Crown hundred crownty crown

I noticed a certain game had a peculiar life counter, which instead of stopping at 999, gained a new digit – the next number was crown hundred or 👑00. After 👑99 came crown hundred crownty (👑👑0) and the last number, after 👑👑9, was crown hundred crownty crown or 👑👑👑, which would be 1110 in decimal.

Your task is to write a program or a function that outputs this counter.

Given an integer from the range [0,1110] (inclusive on both ends), output a three character string where

• every character is from the list 0123456789👑
• the crown (👑) can only appear as the leftmost character or when there's a crown to the left of it
• when this number is read as a decimal number but with the crown counting as 10, you get back the original number

Test cases

0 → "000"
15 → "015"
179 → "179"
999 → "999"
1000 → "👑00"
1097 → "👑97"
1100 → "👑👑0"
1108 → "👑👑8"
1110 → "👑👑👑"

You may use any non-decimal character instead of the crown. To encourage pretty printing, the crown character (UTF8 byte sequence "\240\159\145\145") counts as one byte instead of four. Your program doesn't have to work for numbers outside the valid range.

This is , so the shortest answer, measured in bytes, wins!

• Oh, Super Mario 3D Land! – Deusovi Sep 8 '18 at 9:38
• @Deusovi I was actually thinking about the follow-up game, Super Mario 3D World, but well guessed! – Angs Sep 8 '18 at 10:00
• This should be the IMO number for Boaty McBoatFace. – Mr Lister Sep 8 '18 at 11:19
• The bonus is multiplied by the number of crowns in the code, right? – Erik the Outgolfer Sep 8 '18 at 14:13
• @JeffZeitlin it's a redundant decimal system, where a number may have more than one representation (even disregarding leading zeroes). The crown is then reserved as a surprise element, only used when absolutely needed. – Angs Sep 8 '18 at 20:32

05AB1E, 20 18 bytes

₄‹i₄+¦ëTð.;„1 „  :

Uses spaces for crowns.

Explanation:

₄‹i               # If the (implicit) input is smaller than 1000:
₄+             #  Add 1000 to the (implicit) input
¦            #  And remove the leading 1 (so the leading zeros are added)
#   i.e. 17 → 1017 → "017"
ë               # Else:
Tð.;           #  Replace the first "10" with a space " "
#   i.e. 1010 → " 10"
#   i.e. 1101 → "1 1"
#   i.e. 1110 → "11 "
„1 „  :    #  Replace every "1 " with "  " (until it no longer changes)
#   i.e. " 10" → " 10"
#   i.e. "1 1" → "  1"
#   i.e. "11 " → "   "

JavaScript (Node.js), 50 bytes

f=(n,p=1e3)=>n<p?(n+p+'').slice(1):'#'+f(n-p,p/10)

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TIO was based on Arnauld's answer. Show 👑 as #.

• Nice recursive approach! – Arnauld Sep 8 '18 at 9:43

Shakespeare Programming Language, 763692690 689 bytes

,.Ajax,.Ford,.Page,.Act I:.Scene I:.[Enter Ajax and Ford]Ford:Listen tothy!Ajax:You big big cat.Scene V:.Ajax:Remember the remainder of the quotient between I twice the sum of a cat a big big cat.Ford:You be the quotient between you twice the sum of a cat a big big cat.Ajax:You be the sum of you a pig.Be you nicer zero?If solet usScene V.Ford:You big big cat.[Exit Ajax][Enter Page]Page:Recall.Ford:You be I.Scene X:.Page:Recall.Am I nicer zero?If notopen heart.If notlet usScene L.Ford:You big big big big big cat.Speak thy.Am I worse a cat?If soyou zero.Scene L:.[Exit Page][Enter Ajax]Ford:You be the sum of you a pig.Is you nicer a cat?[Exit Ajax][Enter Page]Ford:If solet usScene X.

Try it online!

Uses " " instead of crowns. At the cost of 4 more bytes, this could be modified to show a "visible" character instead.

Explanation:

,.Ajax,.Ford,.Page,.Act I:.Scene I:.[Enter Ajax and Ford]

Boilerplate, introducing the characters.

Ford:Listen tothy!

Input a value to Ajax.

Ajax:You big big cat.

Set Ford's value to 4 (we will be pushing 4 digits from Ajax onto Ford's personal stack).

Scene V:.Ajax:Remember the remainder of the quotient between I twice the sum of a cat a big big cat.Ford:You be the quotient between you twice the sum of a cat a big big cat.

DIGIT-PUSHING LOOP: Push Ajax's last digit onto Ford's stack; divide Ajax by 10.

Ajax:You be the sum of you a pig.Be you nicer zero?If solet usScene V.

Decrement Ford; loop until Ford is 0.

Ford:You big big cat.

Set Ajax's value to 4 (we will pop 3 digits from Ford's stack in the next loop).

[Exit Ajax][Enter Page]Page:Recall.Ford:You be I.

Pop the top value off Ford's stack, and store that into Page.
Here, Page will contain 0 if there are no crowns to be drawn,
and 1 if there are crowns to be drawn.

Scene X:.Page:Recall.Am I nicer zero?If notopen heart.If notlet usScene L.

DIGIT-DRAWING LOOP: Pop the top value off of Ford's stack and set Ford equal to that value.
If there are no crowns to be drawn, output Ford's literal value here, and skip the crown-drawing section.

Ford:You big big big big big cat.Speak thy.Am I worse a cat?If soyou zero.

Draw crown.
If we are drawing crowns, and Ford contains 0 here, then we are now done drawing crowns, and thus we store 0 into Page.
(Put in one more "big" for the crown to look like an @ symbol.)

Scene L:.[Exit Page][Enter Ajax]Ford:You be the sum of you a pig.Is you nicer a cat?[Exit Ajax][Enter Page]Ford:If solet usScene X.

Decrement Ajax; loop until Ajax is 1 (i.e. 3 times).

JavaScript (ES6), 62 bytes

Outputs crowns as x characters.

n=>(n+1e3+'').replace(/^21*0/,s=>s.replace(/./g,x)).slice(1)

Try it online!

How?

We add $1000$ to the input, coerce it to a string, look for the /^21*0/ pattern and replace each matching character with x. Finally, we remove the leading character.

Examples:

\begin{align}&0 &\rightarrow &\text{ "1000"} &\rightarrow &\text{ "1000"} &\rightarrow &\text{ "000"}\\ &123 &\rightarrow &\text{ "1123"} &\rightarrow &\text{ "1123"} &\rightarrow &\text{ "123"}\\ &1023 &\rightarrow &\text{ "}\color{red}{\text{20}}\text{23"} &\rightarrow &\text{ "xx23"} &\rightarrow &\text{ "x23"}\\ &1103 &\rightarrow &\text{ "}\color{red}{\text{210}}\text{3"} &\rightarrow &\text{ "xxx3"} &\rightarrow &\text{ "xx3"}\\ &1110 &\rightarrow &\text{ "}\color{red}{\text{2110}}\text{"} &\rightarrow &\text{ "xxxx"} &\rightarrow &\text{ "xxx"} \end{align}

• s.replace(/./g,#) is neat... I had Array(s.length+1).join#, and my regex was longer too! Nice work, +1 – Mr. Xcoder Sep 8 '18 at 9:23

Python 2, 75 bytes

Shamelessly steals Arnauld's regex. Uses C as crown.

lambda k:re.sub("^21*0",lambda g:len(g.group())*"C",k+1000)[1:]
import re

Try it online!

Python 2, 51 bytes

This instead implements tsh's recursive method. Saved 2 bytes thanks to ovs.

f=lambda n,p=1000:n/p and'C'+f(n-p,p/10)orn+p[1:]

Try it online!

Retina 0.8.2, 41 bytes

\b((.)|..)\b
$#2$*00$& Td_#(?=....)1+0 Try it online! Uses #s instead of 👑s. Link includes test cases. Explanation: \b((.)|..)\b$#2$*00$&

Pad 1- and 2-digit numbers to three digits.

Td_#(?=....)1+0

Change leading 1s of 4-digit numbers to #s and delete the next 0.

Clean, 87 bytes

Doesn't output crowns (uses c).

import StdEnv,Text
$n#i=3-n/1000-n/1100-n/1110 =("ccc"+lpad(""<+n rem(10^i))i'0')%(i,9) Try it online!$ n                   // function $of n # i = // define i as (the number of digits that aren't crowns) 3 - // three minus n / 1000 - // 1 if first digit is crown n / 1100 - // 1 if second digit is crown n / 1110 // 1 if third digit is crown = ( // the string formed by "ccc" + // prefixing three crowns to lpad ( // the padding of "" <+ n rem (10^i) // non-crown digits of n ) i '0' // with zeroes ) % (i, 9) // and removing the extra crowns Clean, 99 - 3 = 96 bytes This one has crowns. import StdEnv,Text$n#i=3-n/1000-n/1100-n/1110

Try it online!

• The second answer has a score of only 90. – pppery Jul 14 at 0:50

Jelly, 19 bytes - 0 = 19

<ȷ¬ȧDi0ɓ⁶ẋ⁹Ḋ;+ȷDḊṫʋ

A full program printing the result using a space character as the crown.
(As a monadic Link a mixed list of integer digits and space characters is yielded)

Try it online! Or see the test-suite.

...maybe a recursive implementation will be shorter.

How?

<ȷ¬ȧDi0ɓ⁶ẋ⁹Ḋ;+ȷDḊṫʋ - Main Link: integer, N    e.g. 1010       or   10
ȷ                  - literal 1000                  1000            1000
<                   - less than?                    0               1
¬                 - logical not                   1               0
D               - to decimal list               [1,0,1,0]       [1,0]
ȧ                - logical and                   [1,0,1,0]       0
0             - literal zero                  0               0
i              - first index - call this I     2               1  (0 treated as  by i)
ɓ            - new dyadic chain with swapped arguments - i.e. f(N, I)
⁶           - literal space character       ' '             ' '
⁹         - chain's right argument        2               1
ẋ          - repeat                        [' ',' ']       [' ']
Ḋ        - dequeue                       [' ']           []
ʋ - last four links as a dyad - i.e. f(N, I)
+ȷ     -   add 1000                    2010            1010
D    -   to decimal list             [2,0,1,0]       [1,0,1,0]
Ḋ   -   dequeue                     [0,1,0]         [0,1,0]
ṫ  -   tail from index (I)         [1,0]           [0,1,0]
;       - concatenate                   [' ',1,0]       [0,1,0]
- implicit print                " 10"           "010"

Python 2, 52 bytes

lambda n:['%03d'%n,'%3s'%n.lstrip('1')[1:]][n>999]

Try it online!

Japt, 20 bytes

A naïve (and slightly drunk!) port of Arnauld's solution. Uses " for crown.

U+A³ s r"^21*0"_çQÃÅ

Try it

Java 10, 84 83 bytes

n->{for(int p=100,t;p>0;n-=t%12*p,p/=10)System.out.printf("%c",t=n/p>9?46:48+n/p);}

Port of @tsh' C comment.
Uses . instead of crowns.

Try it online.

Alternative approach (84 (87-3) bytes):

n->f(n,1000)String f(int n,int p){return n<p?(n+p+"").substring(1):"👑"+f(n-p,p/10);}

Port of @tsh' JavaScript's answer.

Try it online.

APL (Dyalog Unicode), 32 bytes

1e3∘{⍵<⍺:1↓⍕⍵+⍺⋄'C',(⍵-⍺)∇⍨⍺÷10}

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Prefix direct function.

Port of @tsh's JS Answer.

How:

1e3∘{⍵<⍺:1↓⍕⍵+⍺⋄'C',(⍵-⍺)∇⍨⍺÷10} ⍝ Main function, arguments ⍵ and ⍺ (⍵ → input, ⍺ → 1000).
⍵<⍺:                          ⍝ If ⍵<⍺
1↓                        ⍝ Drop (↓) the first element (1) of
⍕                       ⍝ Format (⍕); 'stringify'
⍵+⍺                    ⍝ ⍵+⍺
⋄                   ⍝ else
'C',               ⍝ Concatenate (,) the literal 'C' with
∇⍨        ⍝ Recursive call (∇) with swapped arguments (⍨)
(⍵-⍺)  ⍺÷10    ⍝ New arguments; ⍵ → ⍵-⍺; ⍺ → ⍺÷10