32
\$\begingroup\$

Shortest answer wins.

It must be sorted, and in 24hr time. The final line does not have a comma.

Output should be as follows:

'00:00',
'00:30',
'01:00',
'01:30',
'02:00',
'02:30',
'03:00',
'03:30',
'04:00',
'04:30',
'05:00',
'05:30',
'06:00',
'06:30',
'07:00',
'07:30',
'08:00',
'08:30',
'09:00',
'09:30',
'10:00',
'10:30',
'11:00',
'11:30',
'12:00',
'12:30',
'13:00',
'13:30',
'14:00',
'14:30',
'15:00',
'15:30',
'16:00',
'16:30',
'17:00',
'17:30',
'18:00',
'18:30',
'19:00',
'19:30',
'20:00',
'20:30',
'21:00',
'21:30',
'22:00',
'22:30',
'23:00',
'23:30'
\$\endgroup\$
12
  • 6
    \$\begingroup\$ Is this 24 hour time? Could you please provide the entire output? \$\endgroup\$
    – xnor
    May 7, 2015 at 8:21
  • 1
    \$\begingroup\$ Must the output be sorted? \$\endgroup\$
    – orlp
    May 7, 2015 at 8:30
  • 45
    \$\begingroup\$ What if it takes my program 23.5 hours to run? \$\endgroup\$
    – tfitzger
    May 7, 2015 at 13:59
  • 9
    \$\begingroup\$ I can't see why that would be a negative. \$\endgroup\$ May 7, 2015 at 14:00
  • 6
    \$\begingroup\$ Just a small suggestion about upvotes. I try to upvote every valid answer to a challenge I create as a small reward for taking the effort. I have plenty of rep and don't really need an upvote here, but please consider other answerers who may give up responding if their answers are ignored after lots of hard thinking and debugging. \$\endgroup\$ May 16, 2015 at 11:21

47 Answers 47

17
\$\begingroup\$

Pyth, 26 bytes

Pjbm%"'%02d:%s0',"d*U24"03

Demonstration.

We start with the cartesian product of range(24) (U24) with the string "03".

Then, we map these values to the appropriate string formating substitution (m%"'%02d:%s0',"d).

Then, the resultant strings are joined on the newline character (jb).

Finally, we remove the trailing comma (P) and print.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Looks like we have a new contender for the prize. Author of pyth, that's almost cheating ;) \$\endgroup\$ May 7, 2015 at 18:54
15
\$\begingroup\$

Befunge-93, 63 bytes

0>"'",:2/:55+/.55+%.":",:2%3*v
 ^,+55,","<_@#-*86:+1,"'".0. <

 

animation (animation made with BefunExec)

\$\endgroup\$
0
14
\$\begingroup\$

Bash: 58 47 46 characters

h=`echo \'{00..23}:{0,3}0\'`
echo "${h// /,
}"
\$\endgroup\$
6
  • \$\begingroup\$ I like this solution :) \$\endgroup\$ May 7, 2015 at 16:35
  • 1
    \$\begingroup\$ Anonymous user suggested echo \'{00..23}:{0,3}0\'|sed 's/ /,\n/g' of 40 characters. Nice. Thanks. But I prefer to make use of bash's own strength. \$\endgroup\$
    – manatwork
    May 8, 2015 at 9:58
  • 1
    \$\begingroup\$ printf "'%s',\n" {00..23}:{0,3}0 \$\endgroup\$
    – izabera
    May 9, 2015 at 12:22
  • \$\begingroup\$ @manatwork oh i didn't read the question well, sorry... \$\endgroup\$
    – izabera
    May 9, 2015 at 12:27
  • \$\begingroup\$ printf "'%s'\n" {00..23}:{0,3}0|sed $\!s/$/,/ is 45 bytes \$\endgroup\$
    – izabera
    May 9, 2015 at 12:31
10
\$\begingroup\$

CJam, 31 30 29 bytes

24,[U3]m*"'%02d:%d0',
"fe%~7<

This is pretty straight forward using printf formatting:

24,                              e# get the array 0..23
   [U3]                          e# put array [0 3] on stack
       m*                        e# do a cartesian product between 0..23 and [0 3] array
                                 e# now we have tuples like [[0 0], [0 3] ... ] etc
         "'%02d:%d0',
"fe%                             e# this is standard printf formatting. What we do here is
                                 e# is that we format each tuple on this string
    ~7<                          e# unwrap and remove comma and new line from last line
                                 e# by taking only first 7 characters

Try it online here

\$\endgroup\$
3
  • \$\begingroup\$ Looks like this is going to take the gold. We'll give it until tomorrow ;) \$\endgroup\$ May 7, 2015 at 16:38
  • \$\begingroup\$ @EspenSchulstad I wish it would actually give me some gold. sigh \$\endgroup\$
    – Optimizer
    May 7, 2015 at 16:39
  • 3
    \$\begingroup\$ Remember, it's not only viritual gold, but also honor. What we do in life echoes in eternity. \$\endgroup\$ May 7, 2015 at 16:40
10
\$\begingroup\$

Python 2, 58 56 bytes

for i in range(48):print"'%02d:%s0',"[:57-i]%(i/2,i%2*3)

Like sentiao's answer but using a for loop, with slicing to remove the comma. Thanks to @grc for knocking off two bytes.

\$\endgroup\$
3
  • \$\begingroup\$ Would this work: "'%02d:%s0',"[:57-i] \$\endgroup\$
    – grc
    May 7, 2015 at 14:16
  • \$\begingroup\$ @grc Ahaha of course, that's much better \$\endgroup\$
    – Sp3000
    May 7, 2015 at 15:03
  • \$\begingroup\$ Better than I could do! \$\endgroup\$
    – Tim
    May 7, 2015 at 18:15
6
\$\begingroup\$

Java - 119 bytes

I started with Java 8's StringJoiner, but that means including an import statement, so I decided to do it the old way:

void f(){String s="";for(int i=0;i<24;)s+=s.format(",\n'%02d:00',\n'%02d:30'",i,i++);System.out.print(s.substring(2));}

Perhaps this can be improved by getting rid of the multiple occuring String and System keywords.

\$\endgroup\$
5
  • \$\begingroup\$ a version in groovy perhaps :) \$\endgroup\$ May 7, 2015 at 10:24
  • \$\begingroup\$ You can shorten this to 159: Remove the space before a, move the i increment, lose the for braces, and move the comma/newline to the beginning (which lets you use the shorter substring and get rid of length()). Since functions are allowed by default, you can get make it even shorter by eliminating boilerplate: void f(){String s="";for(int i=0;i<24;)s+=String.format(",\n'%02d:00',\n'%02d:30'",i,i++);System.out.print(s.substring(2));} Even a bit more if you make it just return the string instead of printing it, but that seems to go against the spirit, if not the letter. \$\endgroup\$
    – Geobits
    May 7, 2015 at 13:00
  • \$\begingroup\$ Damn, that's brilliant! I got to keep those tricks in mind, thanks! \$\endgroup\$ May 7, 2015 at 13:53
  • \$\begingroup\$ Oh, another few: change String.format to s.format. Your compiler/IDE may complain about it, but it works ;) \$\endgroup\$
    – Geobits
    May 7, 2015 at 13:55
  • 1
    \$\begingroup\$ Nice one! format's a static method, so it should be accessed by its class, but indeed, using it this way also works! \$\endgroup\$ May 8, 2015 at 6:18
5
\$\begingroup\$

Ruby, 94 61 56 51

$><<(0..47).map{|i|"'%02d:%02d'"%[i/2,i%2*30]}*",
"

Thanks to @blutorange (again) for his help in golfing!

\$\endgroup\$
4
  • \$\begingroup\$ You can reduce that to 61 bytes: puts (0..47).to_a.map{|h|"'%02d:%02d'"%[h/2,h%2*30]}.join"," (there's a newline after the last comma) \$\endgroup\$
    – blutorange
    May 7, 2015 at 14:37
  • \$\begingroup\$ Thank you again! I really didn't try it much... \$\endgroup\$
    – ror3d
    May 7, 2015 at 14:42
  • \$\begingroup\$ Ruby needs some love. 58 bytes ;) puts Array.new(48){|i|"'%02d:%02d'"%[i/2,i%2*30]}.join',' \$\endgroup\$
    – blutorange
    May 7, 2015 at 16:57
  • \$\begingroup\$ @blutorange @rcrmn you can reduce 4 more bytes (and get to my solution below :)) ) by replacing .join with * :) \$\endgroup\$ May 8, 2015 at 5:54
5
\$\begingroup\$

Perl, 52 50 48 45B

$,=",
";say map<\\'$_:{0,3}0\\'>,"00".."23"

With help from ThisSuitIsBlackNot :)

\$\endgroup\$
0
4
\$\begingroup\$

JAVA 95 94 bytes

I love the fact that printf exists in Java:

void p(){for(int i=0;i<24;)System.out.printf("'%02d:00',\n'%02d:30'%c\n", i,i++,(i<24)?44:0);}

Ungolfed

void p(){
    for(int i=0;i<24;)
        System.out.printf("'%02d:00',\n'%02d:30'%c\n", i,i++,(i<24)?44:0);
}

EDIT Replaced the ',' with 44

\$\endgroup\$
5
  • \$\begingroup\$ 24.times{printf("'%02d:00',\n'%02d:30'%c\n",it,it,(it<24)?44:0)}​ in groovy :) same solution, just that it's down to 65 chr \$\endgroup\$ May 7, 2015 at 16:24
  • \$\begingroup\$ @EspenSchulstad Would that be considered a standalone function, or is there other boilerplate that would be needed to get it to run? \$\endgroup\$
    – tfitzger
    May 7, 2015 at 16:28
  • 1
    \$\begingroup\$ If you have groovy, you could either run that in groovysh or just as a groovy script in a .groovy-file. then you don't need no function. \$\endgroup\$ May 7, 2015 at 16:30
  • \$\begingroup\$ groovyconsole.appspot.com/edit/24001 \$\endgroup\$ May 7, 2015 at 16:33
  • \$\begingroup\$ @EspenSchulstad Interesting. I've only done minimal work with Groovy. \$\endgroup\$
    – tfitzger
    May 7, 2015 at 16:56
4
\$\begingroup\$

Pyth, 32 31 bytes

I golfed something in python but it turned out to be exactly the same as Sp3000's answer. So I decided to give Pyth a try:

V48<%"'%02d:%d0',",/N2*3%N2-54N

It's a exact translation of Sp3000 answer:

for i in range(48):print"'%02d:%d0',"[:57-i]%(i/2,i%2*3)

It's my first go at Pyth, so please do enlighten me about that 1 byte saving.

\$\endgroup\$
1
  • \$\begingroup\$ Nicely done, and welcome to Pyth. \$\endgroup\$
    – isaacg
    May 7, 2015 at 20:25
3
\$\begingroup\$

PHP, 109 bytes

foreach(new DatePeriod("R47/2015-05-07T00:00:00Z/PT30M")as$d)$a[]=$d->format("'H:i'");echo implode(",\n",$a);
\$\endgroup\$
3
\$\begingroup\$

Ruby, 54 51 bytes

puts (0..23).map{|h|"'#{h}:00',
'#{h}:30'"}.join",
"
\$\endgroup\$
2
  • 1
    \$\begingroup\$ You can reduce 3 bytes by changing \n to actual newlines and removing the space between join and ". On the other hand, take note that the specified output has leading zeros for the hours. \$\endgroup\$
    – ror3d
    May 7, 2015 at 16:13
  • \$\begingroup\$ Also, some more bytes by changing puts to $><< (without space) and .join with *. You still have the leading zero problem for the hours, though. \$\endgroup\$
    – ror3d
    May 8, 2015 at 8:57
3
\$\begingroup\$

C, 116,115,101,100,95,74,73, 71

May be able to scrape a few more bytes off this...

main(a){for(;++a<50;)printf("'%02d:%d0'%s",a/2-1,a%2*3,a<49?",\n":"");}
\$\endgroup\$
4
  • \$\begingroup\$ You can save 3 bytes by creating a function rather than a complete program. Just replace "main" with "f", or whatever your favourite letter is. \$\endgroup\$
    – Bijan
    Mar 27, 2017 at 22:14
  • \$\begingroup\$ Oh thats a very nice suggestion.....one I always forget! \$\endgroup\$
    – Sarima
    Mar 28, 2017 at 7:15
  • \$\begingroup\$ Suggest a/49*2+",\n" instead of a<49?",\n":"" \$\endgroup\$
    – ceilingcat
    Jul 13, 2020 at 1:56
  • \$\begingroup\$ It's from 3 years ago - I'm not updating it. \$\endgroup\$
    – Sarima
    Jul 13, 2020 at 10:13
3
\$\begingroup\$

T-SQL, 319 307 305 bytes

WITH t AS(SELECT t.f FROM(VALUES(0),(1),(2),(3),(4))t(f)),i AS(SELECT i=row_number()OVER(ORDER BY u.f,v.f)-1FROM t u CROSS APPLY t v),h AS(SELECT i,h=right('0'+cast(i AS VARCHAR(2)),2)FROM i WHERE i<24)SELECT''''+h+':'+m+CASE WHEN i=23AND m='30'THEN''ELSE','END FROM(VALUES('00'),('30'))m(m) CROSS APPLY h

Un-golfed version:

WITH
t AS(
    SELECT
        t.f
    FROM(VALUES
         (0),(1),(2),(3),(4)
    )t(f)
),
i AS(
    SELECT
        i = row_number() OVER(ORDER BY u.f,v.f) - 1
    FROM t u 
    CROSS APPLY t v
),
h AS(
    SELECT
        i,
        h = right('0'+cast(i AS VARCHAR(2)),2)
    FROM i
    WHERE i<24
)
SELECT
    '''' + h + ':' + m + CASE WHEN i=23 AND m='30' 
                              THEN '' 
                              ELSE ',' 
                         END
FROM(
    VALUES('00'),('30')
)m(m)
CROSS APPLY h
\$\endgroup\$
2
\$\begingroup\$

Pyth, 34 bytes

j+\,bmjk[\'?kgd20Z/d2\:*3%d2Z\')48

This can definitely be improved.

Try it online: Pyth Compiler/Executor

Explanation:

     m                          48   map each d in [0, 1, ..., 47] to:
        [                      )       create a list with the elements:
         \'                              "'"
           ?kgd20Z                       "" if d >= 20 else 0
                  /d2                    d / 2
                     \:                  ":"
                       *3%d2             3 * (d%2)
                            Z            0
                             \'          "'"
      jk                               join by "", the list gets converted into a string
j+\,b                                join all times by "," + "\n"
\$\endgroup\$
2
  • \$\begingroup\$ Always impressive with these golfing languages :) I do somehow appreciate it more when it's done in a non-golfing language. Does anyone know if there is a golf-jar-lib to java for instance? \$\endgroup\$ May 7, 2015 at 10:30
  • \$\begingroup\$ Obligatory port of sentiao's gives 31: j+\,bm%"'%02d:%s0'",/d2*3%d2 48 with string formatting \$\endgroup\$
    – Sp3000
    May 7, 2015 at 12:28
2
\$\begingroup\$

Python 2, 69 bytes

print',\n'.join(["'%02d:%s0'"%(h,m)for h in range(24)for m in'03'])

Quite obvious, but here's an explanation:

  • using double-for-loop
  • alternating between '0' and '3' in string format is shorter than a list
  • %02d does the padding for h
  • mdoesn't need padding as the alternating character is on a fixed position
  • '\n'.join() solves the final-line requirements

I have no idea if it can be done shorter (in Python 2).

by Sp3000, 61 bytes : print',\n'.join("'%02d:%s0'"%(h/2,h%2*3)for h in range(48))

\$\endgroup\$
3
  • 1
    \$\begingroup\$ How about: print',\n'.join("'%02d:%s0'"%(h/2,h%2*3)for h in range(48)) \$\endgroup\$
    – Sp3000
    May 7, 2015 at 11:29
  • \$\begingroup\$ Brilliant, Sp3000! (you should post it) \$\endgroup\$
    – sentiao
    May 7, 2015 at 11:32
  • 1
    \$\begingroup\$ Nah, not different enough to post in my book. All I did was drop the square brackets (which are unnecessary even in your one) and drop m. (Also it's 59 bytes, not 61) \$\endgroup\$
    – Sp3000
    May 7, 2015 at 11:35
2
\$\begingroup\$

Haskell, 85 bytes

putStr$init$init$unlines$take 48['\'':w:x:':':y:"0',"|w<-"012",x<-['0'..'9'],y<-"03"]

Unfortunately printf requires a 19 byte import, so I cannot use it.

\$\endgroup\$
2
\$\begingroup\$

Julia: 65 64 61 characters

[@printf("'%02d:%d0'%s
",i/2.01,i%2*3,i<47?",":"")for i=0:47]

Julia: 64 characters

(Kept here to show Julia's nice for syntax.)

print(join([@sprintf("'%02d:%d0'",h,m*3)for m=0:1,h=0:23],",
"))
\$\endgroup\$
2
\$\begingroup\$

Fortran 96

do i=0,46;print'(a1,i2.2,a,i2.2,a2)',"'",i/2,":",mod(i,2)*30,"',";enddo;print'(a)',"'23:30'";end

Standard abuse of types & requirement only for the final end for compiling. Sadly, due to implicit formatting, the '(a)' in the final print statement is required. Still, better than the C and C++ answers ;)

\$\endgroup\$
2
\$\begingroup\$

JavaScript (ES6), 77 86+1 bytes

Didn't realize there had to be quotes on each line (+1 is for -p flag with node):

"'"+Array.from(Array(48),(d,i)=>(i>19?"":"0")+~~(i/2)+":"+3*(i&1)+0).join("',\n'")+"'"

old solution:

Array.from(Array(48),(d,i)=>~~(i/2).toFixed(2)+":"+3*(i&1)+"0").join(",\n")

ungolfed version (using a for loop instead of Array.from):

var a = [];
// there are 48 different times (00:00 to 23:30)
for (var i = 0; i < 48; i++) {
    a[i] =
        (i > 19 ? "" : "0") +
            // just a ternary to decide whether to pad
            // with a zero (19/2 is 9.5, so it's the last padded number)
        ~~(i/2) +
            // we want 0 to 24, not 0 to 48
        ":" +  // they all then have a colon
        3*(i&1) +
            // if i is odd, it should print 30; otherwise, print 0
        "0" // followed by the last 0
}
console.log(a.join(",\n"));
\$\endgroup\$
1
  • 1
    \$\begingroup\$ 72: Array.from(Array(48),(d,i)=>`'${i>19?"":0}${0|i/2}:${i%2*3}0'`).join`,\n`. Replace \n with an actual newline. \$\endgroup\$ Nov 7, 2015 at 0:22
2
\$\begingroup\$

golflua 52 51 chars

~@n=0,47w(S.q("'%02d:%d0'%c",n/2,n%2*3,n<47&44|0))$

Using ascii 44 = , and 0 a space saves a character.

An ungolfed Lua version would be

for h=0,47 do
   print(string.format("'%02d:%d0'%c",h/2,h%2*3, if h<47 and 44 or 0))
end

The if statement is much like the ternary operator a > b ? 44 : 0.

\$\endgroup\$
2
\$\begingroup\$

C# - 120 bytes

class P{static void Main(){for(var i=0;i<24;i++)System.Console.Write("'{0:00}:00',\n'{0:00}:30'{1}\n",i,i==23?"":",");}}
\$\endgroup\$
2
\$\begingroup\$

Python, 60 58 64 bytes

for i in range(24):print("'%02d:00,\n%02d:30'"%(i,i)+', '[i>22])

Ungolfed:

for i in range(24):
    if i <23:
        print( ('0'+str(i))[-2:] + ':00,\n' + str(i) + ':30,')
    else:
        print( ('0'+str(i))[-2:] + ':00,\n' + str(i) + ':30')

Try it online here.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Why not put it on 1 line, and save another 2 bytes. \$\endgroup\$
    – isaacg
    May 7, 2015 at 18:15
  • \$\begingroup\$ I don't think this works - no leading zero on the hour. \$\endgroup\$
    – isaacg
    May 7, 2015 at 18:17
  • 1
    \$\begingroup\$ @isaacg fixed that! \$\endgroup\$
    – Tim
    May 7, 2015 at 18:25
  • \$\begingroup\$ The output is not the same as in the original question. \$\endgroup\$
    – sentiao
    May 8, 2015 at 8:51
  • \$\begingroup\$ @sentiao what is different? \$\endgroup\$
    – Tim
    May 8, 2015 at 11:41
2
\$\begingroup\$

𝔼𝕊𝕄𝕚𝕟, 39 chars / 67 bytes (non-competing)

↺;Ḁ⧺<Ḱ;)ᵖ`'⦃Ḁ<Ḕ?0:⬯}⦃0|Ḁ/2}:⦃Ḁ%2*3}0',”

Try it here (Firefox only).

Not a single alphabetical character in sight...

\$\endgroup\$
0
2
\$\begingroup\$

PHP, 69 70 62 bytes

for($x=-1;++$x<47;)printf("'%02d:%d0',
",$x/2,$x%2*3)?>'23:30'

Try it online

Outputting '23:30' at the end is a bit lame, and so is closing the php context using ?> without opening or re-opening it. An cleaner alternative (but 65 bytes) would be:

for($x=-1;++$x<48;)printf("%s'%02d:%d0'",$x?",
":'',$x/2,$x%2*3);

Try it online

Thank you @Dennis for the tips. Alternative inspired by the contribution of @ismael-miguel.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Welcome to Programming Puzzles & Code Golf! Your code prints a null byte at the end. I'm not sure if that is allowed. \$\endgroup\$
    – Dennis
    Jan 11, 2016 at 5:44
  • \$\begingroup\$ @Dennis Thank you, you're right... I assumed PHP would see it as end-of-string. Posted a new version. \$\endgroup\$
    – mk8374876
    Jan 11, 2016 at 18:43
  • \$\begingroup\$ <?...?>'23:30' saves three bytes. Also, you can replace \n with an actual newline. \$\endgroup\$
    – Dennis
    Jan 11, 2016 at 19:07
2
\$\begingroup\$

Swift, 74 bytes

Updated for Swift 2/3...and with new string interpolation...

for x in 0...47{print("'\(x<20 ?"0":"")\(x/2):\(x%2*3)0'\(x<47 ?",":"")")}
\$\endgroup\$
1
  • \$\begingroup\$ The final line is specified to not have a comma & your code prints it. \$\endgroup\$
    – Kyle Kanos
    May 7, 2015 at 13:32
1
\$\begingroup\$

Javascript, 89 bytes

for(i=a=[];i<24;)a.push((x="'"+("0"+i++).slice(-2))+":00'",x+":30'");alert(a.join(",\n"))
\$\endgroup\$
8
  • 1
    \$\begingroup\$ Quick question: how many arguments Array.push() supports? ;) \$\endgroup\$
    – manatwork
    May 7, 2015 at 14:31
  • \$\begingroup\$ You're right. That is a polyad. Thanks, I'll make that change after I finish testing my entry for another challenge. \$\endgroup\$ May 7, 2015 at 15:09
  • \$\begingroup\$ A few more characters can be removed with some elementary reorganizations: for(i=a=[];i<24;)a.push((x=("0"+i++).slice(-2))+":00",x+":30");alert(a.join(",\n")) \$\endgroup\$
    – manatwork
    May 7, 2015 at 15:28
  • \$\begingroup\$ There, I fixed it. \$\endgroup\$ May 7, 2015 at 15:38
  • \$\begingroup\$ That reuse of variable x is quite ugly coding habit. If you change the loop control variable to something else (as in my suggestion at 2015-05-07 15:28:25Z), then you can add the opening single quotes to x's value to reduce the two "'"+ pieces to one: for(i=a=[];i<24;)a.push((x="'"+("0"+i++).slice(-2))+":00'",x+":30'");alert(a.join(",\n")) \$\endgroup\$
    – manatwork
    May 7, 2015 at 15:58
1
\$\begingroup\$

Python 2: 64 bytes

print ',\n'.join(['%02d:00,\n%02d:30'%(h,h) for h in range(24)])
\$\endgroup\$
2
  • \$\begingroup\$ The output is not the same as in the original question. \$\endgroup\$
    – sentiao
    May 8, 2015 at 8:51
  • \$\begingroup\$ I don't know what happened to the edit I made. here is the correct version also 64 chars: print',\n'.join("'%02d:00',\n'%02d:30'"%(h,h)for h in range(24)) \$\endgroup\$ May 8, 2015 at 14:04
1
\$\begingroup\$

Ruby - 52 bytes

puts (0..47).map{|i|"'%02d:%02d'"%[i/2,i%2*30]}*",
"
\$\endgroup\$
3
  • \$\begingroup\$ This seems like my solution with just the change of the .join for *... It's common courtesy to instead of just posting a new answer with a minor improvement, to suggest the improvement to the original poster. See meta.codegolf.stackexchange.com/questions/75/… \$\endgroup\$
    – ror3d
    May 8, 2015 at 8:13
  • \$\begingroup\$ @rcrmn I agree I made a mistake and I apologize for it - should've had looked for other ruby answers first. Great work though in your answer with $><<! \$\endgroup\$ May 8, 2015 at 8:45
  • \$\begingroup\$ Don't worry about it, you are new here: I was advising you so that you could avoid this in the future. \$\endgroup\$
    – ror3d
    May 8, 2015 at 8:55
1
\$\begingroup\$

Python 2, 74 65 bytes

We generate a 2 line string for each hour, using text formatting:

print',\n'.join("'%02u:00',\n'%02u:30'"%(h,h)for h in range(24))

This code is fairly clear, but the clever indexing and integer maths in the answer by Sp3000 gives a shorter solution.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ no reason to use formatting for 00 and 30, save 9 bytes with "'%02u:00',\n'%02u:30'"%(h,h) \$\endgroup\$
    – DenDenDo
    May 7, 2015 at 10:42
  • \$\begingroup\$ you could save some bytes by having too loops : print',\n'.join("'%02u:%02u'"%(h,i)for h in range(24)for i in[0,30]) \$\endgroup\$
    – dieter
    May 7, 2015 at 11:01
  • \$\begingroup\$ Thanks DenDenDo. I used this to save some bytes. Dieter, I can see where your idea may help, but I could save more bytes with that idea from DenDenDo. \$\endgroup\$ May 8, 2015 at 11:25

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