 bronze badges

\$\begingroup\$ You don't really need the lambda here, and I would argue the answer is more valid without it. \$\endgroup\$
– VisualMelon
Jan 13, 2017 at 9:26
\$\begingroup\$ @VisualMelon In the meta section, they ask for a function. (If I remember) \$\endgroup\$
– aloisdg
Jan 14, 2017 at 15:24
5

\$\begingroup\$ Aye, but DateTime.IsLeapYear is a function ;) generally the rule is that it has to evaluate to a function if it isn't named, and strictly your current submission is meaningless (join the argument here!), while System.DateTime.IsLeapYear is fully typed (System.Func<int,bool>A=System.DateTime.IsLeapYear compiles fine) without demanding any user inference (or indeed the text in your answer which specifies the input and output types). \$\endgroup\$
– VisualMelon
Jan 14, 2017 at 17:22

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Martin Ender · Accepted Answer · 2016-09-02 17:29:48Z

1

BASH, 68 bytes

((!(y % 4) && ( y % 100 || !(y % 400)))) &&  echo "leap" || echo "no"

edited Sep 2, 2016 at 17:29

Martin Ender

195k65 gold badges441 silver badges962 bronze badges

answered Sep 2, 2016 at 17:07

michael501

1916 bronze badges

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Martin Ender · Accepted Answer · 2016-12-21 16:37:11Z

1

Retina, 24 bytes

[^04]00$

.+
$*
....

^$

Try it online!

answered Dec 21, 2016 at 16:37

Martin Ender

195k65 gold badges441 silver badges962 bronze badges

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Dexa · Accepted Answer · 2017-01-13 12:30:57Z

1

PHP, 45 bytes

<?=date_create($argv[1].'-1-1')->format('L');

Not the shortest, but using built in functions. Totally not code-golf.

answered Jan 13, 2017 at 12:30

Dexa

2341 silver badge3 bronze badges

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Erik the Outgolfer · Accepted Answer · 2017-10-18 19:41:10Z

1

Jelly, 8 bytes

ȷ2*ḍ¥×4ḍ

Try it online!

answered Oct 18, 2017 at 19:41

Erik the Outgolfer

40.3k5 gold badges39 silver badges120 bronze badges

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user41805 · Accepted Answer · 2018-03-25 12:37:15Z

1

Bash, 14 bytes

date -d$1/2/29

Try it online!

Outputs via exit code, 0 for leap year and 1 otherwise.

date -d STRING will display the date indicated by STRING. $1/2/29 represents February 29th of $1, the argument. If STRING is not valid, i.e. Feb 29 does not exist, then date errors out.

answered Mar 25, 2018 at 12:37

user41805

12.9k6 gold badges39 silver badges87 bronze badges

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G B · Accepted Answer · 2018-05-09 07:04:10Z

1

Ruby, 22 bytes

->n{1>n%(n%25<1?16:4)}

Try it online!

The challenge is quite old, but I noticed there was no valid Ruby answer yet. Nothing particularly original, anyway.

answered May 9, 2018 at 7:04

G B

19.7k1 gold badge19 silver badges50 bronze badges

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Adalynn · Accepted Answer · 2018-11-10 20:19:00Z

1

D, 43 42 bytes

T l(T)(T y){return!(y%4)&&y%100+!(y%400);}

Try it online!

Another port of @HatsuPointerKun's C++ answer.

answered Nov 10, 2018 at 20:19

Adalynn

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

C++, 50 43 bytes

-7 bytes thanks to Zacharý

int l(int y){return!(y%4)&&y%100+!(y%400);}

And the test code is :

auto t = {
    1936,1805,1900,2272,2400
};

for (auto&a : t) {
    std::cout << "Year : " << a << " is " << (l(a) ? "" : "NOT") << " a leap year\n";
}

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Mar 26, 2018 at 7:27

HatsuPointerKun

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\$\begingroup\$ You can make it int l(int y){return(y%4==0&&y%100)+(y%400==0);} to save a few bytes \$\endgroup\$
– Adalynn
Mar 29, 2018 at 13:49
\$\begingroup\$ I think making +(y%400==0) +!(y%400) might work as well. \$\endgroup\$
– Adalynn
Apr 1, 2018 at 18:54
\$\begingroup\$ int l(int y){return!(y%4)&&y%100+!(y%400);} \$\endgroup\$
– Adalynn
Nov 10, 2018 at 20:21

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snail_ · Accepted Answer · 2018-12-05 01:14:29Z

1

Kotlin, 30 bytes

{it%4<1&&(it%100>0||it%400>0)}

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answered Dec 5, 2018 at 1:14

snail_

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Renzo · Accepted Answer · 2018-12-22 11:42:25Z

1

Common Lisp, 46 bytes

(=(mod(setq x(read))(if(>(mod x 25)0)4 16))0)

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Based on the David Hammen’s method.

answered Dec 22, 2018 at 11:42

Renzo

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maxb · Accepted Answer · 2019-01-08 07:42:57Z

MathGolf, 9 bytes

4♀`*]÷~≤*

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Explanation

4         Push 4
 ♀        Push 100
  `*      Duplicate top two elements of stack and multiply (400)
    ]     Wrap stack in array ([4, 100, 400])
     ÷    Check input for divisibility with all array items
      ~   Dump array onto stack
       ≤  Check if divisibility by 100 is <= than divisibility with 400
        * Multiply with the divisibility bool for 4, works like logical and

The trick is that ≤ ensures that if a number is divisible by 100, it must also be divisible by 400, but if it's not divisible by 100, anything goes.

Mayube · Accepted Answer · 2019-01-08 14:03:23Z

Clam, 33 31 bytes

=a*rp|e%a*"400"0&%a*"100"e%a*40

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-2 bytes thanks to ASCII-only

Outputs true for leap years. Outputs false for other years

Explanation

=a*rp|e%a*"400"0&%a*"100"e%a*40
=a*r                             read input and store in a*
    p                            Print..
       %a*"400"                    a* % 400
      e                            ==
               0                   0
     |                             OR
                 %a*"100"            a* % 100
                &                    AND
                          %a*4       a* % 4
                         e           ==
                              0      0

Very awkward explanation, but basically what it does is this:

print(year % 400 == 0 || (year % 100 && year % 4 == 0))

Clam follows JS rules for truthy and falsey values, meaning year % 100 is true if it does not equal 0 (and it's shorter than adding an e before it and an 0 after it)

Resulting JS:

myVar = read();
console.log(year % 400 == 0 || (year % 100 && year % 4 == 0));

\$\begingroup\$ ew 8552. use something nicer like 4545 pls \$\endgroup\$
– ASCII-only
Jan 8, 2019 at 6:18 — ASCII-only, Jan 8, 2019 at 6:18
\$\begingroup\$ jk, 31 \$\endgroup\$
– ASCII-only
Jan 8, 2019 at 6:19 — ASCII-only, Jan 8, 2019 at 6:19
\$\begingroup\$ also pls fix your indentation in the explanation :| \$\endgroup\$
– ASCII-only
Jan 8, 2019 at 6:27 — ASCII-only, Jan 8, 2019 at 6:27
\$\begingroup\$ fixed that year % 100 thing \$\endgroup\$
– ASCII-only
Jan 8, 2019 at 6:42 — ASCII-only, Jan 8, 2019 at 6:42

Oliver · Accepted Answer · 2019-02-05 16:21:12Z

1

Japt `-h!`, 8 bytes

Input is taken as a string.

k0 ò o%4

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answered Feb 5, 2019 at 16:21

Oliver

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dfeuer · Accepted Answer · 2019-03-01 07:58:39Z

1

Haskell, 35 bytes

l x|x!25<1=x!16<1
l x=x!4<1
(!)=mod

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answered Mar 1, 2019 at 7:58

dfeuer

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Is it a leap year?

The Catalogue

76 Answers 76

PowerShell, 31 bytes

LOLCODE, 228 202 159 bytes

Mathematica, 15 bytes

PHP, 28 bytes

SmileBASIC, 40 bytes

PHP, 34 30 bytes

Tweaks

Haskell, 43 41 39 bytes

05AB1E, 9 7 bytes

x86 machine code, 14 bytes

Excel, 30 25 bytes

Python 3, 26 bytes

Python 3, 28 bytes

Ruby, 28 bytes

Explanation

Arturo, 5 bytes

SpecBAS - 77

Python, 50 48 bytes

C#, 26 bytes

BASH, 68 bytes

Retina, 24 bytes

PHP, 45 bytes

Jelly, 8 bytes

Bash, 14 bytes

Ruby, 22 bytes

D, 43 42 bytes

C++, 50 43 bytes

Kotlin, 30 bytes

Common Lisp, 46 bytes

MathGolf, 9 bytes

Explanation

Clam, 33 31 bytes

Explanation

Japt -h!, 8 bytes

Haskell, 35 bytes

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Japt `-h!`, 8 bytes