# Goal

Write a program or function that translates a numerical telephone number into text that makes it easy to say. When digits are repeated, they should be read as "double n" or "triple n".

# Requirements

## Input

A string of digits.

• Assume all characters are digits from 0 to 9.
• Assume the string contains at least one character.

## Output

Words, separated by spaces, of how these digits can be read out loud.

• Translate digits to words:

0 "oh"
1 "one"
2 "two"
3 "three"
4 "four"
5 "five"
6 "six"
7 "seven"
8 "eight"
9 "nine"

• When the same digit is repeated twice in a row, write "double number".

• When the same digit is repeated thrice in a row, write "triple number".
• When the same digit is repeated four or more times, write "double number" for the first two digits and evaluate the rest of the string.
• There is exactly one space character between each word. A single leading or trailing space is acceptable.
• Output is not case sensitive.

# Scoring

Source code with the least bytes.

# Test Cases

input        output
-------------------
0123         oh one two three
4554554      four double five four double five four
000          triple oh
00000        double oh triple oh
66667888     double six double six seven triple eight
19999999179  one double nine double nine triple nine one seven nine

• Anyone interested in "speech golf" should note that "double six" takes longer to say than "six six". Of all the numerical possibilities here, only "triple seven" saves syllables. – Purple P Aug 13 '19 at 4:40
• @Purple P: And as I'm sure you know, 'double-u double-u double-u'>'world wide web'.. – Chas Brown Aug 13 '19 at 4:43
• I vote to change that letter to "dub". – Hand-E-Food Aug 13 '19 at 4:45
• I know this is only an intellectual exercise, but I have in front of me a gas bill with the number 0800 048 1000, and I would read that as "oh eight hundred oh four eight one thousand". The grouping of digits is significant to human readers, and some patterns such as "0800" are treated specially. – Michael Kay Aug 14 '19 at 8:29
• @PurpleP Anyone interested in clarity of speech, however, especially when speaking over the phone, might want to use "double 6" since it's clearer that the speaker means two sixes and didn't repeat the number 6 accidentally. People aren't robots :P – user91988 Aug 14 '19 at 20:47

# 05AB1E, 53525150 49 bytes

γε€T2äθ¬MÊi¨₃1ǝR]˜“Šç€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š‹¶½¿“#sè¸»


Try it online!

Explanation:

γ                      # split input in groups of consecutive equal digits
ε              ]      # for each group
€T                   #  add a 10 before each digit (66 -> [10, 6, 10, 6])
2äθ                #  keep only the second half of that list
¬MÊi     ]      #  if the first element is not the maximum
¨           #   drop the last element
₃1ǝ        #   replace the second element with 95
R       #   reverse the list
˜                      # flatten
“...“                 # compressed string: "oh one two ... nine double triple"
#                # split on spaces
sè              # index (wraps around, so 95 yields "triple")
¸»            # join with spaces

• Oh, M also looks inside lists when determining the maximum integer on the stack? Didn't knew that. Sounds like something to remember. :) – Kevin Cruijssen Aug 14 '19 at 6:13

# 8088 Assembly, IBM PC DOS, 164159156 155 bytes

Binary:

00000000: d1ee 8a0c 03f1 53fd ac3a d075 0343 e2f7  ......S..:.u.C..
00000010: 85db 741c 5f8a d043 f6c3 0174 0a57 bd64  ..t._..C...t.W.d
00000020: 0155 83eb 0374 0957 bd5d 0155 4b4b 75f7  .U...t.W.].UKKu.
00000030: 8ad0 2c2f 7213 518a f0b0 24b1 31bf 6a01  ..,/r.Q...$.1.j. 00000040: fcf2 aefe ce75 fa59 57e2 bc5a 85d2 740c .....u.YW..Z..t. 00000050: b409 cd21 b220 b402 cd21 ebef c364 6f75 ...!. ...!...dou 00000060: 626c 6524 7472 6970 6c65 246f 6824 6f6e ble$triple$oh$on
00000070: 6524 7477 6f24 7468 7265 6524 666f 7572  e$two$three$four 00000080: 2466 6976 6524 7369 7824 7365 7665 6e24$five$six$seven$00000090: 6569 6768 7424 6e69 6e65 24 eight$nine$ Build and test executable using xxd -r from above, or download PHONE.COM. Unassembled listing: D1 EE SHR SI, 1 ; point SI to DOS PSP (80H) for input string 8A 0C MOV CL, BYTE PTR[SI] ; load input string length into CX 03 F1 ADD SI, CX ; move SI to end of input 53 PUSH BX ; push a 0 to signal end of output stack CHAR_LOOP: FD STD ; set LODS direction to reverse AC LODSB ; load next char from [SI] into AL, advance SI 3A D0 CMP DL, AL ; is it same as previous char? 75 03 JNZ NEW_CHAR ; if not, it's a different char 43 INC BX ; otherwise it's a run, so increment run length E2 F7 LOOP CHAR_LOOP ; move on to next char NEW_CHAR: 85 DB TEST BX, BX ; is there a run greater than 0? 74 1C JZ GET_WORD ; if not, look up digit name 5F POP DI ; get name for the current digit 8A D0 MOV DL, AL ; save current char in DL 43 INC BX ; adjust run count (BX=1 means run of 2, etc) F6 C3 01 TEST BL, 1 ; is odd? if so, it's a triple 74 0A JZ IS_DBL ; is even, so is a double 57 PUSH DI ; push number string ("one", etc) to stack BD 0164 MOV BP, OFFSET T ; load "triple" string 55 PUSH BP ; push to stack 83 EB 03 SUB BX, 3 ; decrement run count by 3 74 09 JZ GET_WORD ; if end of run, move to next input char IS_DBL: 57 PUSH DI ; push number string to stack BD 015D MOV BP, OFFSET D ; load "double" string 55 PUSH BP ; push to stack 4B DEC BX ; decrement by 2 4B DEC BX 75 F7 JNZ IS_DBL ; if not end of run, loop double again GET_WORD: 8A D0 MOV DL, AL ; save current char into DL 2C 2F SUB AL, '0'-1 ; convert ASCII char to 1-based index 72 13 JB NOT_FOUND ; if not a valid char, move to next 51 PUSH CX ; save outer loop counter 8A F0 MOV DH, AL ; DH is the index to find, use as scan loop counter B0 24 MOV AL, '$'            ; word string is $delimited B1 31 MOV CL, 031H ; search through length of word data (49 bytes) BF 016A MOV DI, OFFSET W ; reset word data pointer to beginning FC CLD ; set DF to scan forward for SCAS SCAN_LOOP: F2/ AE REPNZ SCASB ; search until delimiter '$' is found in [DI]
FE CE       DEC  DH                 ; delimiter found, decrement counter
75 FA       JNZ  SCAN_LOOP          ; if counter reached 0, index has been found
59          POP  CX                 ; restore outer loop position
57          PUSH DI                 ; push string on stack
NOT_FOUND:
E2 BC       LOOP CHAR_LOOP          ; move to next char in input
OUTPUT_STACK:
5A          POP  DX                 ; get string from top of stack
85 D2       TEST DX, DX             ; it is the last?
74 0C       JZ   EXIT               ; if so, exit
B4 09       MOV  AH, 09H            ; DOS display string function
CD 21       INT  21H                ; write string to console
B2 20       MOV  DL, ' '            ; load space delimiter
B4 02       MOV  AH, 02H            ; DOS display char function
CD 21       INT  21H                ; write char to console
EB EF       JMP  OUTPUT_STACK       ; continue looping
EXIT:

D   DB "double$" T DB "triple" W DB "$oh$","one$","two$","three$","four$","five$","six$","seven$","eight$","nine$"


TL;DR:

The input string is read right to left to make it easier to find a triple. The output is pushed onto the x86 stack to simplify reversing out the display order and also facilitate rearranging the "double" and "triple" words to precede the digit's name.

If the next digit is different than the last, the name is looked up in the list of words and pushed on to the stack. Since there's no formal concept of an "indexed array of variable-length strings" in machine code, the list of words is scanned i (the word's index) number of times for the string delimiter ($) to find the corresponding word. Helpfully, x86 does have a pair of short instructions (REPNZ SCASB which is similar to memchr() in C), that simplifies this (thanks CISC!). If the digit is the same as the previous one, the counter for the length of a "run" is incremented and continues looping leftward on the input. Once the run is over, the digit's name is taken from the stack since it will need to placed after the "double" or "triple" for each grouping. If the length of the run is odd (and run length is > 1), the digit's name followed by the string "triple" is pushed to the stack and the run length is reduced by 3. Since the run length will now be even, the step is repeated for "double" until the run length is 0. When the input string has reached the end, the stack is dumped out with each saved string written to the screen in reverse order. I/O: A standalone PC DOS executable, input from command line output to console. Download and test PHONE.COM. • repne scasb is memchr (or strchr if you know there's going to be a hit), not strstr. – Peter Cordes Aug 15 '19 at 23:49 • Is CH=0 on process entry guaranteed by a standard, or is that just what some DOS version happens to do? I notice you assume that mov cl, byte[si] is equivalent to movzx cx, byte [si]. I wonder if using a different reg, like AH, for the count with dec ah / jnz instead of loop would save anything from not having to push/pop CX. Probably not, and you don't have any 16-bit regs left that would allow 1-byte dec. – Peter Cordes Aug 15 '19 at 23:56 • @PeterCordes, for the CH=0 I go by fysnet.net/yourhelp.htm, which for any reasonable release of DOS is always zero'd, same with BX. Good thought about the zero-extend mov, though technically I don't think movzx is available on the 808x (keeping the target platform an IBM PC 5150 and all). I fiddled around with all of the registers as best I could to save the bytes, but if you see something I likely missed, please let me know! – 640KB Aug 16 '19 at 0:04 • It's more accurate to call it memchr IMO. The "string instruction" naming misleads people into thinking they work on implicit-length C strings, but really they work on explicit-length strings like std::string or buffers. Like memcpy, memset (movs/stos), memchr/memrchr (repne scas with DF=0 or 1), and memcmp (repe cmps). The only C equivalent for repe scas is strspn because I don't think there's a mem function for that. You can even describe stosw or stosd as wmemset for example. – Peter Cordes Aug 16 '19 at 0:06 • movzx costs an extra opcode byte, and yes it was only introduced with 386. It was just easier to type to describe the fact that you're doing a low-byte-merge and assuming it's correctly zero extended. If you do know CX or at least CH=0, then yes for golfing always go with mov to CL. But outside of golfing, x86's go-to byte-load instructions are movzx and movsx: they avoid any false dependencies or other partial-register shenanigans. On modern CPUs with a dword destination, they're as fast as dword mov loads. – Peter Cordes Aug 16 '19 at 0:09 # 05AB1E, 61565352 51 bytes γvyDg;LàäRv… ‹¶½¿#ygè“Šç€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š“#yè])áðý  -9 bytes thanks to @Grimy. Explanation: γ # Split the (implicit) input into substrings of equal adjacent characters # i.e. "199999991779" → ["1","9999999","1","77","9"] v # Loop over each substring y: Dg # Get the length of a copy of the substring ; # Halve it L # Create a list in the range [1, length/2], where odd lengths are # automatically truncated/floored # i.e. "1" (length=1) → 0.5 → [1,0] # i.e. "9999999" (length=7) → 3.5 → [1,2,3] à # Pop and push the maximum of this list y ä # Divide the string into that many parts # → ["1"] # → ["999","99","99"] R # Reverse the list # → ["99","99","999"] v # Inner loop over each item y: … ‹¶½¿ # Push dictionary word: " double triple" # # Split it on spaces: ["","","double","triple"] yg # Get the length of the current item y è # And use it to (0-based) index into the list “Šç€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š“ # Push dictionary string "oh two three four five six seven eight nine" # # Split it on spaces: ["oh","two","three",...,"nine"] yè # Use y to index into the string-list (with automatic wrap-around, # so since there are 10 words, it basically indexes with a single digit # due to an implicit modulo-10) # i.e. "77" → "seven" ] # Close both loops ) # Wrap all values on the stack into a list á # Only keep letters, which removes the empty strings from the list ðý # And join the list on spaces # (after which the result is output implicitly)  See this 05AB1E tip of mine (section How to use the dictionary?) to understand why … ‹¶½¿ is " double triple" and “Šç€µ‚•„í†ìˆÈŒšï¿Ÿ¯¥Š“ is "oh two three four five six seven eight nine". • @Grimy Ah, of course.. I added the if(length>=4) before adding the rest, but of course it isn't necessary for integers of size 1,2,3, because the ;Å2¨3ª£ will leave the strings intact (merely wrapped in a list we flatten after the map anyway). Thanks for noticing! And looking forward seeing your answer with Åγ. I indeed had the feeling the first part could be done a lot shorter somehow. – Kevin Cruijssen Aug 13 '19 at 11:49 • Dg;LàäR is still a byte shorter than āÉ¨Å¡JõK, and much more similar to what you originally had. – Grimmy Aug 13 '19 at 15:02 • @Grimy Ah, that's indeed closed to what I initially had indeed, I like it. :) Thanks again! – Kevin Cruijssen Aug 13 '19 at 15:27 • @Grimy Been able to find one more golf myself which I forgot about.. á instead of õK at the end. :) – Kevin Cruijssen Aug 13 '19 at 15:33 • Nice find with á! Here's a 51 and another one. 50 feels possible. – Grimmy Aug 13 '19 at 16:06 # QuadR, 137 bytesSBCS Title case with a leading space. ∊¯2↑¨@(∊∘⎕A)⍵ (.)\1* {⍺←(,¨⎕D)⎕R('OhOneTwoThreeFourFiveSixSevenEightNine'(∊⊂⊣)⎕A)⋄' '∊w←⍺,⊃⍵:⍬⋄1=≢⍵:⍺⍵⋄3=≢⍵:'Triple',w⋄'Double',w,∇2↓⍵}⍵M  Try it online! ∊ϵnlist (flatten) ¯2↑¨ take the last two characters (padding on left with a space) of each characters @ at positions where (∊∘⎕A) characters are members of the uppercase Alphabet ⍵ in the result of the below PCRE Replace operation… (.) any character \1 followed by itself * zero or more times, is replaced by the result of the following… {…}⍵M "dfn"; ⍵ is the Match of the above pattern ('OhOneTwoThreeFourFiveSixSevenEightNine'()⎕A) apply the following anonymous tacit function with the long string and the uppercase Alphabet as left arguments: ∊ membership (of letters in the long string in the uppercase alphabet) ⊂ partitions (with a new partition beginning whenever is-a-member ⊣ the left argument (i.e. the long string) ()⎕R PCRE Replace the following patterns with those words: ⎕D the digits 0 through 9 ,¨ treat each as a separate pattern ⍺← assign this replacement function to ⍺ (for alphabetise) ⋄ then, ⊃⍵ the first character of the match , as a string ⍺ apply ⍺ to it w← assign this to w (for word) ' '∊: if space is a member thereof (i.e. if the match was empty): ⍬ return nothing (becomes the empty string) ⋄ else, 1=≢⍵: if one equals the tally of characters in the match (i.e. its length): ⍺⍵ alphabetise that digit ⋄ else, 3=≢⍵: if three equals the tally of characters in the match (i.e. its length): 'Triple',w prepend "Triple" to the word ⋄ else, 2↓⍵ drop to digits from the match ∇ recurse on that w, prepend the word 'Double', prepend "Double" # JavaScript (ES6), 161 160 152 144 bytes The output includes a single leading space. s=>[n=>' '+'oh/one/two/three/four/five/six/seven/eight/nine'.split/[n],'$1 double$2','triple$2'].map(r=>s=s.replace(/(\S*)( \S+)\2|\d/g,r))&&s


Try it online!

## How?

The conversion is processed in three steps:

1. replace each digit with the corresponding English word, preceded by a space
2. replace each pattern "X X" with "double X"
3. replace each pattern "double X X" with "triple X"

To save bytes, we use the same regular expression for all steps:

/(\S*)( \S+)\2|\d/g


which works as follows:

(\S*)  -> 1st capturing group: any word, or nothing at all
( \S+) -> 2nd capturing group: a space, followed by a word
\2     -> a copy of the 2nd capturing group
|\d    -> or try to capture a digit instead (for step 1)


At step 1, we use a callback function that picks the correct word from a lookup table:

• "799999"" seven nine nine nine nine nine"

At step 2, we replace with "$1 double$2":

• " (seven)( nine)( nine)"" seven double nine"
• "( nine)( nine) nine"" double nine nine"

At step 3, we replace with "triple$2": • " (double)( nine)( nine)"" triple nine" # Wolfram Language (Mathematica), 115 bytes {Switch[L=Tr[1^{##}],1," ",3," triple ",_," double "],If[#<1,"oh",IntegerName@#],If[L>3,#0@##3,""]}&@@@Split@#<>""&  Try it online! Takes a list of digits as input. Output includes a leading space. # Stax, 56 bytes ÇÖ◘⌡¿╒ô╞Γ▓8m☻t7♦3├(Ä⌂≤(┬Ω☻9▲ç╕C╞⌡òσ╗─╣╥─☻╪▼⌡5■ÿ,(┬¥?☺÷•±  Run and debug it # Python 2, 171169 168 bytes s=input() while s:c=s[0];n=(s[1:2]==c)+(s[:3]==c*3!=s[1:4]);print' eellpbiurotd'[-n:0:-2]+'oh one two three four five six seven eight nine'.split()[int(c)],;s=s[1+n:]  Try it online! -1 byte, thanks to Jitse • Beat me again! Save 1 byte like so – Jitse Aug 13 '19 at 9:14 • @Jitse, that doesn't work for 1312 ;) – TFeld Aug 13 '19 at 9:43 • Ah, you're right! – Jitse Aug 13 '19 at 9:45 • How about this one then: ['','double ','triple '][n] to ' eellpbiurotd'[-n:0:-2] for 168 bytes: Try it online! – Jitse Aug 13 '19 at 9:52 • Alternatively, also 168 bytes – Jitse Aug 13 '19 at 10:13 # Perl 5-p, 111 bytes s/(\d)\1/ double$1/g;s/\w+(\d)\1/triple$1/g;s/\d/' '.qw(oh one two three four five six seven eigth nine)[$&]/ge


Try it online!

Explanation:

s/(\d)\1/ double$1/g; # Replace non-overlapping double digits with " double<digit>" s/\w+(\d)\1/triple$1/g; # Replace remaining double digits preceded by "double" with "triple<digit>"
s/\d/' '.qw(oh one two three four five six seven eigth nine)[$&]/ge # Replace digits with " <word>"  • Shaved a few bytes off: 106 – Xcali Aug 14 '19 at 4:43 # Scala, 213 bytes Got it. Somehow the recursive version I was trying to build was strongly more verbose-y than this one (still recursive though, but only in one case). Function f takes as an input string the phone number and outputs its phonetics with a trailing whitespace. var u="oh one two three four five six seven eight nine" split " " "(.)\\1*".r.replaceAllIn(s,x=>{var o=x.matched var k=u(o(0)-48)+" " o.length match{case 3=>"triple "+k case 1=>k case _=>"double "+k+f(o drop 2)}})  Try it online! Edit : -8b thanks to DrY Wit! # Scala, 215 bytes And here comes the leading whitespace version, two bytes longer for some reason (even with massive refactoring). var u="oh one two three four five six seven eight nine" split " " "(.)\\1*".r.replaceAllIn(s,x=>{var o=x.matched var k=u(o(0)-48) " , double , triple ".split(",")(if(o.length>3){k+=f(o drop 2);1}else o.length-1)+k})  Try it online! • You can save 8 bytes by replacing (o(0)+"").toInt with o(0)-48. – Dr Y Wit Aug 14 '19 at 3:40 • Oof well done @DrYWit thanks! – V. Courtois Aug 14 '19 at 6:28 # PHP, 174169166 159 bytes for(;$s=strspn($argn,$d=$argn[$i],$i++);$s==3?($i+=2)+print'triple ':$s<2?:++$i+print'double ',print[oh,one,two,three,four,five,six,seven,eight,nine][$d].' ');


Try it online!

For each digit at index of $i starting from 0: • If span of the same digit starting from location of $i is equal to 3, prints 'triple ' and adds 2 to $i so next iteration will have 2 digits jumped over. • If span of the same digit starting from location of $i is equal to or more than 2 but is not 3, prints 'double ' and adds 1 to $i so next iteration will have 1 digit jumped over. • Prints word for the digit and a space. • $i++.

# Retina 0.8.2, 105 bytes

+(.)\1
=$1 .$&
= =
triple
=
double
9
nine
8
eight
7
seven
6
six
5
five
4
four
3
three
2
two
1
one
0
oh


Try it online! Outputs a leading space. Explanation: I originally tried a regex that automatically matches 2 or 3 digits but @Arnauld's approach of turned out to be golfier. Explanation:

+(.)\1
=$1  Match pairs of identical digits and replace the first with a =. Then repeat, so that for an odd number the second last digit is also replaced with a =. .$&


Space the digits (and =s) out.

= =
triple


Handle the case of three identical digits.

=
double
9
nine
8
eight
7
seven
6
six
5
five
4
four
3
three
2
two
1
one
0
oh


Replace all remaining characters with words.

⁵;,0;$Ɗ€Ẏ;Ø0ṭ;-œṣjƭƒV€‘$ị“¡ıc⁴Ṛ⁽]@ɱ2¦Ż©Ẉḷ$Æ!)ɗ[ı8ɱḃ%ċ»Ḳ¤K  Try it online! A monadic link that takes a string of digit characters as its argument and returns a Jelly string of space-separated words. When called as a full-program, outputs implicitly. # T-SQL 2017, 238 bytes Added some linebreaks to make it readable WHILE''<left(@,1)SELECT @=stuff(@,1,iif(p<4,p,2),'')+ iif(p=1,' ',iif(p=3,' triple ',' double ')) +trim(substring('oh one two threefour five six seveneightnine',left(@,1)*5,5)) FROM(SELECT~-patindex('%[^'+left(@,1)+']%'+'^',@)p)z PRINT @  Try it online # C++, 382 bytes It's not superclever, but someone needed to write a C++ version. The recursive function R goes through the input string and counts repeated values. If there are more than 3 repeats, it pretends there were 2 repeats, then rewinds and tries again. A few more source chars could probably be squeezed out with #define major, but I'm sure a better algo could squeeze out more. #include <iostream> #include <sstream> using namespace std; char*n[]={"oh","one","two","three","four","five","six","seven","eight","nine"}; void R(ostream& s,const char*r,char p='x',int c=0){if(*r==p)R(s,r+1,p,c+1);else {if(c>1){if(c>= 4){s<<"double ";r-=(c-2);}else if(c==3)s<< "triple ";else if(c==2)s<< "double "; }if(c >0)s<<n[p-'0']<<" ";if(!*r)return;R(s,r+1,*r,1);}} void check(const char* in, const char* out) { std::stringstream ss; R(ss,in); if (out == ss.str()) std::cout << "PASS: "; else std::cout << "FAIL! "; std::cout << in << "\n< " << out << "\n> " << ss.str() << std::endl; } int main(int c,char**argv) { if (argv[1] == std::string("test")) { check("0123" ,"oh one two three "); check("4554554" ,"four double five four double five four "); check("000" ,"triple oh "); check("00000" ,"double oh triple oh "); check("66667888" ,"double six double six seven triple eight "); check("19999999179" ,"one double nine double nine triple nine one seven nine "); } else { char* v = argv[1]; R(std::cout,v); std::cout << std::endl; } }  and verification of test cases: pa-dev01$ ./a.out test
PASS: 0123
< oh one two three
> oh one two three
PASS: 4554554
< four double five four double five four
> four double five four double five four
PASS: 000
< triple oh
> triple oh
PASS: 00000
< double oh triple oh
> double oh triple oh
PASS: 66667888
< double six double six seven triple eight
> double six double six seven triple eight
PASS: 19999999179
< one double nine double nine triple nine one seven nine
> one double nine double nine triple nine one seven nine

• Does the golfed part actually need #include <sstream>? Or could you move that down after the golfed part for the test function? I think typing std::ostream&s would take less space than using namespace std;, unless there are other places where you'd need a std::. – Peter Cordes Aug 16 '19 at 0:35
• 253 bytes – ceilingcat Sep 6 '19 at 22:35

{S:g/(.)$0?[$0{}<!$0>]?/{(<triple double>X$)[3-$/.comb]}{+$/??uniname(~$0).words[1]!!'oh'} /}  Try it online! This is an anonymous code block that takes a number and returns a string with the numbers in uppercase, e.g. 0123 => oh ONE TWO THREE with a single trailing space. This was deleted for a while until I found out how to use captures in a lookahead, but it should be fixed now. # Red, 242 bytes func[s][b:[copy t skip t]parse s[any[change[b t ahead not t](rejoin["triple "t])| change b(rejoin["double "t])| skip]]foreach c s[prin either i: find"0123456789"c[rejoin[pick[:oh:one:two:three:four:five:six:seven:eight:nine]index? i" "]][c]]]  Try it online! # Scala, 253 bytes def g(s:String):String={val t="oh one two three four five six seven eight nine".split(" ")(s(0)-48) s.length match{case 3=>"triple "+t;case 2=>"double "+t;case 1=>t;case _=>"double "+t+" "+g(s drop 2)}} s=>"(.)\\1*".r.findAllIn(s).map(g(_)) mkString " "  Try it online! # Oracle SQL, 578 bytes (in formatted form) Solution is not concise by any means so posting it in formatted way. with r(s) as (select x from t union all select case when length(regexp_substr(s, '(.)(\1)+')) = 3 then regexp_replace(s, '^...') else regexp_replace(s, '^(.)\1|^.') end from r where s is not null) select listagg(decode(length(r), 2, 'double ', 3, 'triple ') || decode(substr(r, 1, 1), 0, 'oh', to_char(to_date(substr(r, 1, 1), 'j'), 'jsp')), ' ') within group (order by rownum) from (select regexp_replace(s, lag(s) over (order by length(s)) || '$') r
from r order by length(s) desc);


Test in SQL*Plus

SQL> create table t(x) as select /*'45547777777774'*/ '1999999910079' from dual;

Table created.

SQL> set pages 0
SQL> with r(s) as
2  (select x from t
3    union all
4   select case
5             when length(regexp_substr(s, '(.)(\1)+')) = 3
6             then regexp_replace(s, '^...')
7             else regexp_replace(s, '^(.)\1|^.')
8          end
9     from r
10    where s is not null)
11  select listagg(decode(length(r),  2, 'double ',  3, 'triple ') ||
12                 decode(substr(r, 1, 1), 0, 'oh', to_char(to_date(substr(r, 1, 1), 'j'), 'jsp')), ' ')
13                 within group (order by rownum)
14    from (select regexp_replace(s, lag(s) over (order by length(s)) || '\$') r
15            from r order by length(s) desc);
one double nine double nine triple nine one double oh seven nine


The main trick is that digits converted into words using Oracle format models instead of hard-coded literals "one" ... "nine".

• any chance of this getting golfed ? it seems you can remove a bunch of spaces. I also imagine you can rewrite WHERE s is not null to WHERE s>'' – t-clausen.dk Aug 14 '19 at 14:04
• You can save a few characters by replacing what’s after union all with select regexp_replace(s,case when length(regexp_substr(s, '(.)(\1)+')) = 3 then '^...' else '^(.)\1|^.' end) from r. – Steve Kass Aug 15 '19 at 16:41

# JavaScript, 142 bytes

s=>s.replace(/(.)(\1\1(?!\1)|\1|)/g,t=>(t[2]?' triple ':t[1]?' double ':' ')+'oh one two three four five six seven eight nine'.split [t[0]])


Try it online!

# (Roblox) Lua 5.1, 166 bytes

for I,N in('111 triple 11 double 1 '):gmatch'(%d+)(%D+)'do for i,n in('0oh1one2two3three4four5five6six7seven8eight9nine'):gmatch'(.)(%l+)'do s=s:gsub(i*I,N..n)end end


Ensure s is a predefined string value populated only with digits; that will be the variable to be modified. The result will include a leading space [\u20] character.

• Welcome to the site! As Lua can take input via standard methods, it's against the rules to require s to already have the input. Aside from that, you've got a good first post! I would recommend you include a link to an online testing site e.g. tio.run/#lua so that others can test your solution – caird coinheringaahing Aug 13 '19 at 9:30
• Hi. The variant of Lua I was testing on (Rbx.Lua) doesn't contain input methods, although the sandbox has print, warn, and error output methods. – VisualPlugin Rōblox Aug 14 '19 at 17:36