What year will it be in n seconds?

Question

You are given an integer n as input, and, regardless of the current date, must return the calendar year (Gregorian calendar, other calendars not allowed) taking place n seconds from now.

Rules

• You may not assume that the current year is 2021. In other words, imagine a user testing your solution in five years.
• Standard loopholes forbidden.
• Shortest answer in bytes wins.
• Built-in date functions are allowed.
• You must also support negative numbers.
• On the topic of the year zero: Your answer must follow the ISO specifications, meaning there is a year zero.
• You may assume that the input is always an integer within bounds of your language number system.
• If your language's date functions take leap seconds into account, then follow your language, otherwise do not take leap seconds into account.

Examples (in this case, measured at the time of writing, which is February 15, 2021)

100000000000 -> 5190
1000 -> 2021
54398643 -> 2022
54398643234 -> 3744
-100000000000 -> -1148

• Arnauld: Is it "local date at 00:00:00" + n seconds or "local date at local time" + n seconds?
• user100690: In my case, putting new Date() gives the date at my local time, but that might not be true in other languages. It depends on the language
• Peter Cordes: Was the negative-number requirement to avoid solutions like sleep $n;date being a valid bash answer? I notice there's no performance requirement :P
• Peter Cordes: It's generally accepted that CodeGolf answers that provably work for large test cases are fine, even if nobody has time to wait for them to finish, so for example O(e^n) or n! time and/or space complexity is fine for a problem that would normally take much shorter, like log(n) time. But when the "run-time" couldn't be sped up with a faster computer, so it can never produce an answer before the specified time, yeah that's stretching it in my opinion, too. And a different sort of reason for being slow than the usual excuse.
• user100690: @LuisMendo Edited the question albeit a bit late

Answer 1

Ohm v2, 5 bytes

υ!+υy

Try it online!

Explanation:

υ!+υy
υ! get the current timestamp in seconds
 + sum with the implicit input
 υy get the year from the resulting timestamp

• timthelion: υ is a two byte unicode charicter, so this is 7 bytes. "echo 'υ!+υy' | wc -c" returns 8 including the newline character.
• Cinaski: @timthelion Ohm use a custom code-page as many other languages do, you can read more about it here codegolf.meta.stackexchange.com/a/10187/72733
• Nick Clifford: Wow, haven't seen this language used in a long time! This kind of challenge is exactly what I added the time builtins for, very nice solution!

Answer 2

Bash, 16 bytes

date -d1ドルsec +%Y

Try it online!

Credits

• Saved 12 bytes thanks to @manatwork

• manatwork: Why not date -d1ドルsec +%Y? Try it online!
• user344: Sorry, disregard my comment, forgot it's n seconds and not 1 second.

Answer 3

PHP, 27 bytes

<?=date(Y,time()+$argv[1]);

Try it online!

• Kaddath: Oh my, of course! (deletes his own answer in shame)
• 640KB: 25 bytes as a function
• oktupol: @640KB Nice! It's been a while since I did PHP. Back when I stopped, the Arrow Function RFC was still open. Glad to see it being part of the language now.
• Kaddath: @oktupol yeah but they have strange limitations, such as you cannot do loops in them, or use recursion (without lengthy tricks)
• Christoph: <?=date(Y,time()+$argn); for 24? What's the current consent for $argn? +2 for the flag?

Answer 4

05AB1E, 116 bytes

"`т‰0Kθ4ÖUD2Qi28円X+ë<7%É31α}"Vžežfžg)IŽa+·÷ÄF©IdiY.V‹i®¬>0ë®13⁄4ǝDÅsD12‹i>1ë1円Dǝ¤>2}}ǝëć©ić©i<D_-12š31šë®<šDY.Všë®<š]θ

Here we go again..

Try it online or try it only with custom specified starting date. (Pretty slow for large inputs, so won't be able to output the larger test cases with the current date.)

Explanation:

Since 05AB1E doesn't have any date builtins (except for the current day/time), I've calculated things manually before. I've used the code of going to the next day from this answer of mine, which in turns also uses the leap year calculation of this answer of mine.
Since this challenge also asks to go back in time, I've modified the program accordingly to support that as well.

Step 1: Create a function to calculate the amount of days in a month for a given year/month:

"`т‰0Kθ4ÖUD2Qi28円X+ë<7%É31α}" # Push this string
 V # Pop and store it in variable `Y`
 ` # Pop the list and push all values separated to the stack
 # (the year will be at the top; month below it)
 т‰ # Take the divmod-100 of the year
 0K # Remove the 0s
 θ # Pop and push its last item
 4Ö # Check if it's divisible by 4
 U # Pop and store this in variable `X`
 # (X=1 for leap years; X=0 for non-leap years)
 D # Duplicate the current month
 2Qi # If it's equal to 2 (thus February):
 \ # Discard the duplicated month
 28X+ # Push 28 + isLeapYear from variable `X`
 ë # Else:
 < # Decrease the month by 1
 7% # Modulo-7
 É # Modulo-2
 31α # Absolute difference with 31
 } # Close the if-else statement

Step 2: Now we determine the current date, and loop an amount of times depending on the input:

že # Push the current day
 žf # Push the current month
 žg # Push the current year
 ) # Wrap all three into a list
 I # Push the input-integer
 Ža+ # Push compressed integer 42300
 · # Double it to 84600 (60*60*24)
 ÷ # Integer-divide the input by 84600
 Ä # Take its absolute value
 F # And loop that many times:

Step 3: If the input was positive, calculate the next date:

© # Store the current date in variable `®` (without popping)
 Idi # If the input is non-negative:
 Y.V # Execute string `Y` as 05AB1E code to get the amount
 # of days for the current month/year
 ‹i # If this is smaller than the current days:
 ® # Push the current date again
 ¬ # Get its first item (the days)
 > # Increase the day by 1
 0 # Push index 0
 ë # Else:
 ® # Push the current date again
 1 # Push day=1
 3⁄4 # Push index 0
 ǝ # Insert days=1 at index 0 into the date
 D # Duplicate it
 Ås # Pop and push its middle item (the month)
 D # Duplicate that month
 12‹i # If it's smaller than 12:
 > # Increase the month by 1
 1 # Push index 1
 ë # Else:
 \ # Discard the duplicated month
 1 # Push month=1
 D # Push index 1
 ǝ # Insert month=1 at index 1 into the date
 ¤ # Push its last item (the year)
 > # Increase the year by 1
 2 # Push index 2
 } # Close the if-else statement
 } # Close the if-else statement
 ǝ # Insert the value at the given index into the date

Step 4: If the input was negative instead, calculate the previous date:

ë # Else (the input is negative):
 ć # Extract head; push [month,year] and day separated
 © # Store the day in variable `®` (without popping)
 i # If day == 1:
 ć # Extract head; push [year] and month separated
 © # Store the month in variable `®` (without popping)
 i # If month == 1:
 < # Decrease the [year] by 1
 D # Duplicate it
 _ # Check if it's equal to 0 (1 if 0; 0 otherwise)
 - # Subtract that (so we go from year 1 to -1)
 12š31š # Prepend 12 and 31: [31,12,year-1-(year-1==0)]
 ë # Else:
 ®<š # Prepend month-1 to the [year] list
 D # Duplicate it
 Y.V # Execute string `Y` as 05AB1E code to get the amount
 # of days for this month/year
 š # Prepend this to the [month-1,year] list
 ë # Else:
 ®<š # Prepend day-1 to the [month,year] list

Step 5: After the loop, extract the year to output:

] # Close both the if-else and loop
 θ # Pop the date, and only leave its last item (the year)
 # (after which it is output implicitly as result)

See this 05AB1E tips of mine (section How to compress large integers?) to understand why Ža+ is 42300.

• Luis Mendo: Oh my. Debugging this must have been fun :-)
• Kevin Cruijssen: @LuisMendo Haha, most of it has been done before in other challenges. But I indeed debugged going to the previous date, since that is new. :)
• user100690: Nice, extremely creative. This must have taken a while to write
• Kevin Cruijssen: @expressjs123 Probably longer than most of the answers indeed. 05AB1E is a code-golf language, so in most challenges one of the shortest. With date-related challenges on the other hand, it's usually one of the longest, since it has to do almost everything manually. But I kinda like doing those kind of challenges in 05AB1E regardless. ;)
• user: Looks like the tables have turned :P. +1 for doing something like this in a golfing language

Answer 5

Ruby, 21 bytes

->n{[*Time.now+n][5]}

Try it online!

Shorter by a byte than the more obvious ->n{(Time.now+n).year}.

• user100690: The shortest answer I can actually understand! (the APL one is shorter but it's harder to make something of it)
• Adám: @expressjs123 I've added an explanation. Let me know if I need to expand on anything.
• engineersmnky: Nice. Never even thought about decomposing a Time object.

Answer 6

Octave / MATLAB, 27 bytes

@(x)datestr(x/86400+now,10)

Try it online!

How it works

@(x) % Define anonymous function with input x
 x/86400 % Divide input by 86400, to convert to days
 now % Current time in days since "January 0, 0000"
 + % Add 
 datestr( ,10) % Convert to string with format 10, which is year

• Luis Mendo: Would the downvoter be so kind to explain how I can improve my answer (other than posting an explanation, which I haven't had the time for yet)?
• Razetime: Many answers here are getting downvoted without explanation :/
• Luis Mendo: @Razetime I've seen that happen, sometimes, to answers in golfing languages. Some people don't like those languages and decide that's enough reason to downvote. But Octave?
• Agos: not a downvote but it's easy to find a failing input since not all days are 86400 seconds long
• Luis Mendo: @Agos Good point. I have asked the OP if leap seconds need to be taken into account. Doing it would complicate answers a lot

Answer 7

JavaScript (ES6), 42 bytes

n=>new Date(+new Date+n*1e3).getFullYear()

Try it online!

• user100690: Nice answer, but not accepting it yet (waiting a week to accept any answer)
• Arnauld: @expressjs123 For code-golf challenges, we usually do not accept any answer. If you really want to accept one, you should indeed wait at least one week before doing so and pick the shortest one (which is not going to be this one).

Answer 8

jq, (削除) 20 (削除ここまで) 15 characters

now+.|gmtime[0]

Sample run:

bash-5.0$ jq 'now+.|gmtime[0]' <<< 100000000000
5190

Try it online!

• user344: Sorry, I totally misinterpreted the question and missed that it's n seconds instead of 1 second...

Answer 9

PowerShell, (削除) 32 (削除ここまで) (削除) 26 (削除ここまで) 23 bytes

date|% *dds* @args|% y*

Try it online!

-3 bytes thanks to mazzy and ZaelinGoodman

• Zaelin Goodman: Darnit, I didn't see that somebody already did Powershell before me! I came to almost the same solution, so here's a few bytes for you from mine! Try it online!
• mazzy: so that the array of args does not confuse anyone Try it online! :)

Answer 10

Python 3, (削除) 74 (削除ここまで) (削除) 72 (削除ここまで) 65 bytes

lambda x:(date.today()+timedelta(0,x)).year
from datetime import*

Try it online!

-7 bytes thanks to @Eric

• movatica: According to the rules, you can get rid of the f=: Try it online!
• avarice: @Movatica thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
• Eric Duminil: You need to return a year, not print it. So it's allowed to move print() to the footer, for 7 fewer bytes. tio.run/##FcrBCoQgEADQe1/hcaYNibSlXehPukyoJGSKzCG/…
• avarice: @EricDuminil thanks!!!

Answer 11

Japt, ^{(削除) 12 (削除ここまで)} 10 bytes

ÐKj +Ue3)i

Try it

-2 bytes from AZTECCO.

This transpiles to the following JS:

new Date(K.j() + U * 1e3).i()

where K.j() is the current time converted to milliseconds, U is the input, and .i() gets the year from the date constructor.

• AZTECCO: e can be a number method for 10
• timthelion: As Ð is a two byte unicode character, this is actually 11 bytes.
• Razetime: @timthelion Japt has a custom codepage, just like the 05AB1E answer and Ohm answers do. Click on the "bytes" link to see its encoding.

Answer 12

R, 12 bytes

Sys.time()+n

This returns a full date, the year is at the beginning of the string. This is according to the rules, which didn't say to print just the year; and this is not in the list of "standard loopholes" by the time of this answer.

To print just a year, R would need 25 bytes:

format(Sys.time()+n,"%Y")

• user100690: Even if you used the format solution, you still would have the shortest R answer.
• JDL: Actually, I think the ISO specification only requires 4-digit years, so you can use substr(...,1,4) which is a byte shorter.
• Tomas: @JDL thanks for the idea! :-) But not sure we can rely that the year will be 4-digit after addition/subtraction? More general sub("-.*","",Sys.time()+n) would work but is one char longer.
• JDL: @Tomas I don't think that matters if you interpret the rules as meaning the output is undefined if it would give rise to a year > 9999. From the wikipedia page: "ISO 8601 prescribes, as a minimum, a four-digit year [YYYY] to avoid the year 2000 problem. It therefore represents years from 0000 to 9999"

Answer 13

IBM/Lotus Notes Formula Language, (削除) 34 (削除ここまで) 32 Bytes

@Year(@Adjust(@Now;0;0;0;0;0;i))

Takes input from a field named i. Unfortunately, Formula only supports 32 bit signed integers so it is restricted to a maximum input of +/- 2,147,483,647.

There is no TIO for Formula so here are a couple of screenshots:

enter image description here enter image description here enter image description here enter image description here

Answer 14

MATL, 11 bytes

12L/Z'+10XO

Try it out at MATL Online

Explanation

 % Implicitly grab input as an integer (seconds)
12L % Push the literal 86400 to the stack (seconds in a day)
/ % Divide the input (in seconds) by the seconds in a day to convert to days
Z' % Get the current time in days as a float
+ % Add the current time and the seconds in the future together
10XQ % Determine the year of this date
 % Implicitly display the result

• Agos: not all days are 86400 seconds long!
• Luis Mendo: Hey, @Suever, it's been long! :-D Regarding what Agos says, I have asked the OP if leap seconds need to be taken into account (which would be very difficult)
• Luis Mendo: OP has confirmed it is not necessary

Answer 15

Julia, 38 bytes

using Dates
f(x)=Year(now()+Second(x))

Try it online!

Answer 16

Raku, 28 bytes

{DateTime.new($_+time).year}

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Answer 17

Python 3, 51 bytes

lambda x:gmtime(time()+x).tm_year
from time import*

Try it online!

Answer 18

AWK, 30

1ドル=strftime("%Y",1ドル+systime())

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• 5 bytes saved thanks to @manatwork.

• manatwork: As strftime() returns string and in AWK even 0 is truthy as long as it's string, you can reduce it to 1ドル=strftime("%Y",1ドル+systime()), because will always produce output..
• Digital Trauma: @manatwork thanks! that saves 5 bytes!

Answer 19

GNU coreutils (16 bytes)

date +%Y -d1ドルsec

Saved 4 bytes, thanks to @ovs.

• ovs: You can remove the onds from seconds: Try it online!
• rexkogitans: @ovs ah, I tried s and secs but missed sec somehow, thanks
• CSM: Unfortunately, this is a dupe of 2x-1's answer (label for Bash) codegolf.stackexchange.com/a/219053/55953
• rexkogitans: @CSM really oops. I only looked if someone posted for GNU coreutils, bash slipped my attention.
• Dingus: @CSM Duplicate answers are allowed.

Answer 20

MySQL, 35 bytes

select year(now()+interval [input] second)

Answer 21

Java (JDK), 57 bytes

n->java.time.LocalDateTime.now().plusSeconds(n).getYear()

Try it online!

Beats the Date API (71 bytes):

n->new java.util.Date(System.currentTimeMillis()+n*1000).getYear()+1900

Try it online!

Note that this one returns the year in absolute form (no negative year).

And I don't even speak about the Calendar API.

Answer 22

C# (.NET Core), 98 bytes

public class E{public static void m(double x){Console.WriteLine(DateTime.Now.AddSeconds(x).Year);}

Try it online!

Didn't count the using System; and calling function in Main(), if they needed to be added, please inform me!

• user100690: Still a fine solution, but I kinda doubt whether C# is an appropriate golfing language.
• avarice: @expressjs123 C# is a valid golfing language and has been used in many questions, still they are harder to golf
• Caius Jard: Can probably cut this down using the C# interactive window..
• Turksarama: You can golf in any language. I don't normally consider getting the best overall score all that interesting (since many golfing languages can solve a problem in only a couple of bytes) but prefer seeing the best score possible in a language that people use for everyday programming.

Answer 23

Visual Basic .NET (VBC), 89 bytes

public class E
Shared sub m(x)
Console.WriteLine(DateTime.Now.AddSeconds(x).Year)
end Sub

Try it online!

Answer 24

VBScript, 52 bytes

MsgBox Year(DateAdd("s",Wscript.Arguments(0),Now()) 

Answer 25

C# (Visual C# Interactive Compiler), 34 bytes

x=>DateTime.Now.AddSeconds(x).Year

Try it online!

Answer 26

AppleScript function, 46

to y(s)
(current date+s/60*minutes)'s year
end

Test driver:

repeat with sec in {100000000000, 1000, 54398643, 54398643234, -100000000000}
 log y(sec)
end repeat

Save the function and the test driver as text to nseconds.applescript, then run in the terminal:

osascript nseconds.applescript

Or run in the AppleScript Editor app.

Answer 27

T-SQL, 42 bytes

select year(dateadd(s,a,getdate()))from t;

Uses the standard input method for T-SQL of taking input from a table. You can create a suitable table like this:

create table t (a int not null);
insert into t values (1000000000), (1000), (54398643), (-1000000000);

Note that T-SQL is limited to 32-bit integers for the number of seconds. To avoid that and allow the resulting year to vary between 0 and 9999 costs 6 bytes:

select year(dateadd(d,a/86400,getdate()))from t;

Don't forget to declare a as bigint if you do this. (Note that hours is represented by hh so it's no shorter, and minutes won't let you reach 9999.)

Answer 28

APL (Dyalog Unicode), 20 bytes (SBCS)

Full program, prompting for n from stdin.

⊃⊃20 ̄1⎕DT⎕+20⎕DT'J'

20⎕DT'J' get current local* (Juliett) DateTime as UNIX time (code 20; seconds from beginning of 1970)

⎕+ prompt for n and use that to increment the UNIX time.

20 ̄1 interpret as UNIX times (code 20) and convert to an array of time stamps (code -1)

⊃ extract the first (and only) time stamp

⊃ extract the first element (the year)

Try APL! (first and last case scaled down by a factor of 10 to avoid hitting date-time limits, 1 January 0001 though 28 February 4000; turned into function for ease of use)

* could also have 'Z' (Zulu) to use the current UTC time instead of local time.

• user100690: I don't really understand APL, so I don't entirely know what the problem is, but when I paste your code into tryapl it gives me an error?
• Razetime: @expressjs123 tryAPL uses a safe version of APL which prevents people from running malicious code, so it'll give you an error for using ⎕.
• timthelion: Lots of multibyte unicode characters here, "echo '⊃⊃20 ¯1⎕DT⎕+20⎕DT\'J\'' | wc -c" gives me 32 which after removing the newline makes 31 bytes.
• Adám: @timthelion Sorry about that. Fixed now.

Answer 29

Perl 5 -p, 23 bytes

$_=gmtime$^T+$_;s/.* //

Try it online!

Answer 30

Mathematica, 43 bytes

IntegerPart@DatePlus[#/86400]["YearExact"]&

Try it online!

Thanks to @LegionMammal978 for hinting that DatePlus[#/86400][[1,1]]& fails for negative years

• LegionMammal978: This fails on negative year numbers.
• sanchez: @LegionMammal978 oh... you're absolutely right! thanks, guess i'll have to add a few more bytes :-/
• hakr14: 36 bytes