# Challenge:

Given a list of integer, sort descending by their single largest digit(s). The order for numbers with the same largest digit are then sorted by second largest digit, etc.
We ignore duplicated digits in numbers. And if all digits in a number are the same, the order of those numbers in the list can be in any way you'd like.

Example:

Input:            [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]

Why? Here are the relevant digits the numbers were sorted on:

Output:
[8491,  -904,  62778,   478,     -7738,   6458,  373,   -73,   3120,      123,     0  ]

Relevant digits they were sorted on:
[[9,8], [9,4], [8,7,6], [8,7,4], [8,7,3], [8,6], [7,3], [7,3], [3,2,1,0], [3,2,1], [0]]

## Challenge rules:

• We ignore duplicated digits, so 478 and -7738 will be ordered as 478, -7738, because the largest digits are [8,7,4] and [8,7,3], and not [8,7,4] and [8,7,7,3].
• If multiple numbers have the same digits, the order of those can be either way. So 373 and -73 can be sorted as both 373, -73 or -73, 373 (digits are [7,3] for both of these numbers).
• If a number contains no more digits to check, it will be placed at the back of the relevant numbers. So 123 and 3120 will be sorted as 3120, 123, because the largest digits [3,2,1] are the same, but 0 comes before none.
• You can assume all numbers in the input are in the range [-999999,999999].
• Just one of the possible outputs is enough as result, but you are allowed to output all possible outputs where sublists can be in any permutation if you want (although I doubt it would save bytes in any language).

## General rules:

• This is , so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.

## Test cases:

Input:            [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]

Input:            [11, -312, 902, 23, 321, 2132, 34202, -34, -382]
Possible outputs: [902, -382, 34202, -34, -312, 321, 2132, 23, 11]
[902, -382, 34202, -34, 2132, -312, 321, 23, 11]
etc. The sublist [-312, 321, 2132] can be in any permutation

Input:            [9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0]
Possible outputs: [29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0]
[29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0]
etc. The sublists [4, 44] and [2212, 21] can be in any permutation

Input:            [44, -88, 9, 233, -3, 14, 101, 77, 555, 67]
Output:           [9, -88, 67, 77, 555, 14, 44, 233, -3, 101]

# 05AB1E, 5 bytes

ΣêR}R

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Explanation

Σ  }    # sort input by
ê      # its sorted unique characters
R     # reversed (to sort descending)
R   # reverse the result (to sort descending)

# R, 97 95 bytes

function(x)x[rev(order(sapply(Map(sort,Map(unique,strsplit(paste(x),"")),T),Reduce,f=paste0)))]

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This challenge seems to have been pessimized for R. Explanation of the original version (start at 1. and work up):

f <- function(x) {
x[                                                  # 8. Input vector in
rev(                                              # 7. Reversed
order(                                        # 6. Lexicographical order
sapply(                                 # 5. Paste....
Map(sort,                        # 4. Sort each using...
Map(unique,             # 3. Deduplicate each
strsplit(           # 2. Split each string into characters
paste(x),  # 1. Coerce each number to string
"")),
T),                          # 4. ...descending sort.
paste,collapse="")               # 5. ...back into strings
)
)
]
}

# Perl 6, 363433 31 bytes

-1 byte thanks to Jo King
-2 bytes thanks to Phil H

*.sort:{sort 1,|set -<<m:g/\d/}

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### Explanation

{                      }  # Map each number, e.g. -373
m:g/\d/  # Extract digits: (3, 7, 3)
-<<  # Negate each digit: (-3, -7, -3)
set  # Convert to set to remove duplicates
|  # Pass as list of pairs: (-3 => True, -7 => True)
1,  # Prepend 1 for "none": (1, -3 => True, -7 => True)
sort  # Sort (compares 1 and pair by string value): (-7 => True, -3 => True, 1)
*.sort:  # Sort lexicographically
• Nice! -2 bytes for swapping m:g/\d./ for .abs.comb: tio.run/… – Phil H Nov 16 '18 at 14:43

# Python 2, 6055 54 bytes

-1 byte thanks to Jonas Ausevicius.

def f(l):l.sort(cmp,lambda n:sorted(set(n))[::-1],1)

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## Ungolfed

def f(l):
l.sort(        # Sort the list in place
cmp = cmp,   # ... compare with the builtin function cmp
key = k,     # ... on the function k
reverse = 1  # ... in reverse
)              # As the arguments are used in the right order, no names are necessary.

k = lambda n:sorted( # sort
set(n)           # ... the set of digits
)[::-1]            # reverse the result
# As '-' is smaller than the digits,
# it will be sorted to the back and ignored for sorting

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• None can be replaced with cmp in sort function – Jonas Ausevicius Nov 15 '18 at 10:14
• The [::-1] can be exchanged for a double negation, I think. – DonQuiKong Nov 16 '18 at 9:30
• @DonQuiKong that would be quite a bit longer though, as the digits are all strings, and would need to be converted to ints for this. – ovs Nov 17 '18 at 14:05
• @JonasAusevicius Thanks a lot. – ovs Nov 17 '18 at 14:12

# Python 2, 58 60 bytes

lambda a:sorted(a,key=lambda x:sorted(set(x))[::-1])[::-1]

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# Brachylog, 9 bytes

{ȧdṫo₁}ᵒ¹

Note: due to how ordering works in brachylog, it does not work on number correctly. This is fixed by casting the number to a string () at the cost of 1 byte.

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• What do you mean by "Due to how ordering works in brachylog, it does not work as intended."? I've tried all four test cases, and its giving the correct results (unless I accidentally looked past something). – Kevin Cruijssen Nov 15 '18 at 10:12
• @KevinCruijssen The (to string) fixes the issue. Ordering digits in a number descending works as follows. Order from smallest to largest then reverse. The problem is that the number 3120 ordered from smallest to largest is 0123 which is equal to 123which reversed is 321and not 3210 – Kroppeb Nov 15 '18 at 11:00
• Ah ok, so your current code is working due to the added toString (). As mentioned by @Arnauld, I thought your comment meant your current code doesn't work. It might be better to mention it like: "This could have been 8 bytes by removing the (toString), but unfortunately it does not work as intended due to how ordering works in Brachylog." – Kevin Cruijssen Nov 15 '18 at 11:12
• Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it. – Kroppeb Nov 15 '18 at 11:31

# Pyth, 7 6 bytes

-1 byte by @Sok

_o_{S`

Pyth, which uses only printable ASCII, is at a bit of a disadvantage here. Optimally encoded this would be 6*log(95)/log(256) = 4.927 bytes, beating 05AB1E.

Explained:

o              Sort the implicit input by lambda N:
_               reversed
{               uniquified
S               sorted
'               string representation [of N]
_               then reverse the result.

Try it here.

• The trailing N can be left out to save 1 byte - all lambda-type functions infer the presence of the principle lambda variable if any arguments are missing from the end. Examples include m inferring d, f inferring T, u inferring G... – Sok Nov 15 '18 at 12:08

# Jelly, 8 bytes

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### How it works

µÞU  Sort by (reversed):
AD        Absolute value converted to decimal digits
ṢUQ     Sort, reverse, take unique values
• I just implemented this then found your post. I went with normal reverses, , rather than the upends, U. Note, however, that you do not need the D since sort, , is implemented with an iterable(z, make_digits=True) call inside. So that was AṢQṚµÞṚ for 7. – Jonathan Allan Nov 15 '18 at 18:16

# MathGolf, 7 6 bytes

áÉ░▀zx

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## Explanation

After looking at Emigna's 05AB1E solution, I found that I didn't need the absolute operator (and my previous answer was actually incorrect because of that operator). Now the main difference is that I convert to string and get unique characters instead of using the 1-byte operator in 05AB1E.

áÉ      Sort by the value generated from mapping each element using the next 3 instructions
░     Convert to string
▀    Get unique characters
z   Sort reversed (last instruction of block)
x  Reverse list (needed because I don't have a sort-reversed by mapping)

# Japt, 12 bytes

ñ_a ì â ñnÃw

All test cases

Explanation:

ñ_        Ãw    :Sort Descending by:
a             : Get the absolute value
ì           : Get the digits
â         : Remove duplicates
ñn      : Sort the digits in descending order

import Data.List
f=r.sortOn(r.sort.nub.show);r=reverse

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• Defining r=reverse saves two bytes. We also allow anonymous functions, so the f= does not need to ve counted. – Laikoni Nov 15 '18 at 23:33
• I moved the import and f= to the TIO header. Is that OK? – Martin Lütke Nov 15 '18 at 23:34

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# Julia 1.0, 50 bytes

x->sort(x,by=y->sort(unique([digits(-abs(y));1])))

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# APL(NARS), 366 chars, 732 bytes

_gb←⍬

∇a _s w;t
t←_gb[a]⋄_gb[a]←_gb[w]⋄_gb[w]←t
∇

∇(_f _q)w;l;r;ls;i
(l r)←w⋄→0×⍳l≥r⋄l _s⌊2÷⍨l+r⋄ls←i←l⋄→3
→3×⍳∼0<_gb[i]_f _gb[l]⋄ls+←1⋄ls _s i
→2×⍳r≥i+←1
l _s ls⋄_f _q l(ls-1)⋄_f _q(ls+1)r
∇

∇r←(a qsort)w
r←¯1⋄→0×⍳1≠⍴⍴w⋄_gb←w⋄a _q 1(↑⍴w)⋄r←_gb
∇

f←{∪t[⍒t←⍎¨⍕∣⍵]}

∇r←a c b;x;y;i;m
x←f a⋄y←f b⋄r←i←0⋄m←(↑⍴x)⌊(↑⍴y)⋄→3
→0×⍳x[i]<y[i]⋄→3×⍳∼x[i]>y[i]⋄r←1⋄→0
→2×⍳m≥i+←1⋄r←(↑⍴x)>(↑⍴y)
∇

For the qsort operator, it is one traslation in APL of algo page 139 K&R Linguaggio C. I think in it there is using data as C with pointers... Test

c qsort 123, 478, ¯904, 62778, 0, ¯73, 8491, 3120, 6458, ¯7738, 373
8491 ¯904 62778 478 ¯7738 6458 ¯73 373 3120 123 0
c qsort 11, ¯312, 902, 23, 321, 2132, 34202, ¯34, ¯382
902 ¯382 34202 ¯34 321 ¯312 2132 23 11
c qsort 9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0
29384 192 9 6 6 4 44 2212 21 2 1 0
c qsort 44, ¯88, 9, 233, ¯3, 14, 101, 77, 555, 67
9 ¯88 67 77 555 14 44 233 ¯3 101

# Powershell, 44 bytes

$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d

Test script:

$f = {$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d } @( ,( (123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373), (8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0), (8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0) ) ,( (11, -312, 902, 23, 321, 2132, 34202, -34, -382), (902, -382, 34202, -34, -312, 321, 2132, 23, 11), (902, -382, 34202, -34, 2132, -312, 321, 23, 11) ) ,( (9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0), (29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0), (29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0), (29384, 192, 9, 6, 6, 44, 4, 21, 2212, 2, 1, 0) ) ,( (44, -88, 9, 233, -3, 14, 101, 77, 555, 67), ,(9, -88, 67, 77, 555, 14, 44, 233, -3, 101) ) ) | % {$a, $expected =$_
$result = &$f @a
$true-in($expected|%{"$result"-eq"$_"})
"$result" } Output: True 8491 -904 62778 478 -7738 6458 -73 373 3120 123 0 True 902 -382 34202 -34 2132 -312 321 23 11 True 29384 192 9 6 6 44 4 21 2212 2 1 0 True 9 -88 67 77 555 14 44 233 -3 101 # PHP, 87 86 84 bytes while(--$argc)$a[_.strrev(count_chars($n=$argv[++$i],3))]=$n;krsort($a);print_r($a); Run with -nr or try it online. Replace ++$i with $argc (+1 byte) to suppress the Notice (and render -n obosolete). breakdown while(--$argc)  # loop through command line arguments
$a[ # key= _. # 3. prepend non-numeric char for non-numeric sort strrev( # 2. reverse =^= sort descending count_chars($n=$argv[++$i],3)   # 1. get characters used in argument
)
]=$n; # value=argument krsort($a);     # sort by key descending
print_r(\$a);    # print

- is "smaller" than the digits, so it has no affect on the sorting.

# Common Lisp, 88 bytes

Try it online!

Good old verbose Common Lisp!

Explanation:

(sort                   ; sort
(read)                 ; what to sort: a list of numbers, read on input stream
'string>               ; comparison predicate (remember: this is a typed language!)
:key (lambda (x)       ; how to get an element to sort; get a number
(sort (remove-duplicates  ; then sort the unique digits (characters)
(format() "~d" (abs x))) ; from its string representation
'char>)))  ; with the appropriate comparison operator for characters

# C# (Visual C# Interactive Compiler), 75 74 bytes

-1 thanks @ASCII-only

x=>x.OrderByDescending(y=>String.Concat((y+"").Distinct().OrderBy(z=>-z)))

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In C#, strings are considered "enumerables" of characters. I use this to my advantage by first converting each number to a string. LINQ is then leveraged to get the unique characters (digits) sorted in reverse order. I convert each sorted character array back into a string and use that as the sort key to order the whole list.

• Looks like you'll be able to get away with not adding -, looks like order of those doesn't really matter? – ASCII-only Nov 18 '18 at 9:47
• Without the -, test case #2 returns ... 321 2132 ... which seems incorrect? – dana Nov 18 '18 at 16:11
• nah, read the example more carefully – ASCII-only Nov 18 '18 at 21:52
• OK - I think your right. Thanks for the tip! – dana Nov 18 '18 at 22:24