Imagine the following 24-hour clock that can be controlled by arrow keys:

╔══╗ ┌──┐
╚══╝ └──┘
 HH   mm

Pressing the up arrow twice (↑↑) will increase the currently focused hour input:

╔══╗ ┌──┐
╚══╝ └──┘
 HH   mm

Pressing the right arrow () will focus the other input.

┌──┐ ╔══╗
└──┘ ╚══╝
 HH   mm

Pressing the down arrow thrice (↓↓↓) will now decrease this input.

┌──┐ ╔══╗
└──┘ ╚══╝
 HH   mm

Shortly put:

  • The up arrow () will increase the currently active input.
  • The down arrow () will decrease the active input.
  • The right arrow () will move focus to the right input.
  • The left arrow () will move focus to the left input.
  • Up and down movement will loop around as expected for a time input.
  • Left and right movement don't loop around.

The challenge

The clock starts out at 00:00 with the hour input active (see first schematic). Given a list of input commands, output the resulting time in HH:mm format.
Input can be either a string or a list (or your language equivalent), where the different input directions can be one of the below options:

  • ↑↓←→
  • udlr
  • ^v<>
  • actual arrow key presses if your program has a GUI

Standard loopholes apply.

Test cases

↑↑→↓↓↓ = 02:57
↓→↑←↑→↓ = 00:00
↓→→↓ = 23:59
←←←←→↑ = 00:01
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓→↓ = 23:59
  • 1
    \$\begingroup\$ @JonathanFrech One of the given options, choosing any four unique values (for example 0123) would make the challenge a lot easier in certain languages while not benefitting others. \$\endgroup\$
    – Etheryte
    Commented Jul 18, 2018 at 11:21
  • 1
    \$\begingroup\$ @LuisfelipeDejesusMunoz Yes, that's actually written out under the input rules. \$\endgroup\$
    – Etheryte
    Commented Jul 18, 2018 at 11:45
  • 3
    \$\begingroup\$ I think this would have been more challenging if it included seconds. This would have more logic behind which one is currently in focus \$\endgroup\$
    – Jo King
    Commented Jul 18, 2018 at 12:40
  • 3
    \$\begingroup\$ Missing a special rule to handle Konami code. \$\endgroup\$
    – coredump
    Commented Jul 20, 2018 at 14:02
  • 1
    \$\begingroup\$ @coredump Considered it, but it would probably take more space than the core of the answer in most languages. \$\endgroup\$
    – Etheryte
    Commented Jul 20, 2018 at 18:51

24 Answers 24


HTML on Google Chrome 67 in Chinese (Simplified), 39 bytes

<input type=time value=00:00 autofocus>


Chrome show different UI component in different language. Even a simple time input: AM/PM will be shown if you are using English (US). If you want test this by changing your Chrome's language. Do not mass up how to change it back.

  • 3
    \$\begingroup\$ dudee!! hahah I think is not valid tho \$\endgroup\$ Commented Jul 18, 2018 at 12:35
  • 3
    \$\begingroup\$ Hitting right twice goes to AM/PM for me \$\endgroup\$
    – Jo King
    Commented Jul 18, 2018 at 12:35
  • 1
    \$\begingroup\$ @JoKing I think it's locale/settings dependent? \$\endgroup\$
    – Etheryte
    Commented Jul 18, 2018 at 12:36
  • 1
    \$\begingroup\$ @JoKing That's locale dependent. Maybe have a try by changing your Chrome's language to Chinese Simplify? (Do not mass up how to change it back.) \$\endgroup\$
    – tsh
    Commented Jul 18, 2018 at 12:44
  • 1
    \$\begingroup\$ It works on firefox 61.0.1 \$\endgroup\$ Commented Jul 18, 2018 at 14:10

C (gcc), 117 107 bytes

  • Saved ten bytes thanks to l4m2.

Try it online!

  • 5
    \$\begingroup\$ Nice variable naming. \$\endgroup\$
    – Etheryte
    Commented Jul 18, 2018 at 11:44
  • \$\begingroup\$ # C (gcc), 107 bytes <!-- language-all: lang-c --> t,i,m[8];e(char*_){for(*m=i=2[m]=0;t=*_++;t<63?i=t%4:(i[m]+=t&8?1:119));printf("%02d:%02d",*m%24,2[m]%60);} Try it online! \$\endgroup\$
    – l4m2
    Commented Jul 18, 2018 at 12:06

Stax, 36 35 33 32 bytes


Run and debug it

Uses lrud.


'l/{'r/Bs$2lmM{${:14-m|+i36*24+%2|zm':* Full program,
'l/                                     Split the string on "l"
   {        m                           Map over the resulting array
    'r/                                   Split at "r"
       B                                  Uncons left, first on TOS (top of stack)
        s                                 Swap to get tail to top
         $                                Flatten; this removes multiple 'r's
          2l                              Listify two items, BOS (bottom of stack) is first element
             M                          Transpose: get [hour commands, minute commands]
              {                    m    Map:
               $                          Flatten
                {    m                    Map over single commands:
                 :1                         Number of set bits: 5 for 'u', 3 for 'd'
                   4-                       Subtract 4: u -> 1, d -> -1
                      |+                  Sum
                        i                 Iteration index: hours -> 0, minutes -> 1
                         36*24+           Multiply by 36, add 24: 0 -> 24, 1 -> 60
                               %          Modulo, this does -5 % 60 = 55
                                2|z       Stringify, left-padding with "0" to length 2
                                    ':* Join on ":"
                                        Implicit output

Python 2, 105 bytes

for c in map(' ^<>'.find,input()):w=1/c;m+=w*p;h+=w-w*p;p=[c-2,p][w]

Try it online!


JavaScript (Node.js), 103 bytes

Takes input as a string, using udlr.


Try it online!


C# (.NET Core), 149 132 bytes

s=>{var p=0;int[]h={0,0};foreach(var c in s)h[p=c<63?c/2%2:p]+=c>62?c>95?-1:1:0;return$"{(h[0]%24+24)%24:D2}:{(h[1]%60+60)%60:D2}";}

Try it online!

Using ^v<>.

This one made me realize that the modulo operator in C# does not work as expected, because in C# -1 % 60 = -1, so I need to do that weird operation at the end.

  • \$\begingroup\$ Can't (h[1]%60+60)%60 be replaced with (h[1]+60)%60 ? \$\endgroup\$
    – IanF1
    Commented Jul 18, 2018 at 18:31
  • 2
    \$\begingroup\$ @IanF1 no you can't. What if the user presses the down button 100 times? Or 1000 times? \$\endgroup\$
    – Charlie
    Commented Jul 18, 2018 at 19:07
  • \$\begingroup\$ thanks for clarifying :) it's surprising to me that this way is shorter than applying the modulo on the fly (with 59 in place of -1). \$\endgroup\$
    – IanF1
    Commented Jul 18, 2018 at 19:12

Lua (love2d framework),311 308 bytes

l,b,d,t,f,a=love,{24,60},{1,-1},{0,0},1,{"left","right","up","down"}function c(n,i)t[f]=(n+d[i])%b[f]end function l.draw()h,m=t[1],t[2]l.graphics.print((h<10 and 0 ..h or h)..":"..(m<10 and 0 ..m or m),0,0)end function l.keypressed(k)for i,n in pairs(a)do f=k==n and(i>2 and(c(t[f],i-2)or f)or i)or f end end

Unscrumbeled version:

--initialize all needed values

--increase the numbers depending on the focus and up or down
function c(n,i)

--draw the time to the screen
function l.draw()
  l.graphics.print((h<10 and 0 ..h or h)..":"..(m<10 and 0 ..m or m),0,0)

--get the keys and check if it is an arrow key
function l.keypressed(k)
  for i,n in pairs(a)do
    f=k==n and(i>2 and(c(t[f],i-2)or f)or i)or f 

Probably still not 100% easy to read because all the ifs are interchanged with an trinary statement ( ..and ..or) :)

if started in an main.lua with love then it will pop up a window and you can press the arrowkeys to change the numbers

  • \$\begingroup\$ could you also post expanded version for readability \$\endgroup\$ Commented Jul 18, 2018 at 18:20
  • \$\begingroup\$ sure, I added a expanded version no problem :) \$\endgroup\$
    – Lycea
    Commented Jul 19, 2018 at 7:24

MATL, 57 56 55 bytes

1Oi9\"@5<?y@3-ZS*+}wx7@-X^w]]wx&Zjh24 60h\'%02d:%02d'YD

Try it online!

Represents hour and minutes using complex numbers, with the real part being hours and the imaginary part minutes.

With comments:

1     % Push 1 on the stack
      % represents which timer box we're in, starts at hour box
      % imaginary number j would represent minutes box
O     % Push initial hour and minutes 0+0j
i9\   % Fetch input, mod each character's ASCII value by 9.
      % Gives 4 1 8 6 for ^ v > < respectively
"     % iterate through (modded) input
  @5<?     % Push current input, see if it's < 5 
           % if so, it's an up or down time change
    y        % so copy out the box indicator (1 or j)
    @3-      % Subtract 3 from the current input
    ZS       % Take the result's sign (-1 for v, 1 for ^)
    *        % Multiply indicator with that
    +        % Add the result to the time value
  }        % else, it's a right or left arrow
    wx       % so bring out the box indicator and delete it
    7@-      % Subtract current input from 7. 1 for < and -1 for >
    X^       % Take the square root of that. 1 for < and j for >
    w        % switch stack to bring time value on top again
  ]       % end if
]     % end loop
wx    % bring box indicator out, delete it
&Zj   % split the complex time value to real and imaginary
h     % then concatenate them into an array
24 60h\ % mod hour and minute values by 24 and 60 respectively
'%02d:%02d'YD % sprintf the time array with 0-padding

PHP, 145 134 133 bytes

(-11 bytes by more golfing)

(-1 byte by using Davіd's loop method)


To run it:

php -n -d error_reporting=0 <filename> <command_1> <command_2> ... <command_n>


php -n -d error_reporting=0 time_setter.php u u r d d d l d

Or Try it online!


  • To save some bytes, I have used strings without single/double quotations as the string wrapper. Thus, the error_reporting=0 option is used to not output warnings.
  • Input commands: u d l r
  • \$\begingroup\$ 128 bytes, -6: Try it online! (Nice solution, btw :) \$\endgroup\$
    – Ethan
    Commented Jul 19, 2018 at 6:13
  • \$\begingroup\$ @Davіd: Thanks, but your update has two issues. First on is that $h is not set, so decreasing it on start fails: Try it online! \$\endgroup\$
    – Night2
    Commented Jul 19, 2018 at 8:30
  • \$\begingroup\$ @David: And the second issue happens when we switch hour/minute up or down more than 24/60 times: Try it online! \$\endgroup\$
    – Night2
    Commented Jul 19, 2018 at 8:30
  • \$\begingroup\$ @Davіd: But thanks to your loop method, I could reduce 1 more byte: Try it online! \$\endgroup\$
    – Night2
    Commented Jul 19, 2018 at 8:33
  • \$\begingroup\$ ah, alright, sorry it didn't completely work :) \$\endgroup\$
    – Ethan
    Commented Jul 19, 2018 at 9:03

JavaScript, 104 103 bytes

Takes input as an array of characters, using <>^v.


Try it online


Haskell, 236 bytes

f=u 0 0
k _ _ _ _ _ h m[]=z h++':':z m
k a b c d e h m(q:s)=case q of{'^'->e(a h)(b m)s;'v'->e(c h)(d m)s;'>'->v h m s;'<'->u h m s}
u=k(o(+)24)id(o(-)24)id u
v=k id(o(+)60)id(o(-)60)v
o f m x=mod(f x 1)m
z n|n<10='0':show n
z n=show n

f is the main function, and has type String -> String:

*Main> f "^^>vvv"
*Main> f "v>^<^>v"
*Main> f "v>>v"
*Main> f "<<<<>^"
*Main> f "vvvvvvvvvvvvvvvvvvvvvvvvv>v"

Essentially u and v are mutually recursive functions of type Integer -> Integer -> String -> String. They take the hour, the minute and a list of characters over the set {v,^,<,>}, and return the time string. u acts as if the hour dial is highlighted, recursively calling u if the head of the list is in {v,^}, and v if the head of the list is in {<,>}. v is similar but for the minute dial.

Everything else is just saving characters.


Lua, 132 bytes

loadstring's,t,m=1,{0,0},{24,60}for c in(...):gmatch"."do t[s]=(t[s]+(("d u"):find(c)or 2)-2)%m[s]s=("lr"):find(c)or s end return t'

Try it online!


This is an anonymous function (a way to use it is shown on the link).

s=1 -- s will control the selection (1 is hour and 2 min)
t={0,0} -- is the time itself
m={24,60} -- is the maximum for each 'box' (hour or min)
-- I've actually used Lua's multiple variable assignment: s,t,m=1,{0,0},{24,60}

for c in (...):gmatch(".") do -- go through each character of the input
  t[s] = (t[s] + (("d u"):find(c) or 2)-2) % m[s] -- set the current 'box' as
          t[s] +   -- itself plus ...
                  ("d u"):find(c) or 2   -- it's index on the string "d u" (that means it's going to be 1 or 3)
                                         -- or 2 if it wasn't found (if the current character doesn't sum or subtract from the box)
                                       -2   -- this adjusts the result 1, 2 or 3 to being -1, 0 or 1
                                            -- making the inputs 'd' and 'u' as -1 and +1 respectively, and an input different from both as 0
         (                               ) % m[s]   -- modulo of the maximum of the selected 'box'

  s=("lr"):find(c) or s
    ("lr"):find(c)   -- if the current input character is l or r, then set 's' (the 'box' selection) to being 1 or 2.
                   or s   -- else let it as is
return t -- returns 't', a table with hour and minutes respectively
  • \$\begingroup\$ The output should be in the HH:mm format, rather than a table \$\endgroup\$
    – Jo King
    Commented Jul 29, 2018 at 10:40

Java 8, 121 bytes

c->{int i=0,m[]={0,0,0};for(int t:c)if(t<63)i=t%4;else m[i]+=(t&8)>0?1:119;return"".format("%02d:%02d",m[0]%24,m[2]%60);}

Port of Jonathan Frech's C answer. Accepts ^v<>. Try it online here.


Jelly, 36 bytes

I believe O%5;4ṣ3œṡ€4Z%3’§§%"“ð<‘DŻ€ṫ€-j”: should work for 32, but œṡ seems to currently have a bug.


A full program printing the result to STDOUT (as a monadic link it actually returns a mixed list of integers (albeit single digit ones) and characters (the :).

Uses the udlr option for input.

Try it online! Or see a test-suite.


O%5;4ṣ3i€4$œṖ"$Z%3’§§%"“ð<‘DŻ€ṫ€-j”: - Link: list of characters (in 'udlr')
O                                    - to ordinals
 %5                                  - modulo five  ...maps u:2, d:0, l:3, r:4
   ;4                                - concatenate a 4 (to always end up with both hrs & mins - even when no r is ever pressed)
     ṣ3                              - split at threes (the l presses)
       i€4$œṖ"$                      - a replacement for œṡ€4 (split each at first occurrence of)...
              $                      - | last two links as a monad:
          $                          - |   last two links as a monad:
         4                           - |     literal four
       i€                            - |     for €ach get first index of (4) else yield 0
             "                       - |   zip with:
           œṖ                        - |     partition at indices
               Z                     - transpose (to get a list of two lists of lists)
                %3                   - modulo by three. To replace any 4(r) with 1
                                     -  ...while keeping any 0(d) as 0, or 2(u) as 2
                  ’                  - decrement. All r are now 0, d are -1 and u are 1
                   §                 - sum each
                    §                - sum each. Now we have the total increase value as
                                     -    ...integers for each of hrs and mins
                       “ð<‘          - code-page indices list = [24,60]
                      "              - zip with:
                     %               -   modulo
                           D         - to decimal lists
                            Ż€       - prepend each with a zero (to cater for values less than ten)
                              ṫ€-    - tail each from index -1. Keeps rightmost two digits of each only)
                                  ”: - literal character ':'
                                 j   - join
                                     - as full program implicit print (smashes the digits and characters together)

APL (Dyalog Classic), 97 84 bytes

5↑∊{¯3↑'0',':',⍨⍕⍵}¨24 60|A⊣⍎¨'⎕IO←1' '⎕IO←0' 'A[1]+←1' 'A[1]-←1'['←→↑'⍳⍞,'←']⊣A←0 0

Try it online!

Requires ⎕IO←1


QBasic, 229 bytes

A script that takes input as keystrokes and outputs to the console.

Note: terminal " are included for syntax highlighting only, and do not contribute to the bytecount

CASE z$+"K"
CASE z$+"M"
CASE z$+"H"
CASE z$+"P"
IF r THEN m=(m+y)MOD 60ELSE h=(h+x)MOD 24


z$=CHR$(0)                                      ''  Set var to null char
DO                                              ''
    x=0                                         ''  Set Hours Shift to 0 
    y=0                                         ''  Set Minutes Shift to 0 
    SELECT CASE INKEY$                          ''  Take keystroke input
        CASE z$+"K"                             ''  If is Left Arrow
            r=0                                 ''    Bool to modify right (minutes) 
        CASE z$+"M"                             ''  If is Right Arrow
            r=1                                 ''    Bool to modify left (hours)
        CASE z$+"H"                             ''  If is Up Arrow
            x=1                                 ''    Set Hours Shift to 1 
            y=1                                 ''    Set Minutes Shift to 1
        CASE z$+"P"                             ''  If is Down Arrow
            x=23                                ''    Set Hours Shift to 23 
            y=59                                ''    Set Minutes Shift to 23 
    END SELECT                                  ''
    IF r THEN m=(m+y)MOD 60ELSE h=(h+x)MOD 24   ''  Shift Minutes If `r=1` Else Shift Hours
    CLS                                         ''  Clear Screen
    ?RIGHT$("00000"+LTRIM$(STR$(h*1000+m)),5)   ''  Use math to concat Hours and Minutes 
                                                ''  then Convert to String and prepend 0s 
                                                ''  to a length of 5
    LOCATE 1,3                                  ''  Cursor to the the third digit
    ?":"                                        ''  Overwrite that digit with a `:`
LOOP                                            ''  Loop
  • 1
    \$\begingroup\$ Shouldn't that be (m+y)? \$\endgroup\$
    – Neil
    Commented Jul 18, 2018 at 15:15
  • \$\begingroup\$ In the note, should not does be do? \$\endgroup\$ Commented Jul 18, 2018 at 15:40
  • \$\begingroup\$ @JonathanFrech - Yep it should be. Thanks for keeping my grammar in check \$\endgroup\$ Commented Jul 18, 2018 at 16:26
  • \$\begingroup\$ Sorry, I thought m was for minutes for some reason... I see your commented version is more readable. \$\endgroup\$
    – Neil
    Commented Jul 18, 2018 at 18:49

Powershell, 109 103 bytes

-6 byte thanks AdmBorkBork

$args|%{$t[+$i]+=. @{l={$i=0};r={$i=1};u={1};d={119}}.$_}

Test script:

$f = {

$args|%{$t[+$i]+=. @{l={$i=0};r={$i=1};u={1};d={119}}.$_}


) | % {
    $e, $c = $_
    $r = &$f @c
    "$($r-eq$e): $r"


True: 02:57
True: 00:00
True: 23:59
True: 00:01
True: 23:59


Basic idea is to use a [hashtable], which keys are control commands and values are scriptblocks. The code execute the scriptblock for each command from arguments.

  • 1
    \$\begingroup\$ You can get rid of $i=0 by casting your array index like $t[+$i] to save some bytes. Try it online! \$\endgroup\$ Commented Jul 20, 2018 at 19:35

Perl 6, 101 91 89 86 bytes

{$/=[];$!=0;$_>2>($!=$_-3)||($/[$!]+=$_-1)for .ords X%5;($0%24,$1%60).fmt("%02d",":")}

Try it online!

Anonymous code block that takes a string of uldr characters and returns in the given format


perl -F// -E, 72 bytes


Python, 120 bytes

for s in list(input()):i=(i+(s=='r')-(s=='l')>=1);o[i]+=(s=='u')-(s=='d')
  • \$\begingroup\$ This looks like a snippet taking input in a variable. As a general rule, we require answers to present either a full program (taking input from program arguments or standard input) or a function (taking input from function parameters). \$\endgroup\$ Commented Jul 19, 2018 at 9:51
  • 1
    \$\begingroup\$ Also, won't this run into a wall when an input of, say, ld or rru causes i to leave the range (0,1) and o[i] is accessed afterwards? \$\endgroup\$ Commented Jul 19, 2018 at 11:49
  • \$\begingroup\$ @O.O.Balance oh thanks for reminding me that you need function or unput(). From requirements I figured that L and R actions will never loop around (i.e. no LL) \$\endgroup\$ Commented Jul 19, 2018 at 14:14
  • \$\begingroup\$ @aaaaaa No looping means lll isn't the same as r. Having ll or rr is valid input, it's also in the test-cases, see the third one for example. \$\endgroup\$
    – Etheryte
    Commented Jul 19, 2018 at 14:21
  • \$\begingroup\$ This answer currently has an IndexError on the 3rd test case instead of outputting 23:59. Try it online! \$\endgroup\$
    – 0 '
    Commented Jul 19, 2018 at 14:31

Haskell, 186 bytes

f t i('^':r)=f(i#t$1)i r
f t i('v':r)=f(i#t$ -1)i r
f t i(x:r)=f t x r
f(h,m)_ _=s h++':':s m
s n=['0'|n<10]++show n

Try it online!


R, 368 355 bytes

if(v=="L")z=1 else if(v=="R")z=0
if(v=="U")if(z)x=i(x==24,1,x+1)else y=i(y==60,1,y+1)
if(v=="D")if(z)x=i(x==1,24,x-1)else y=i(y==1,60,y-1)}}

Definitely not the best approach, but works.

Functionality: Run function, type each letter to (in/de)crease or move left/right, typing "s" ends the "game". The catch is that it will accept one and only one letter at a time.

-13 bytes Consolidated some values into one row, overwrote T as F instead of using break, found several spaces to eliminate, and a string stored in a variable instead

f=function(){C=as.character                             # Abbreviate functions
r=1:10                                                  # Initialize and format values
while(T){print(p(h[x],":",m[y]))                        # Begin while loop and print time
v=1                                                     # Initial value reset each iteration to retrieve a new direction
n="[UDLRS]"                                             # Used for verification and request
while(!grepl(n,v))v=toupper(readline(n))                # Will only accept proper directions or stopping rule
if(v=="L")z=1 else if(v=="R")z=0                        # Evaluate for hour or minute
if(v=="S")T=F                                           # Stopping rule, overwrite True to False
if(v=="U")if(z)x=i(x==24,1,x+1)else y=i(y==60,1,y+1)    # Rules for Up
if(v=="D")if(z)x=i(x==1,24,x-1)else y=i(y==1,60,y-1)}}  # Rules for Down

I am also editing an alternate format to accept an R string and/or vector, will post next week.


SmileBASIC, 123 bytes

B=BUTTON(2)D=(B==1)-(B==2)S=S+!S*(B>7)-S*(B==4)H=(H+D*!S+24)MOD 24WAIT
M=(M+D*S+60)MOD 60?FORMAT$("%02D:%02D",H,M)GOTO@L

BUTTON() returns an integer where each bit represents a button

1 = up
2 = down
4 = left
8 = right

BUTTON(2) returns only the buttons that were just pressed (not being held)

WAIT is required because BUTTON only updates once per frame (1/60 of a second). Otherwise the same button press would be detected multiple times.

This can definitely be shorter


05AB1E, 38 37 32 bytes


Uses udlr for the directions, but could also use ^v<> for the same byte-count (the characters ↑↓←→ are not part of 05AB1E's codepage, so using those would increase the byte-count by a lot, since the encoding should be changed).

Try it online or verify all test cases.


'l¡          '# Split the (implicit) input on "l"
              #  i.e. "lllrurulddd" → ["","","","ruru","ddd"]
     δ        # Map each item to:
   'r ¡      '#  Split the item on "r"
              #   → [[""],[""],[""],["","u","u"],["ddd"]]
       0ζ     # Zip/transpose; swapping rows/columns, with 0 as filler
              #  → [["","","","","ddd"],[0,0,0,"u",0],[0,0,0,"u",0]]
ć             # Head extracted: pop and push the remainder and head-item to the stack
              #  → [[0,0,0,"u",0],[0,0,0,"u",0]] and ["","","","","ddd"]
 ‚R           # Pair the head and remainder back together, and reverse the pair
              # (so the remainder portion is basically wrapped in a list now)
              #  → [["","","","","ddd"],[[0,0,0,"u",0],[0,0,0,"u",0]]]
   €S         # Convert each item to a list of characters
              # (implicitly flattens and removes empty strings)
              #  → [["d","d","d"],[0,0,0,"u",0,0,0,0,"u",0]]
      „0usk   # Index each character into the string "0u"
              # (0 remains 0; "u" becomes 1; "d" becomes -1 since it's not in this string)
              #  → [[-1,-1,-1],[0,0,0,1,0,0,0,0,1,0]]
           O  # Then take the sum of each inner list
              #  → [-3,2]
Ž9¦           # Push compressed integer 2460
   2ä         # Split into two parts: [24,60]
     %        # Modulo the two lists
              #  → [21,2]
      T‰      # Divmod each with 10
              #  → [[2,1],[0,2]]
        J     # Join each inner list together
              #  → ["21","02"]
         ':ý '# Join the list with ":" delimiter
              #  → "21:02"
              # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ž9¦ is 2460.


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