25
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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.
  • If possible, please add a link with a test for your code (i.e. TIO).
  • Also, adding an explanation for your answer is highly recommended.

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]
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26 Answers 26

Reset to default
7
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05AB1E, 5 bytes

ΣêR}R

Try it online! or as a Test suite

Explanation

Σ  }    # sort input by
 ê      # its sorted unique characters
  R     # reversed (to sort descending)
    R   # reverse the result (to sort descending)
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7
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R, 97 95 bytes

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

Try it online!

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
              )
        )
    ]
}
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0
7
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Perl 6, 36 34 33 31 bytes

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

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

Try it online!

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
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\$\endgroup\$
1
  • 1
    \$\begingroup\$ Nice! -2 bytes for swapping m:g/\d./ for .abs.comb: tio.run/… \$\endgroup\$
    – Phil H
    Nov 16, 2018 at 14:43
6
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Python 2, 58 60 bytes

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

Try it online!

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6
\$\begingroup\$

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 (ṫ) at the cost of 1 byte.

Try it online!

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4
  • 2
    \$\begingroup\$ 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). \$\endgroup\$
    – Kevin Cruijssen
    Nov 15, 2018 at 10:12
  • \$\begingroup\$ @KevinCruijssen The ṫ (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 \$\endgroup\$
    – Kroppeb
    Nov 15, 2018 at 11:00
  • 2
    \$\begingroup\$ Ah ok, so your current code is working due to the added toString (ṫ). 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 ṫ (toString), but unfortunately it does not work as intended due to how ordering works in Brachylog." \$\endgroup\$
    – Kevin Cruijssen
    Nov 15, 2018 at 11:12
  • \$\begingroup\$ Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it. \$\endgroup\$
    – Kroppeb
    Nov 15, 2018 at 11:31
6
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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.

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1
  • 2
    \$\begingroup\$ 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... \$\endgroup\$
    – Sok
    Nov 15, 2018 at 12:08
6
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Python 2, 60 55 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

Try it online!

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4
  • 5
    \$\begingroup\$ None can be replaced with cmp in sort function \$\endgroup\$
    – Hori
    Nov 15, 2018 at 10:14
  • \$\begingroup\$ The [::-1] can be exchanged for a double negation, I think. \$\endgroup\$
    – DonQuiKong
    Nov 16, 2018 at 9:30
  • \$\begingroup\$ @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. \$\endgroup\$
    – ovs
    Nov 17, 2018 at 14:05
  • \$\begingroup\$ @JonasAusevicius Thanks a lot. \$\endgroup\$
    – ovs
    Nov 17, 2018 at 14:12
5
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Jelly, 8 bytes

ADṢUQµÞU

Try it online!

How it works

ADṢUQµÞU  Main link (monad). Input: integer list
     µÞU  Sort by (reversed):
AD        Absolute value converted to decimal digits
  ṢUQ     Sort, reverse, take unique values
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1
  • 2
    \$\begingroup\$ I just implemented this then found your post. I went with normal reverses, Ṛ, rather than the upends, U. Note, however, that you do not need the D since sort, Ṣ, is implemented with an iterable(z, make_digits=True) call inside. So that was AṢQṚµÞṚ for 7. \$\endgroup\$
    – Jonathan Allan
    Nov 15, 2018 at 18:16
4
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MathGolf, 7 6 bytes

áÉ░▀zx

Try it online! or as a test suite.

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)
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4
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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
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0
4
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Haskell, 54 52 bytes

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

Try it online!

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6
  • \$\begingroup\$ Defining r=reverse saves two bytes. We also allow anonymous functions, so the f= does not need to ve counted. \$\endgroup\$
    – Laikoni
    Nov 15, 2018 at 23:33
  • \$\begingroup\$ I moved the import and f= to the TIO header. Is that OK? \$\endgroup\$
    – Martin Lütke
    Nov 15, 2018 at 23:34
  • \$\begingroup\$ Same byte count, but maybe of some interest: f=r$r id.nub.show;r=(reverse.).sortOn. \$\endgroup\$
    – Laikoni
    Nov 15, 2018 at 23:43
  • 1
    \$\begingroup\$ The import actually needs to be counted. \$\endgroup\$
    – Laikoni
    Nov 15, 2018 at 23:45
  • 2
    \$\begingroup\$ You may want to have a look at our Guide to golfing rules in Haskell. \$\endgroup\$
    – Laikoni
    Nov 15, 2018 at 23:52
4
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Stax, 6 7 bytes

èó≥ü≤♥¥

Run and debug it

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3
  • \$\begingroup\$ This seems to give incorrect results. For example, in the test case of your TIO it outputs -904 8491 478 62778 6458 -7738 -73 373 123 3120 0 instead of the intended 8491 -904 62778 478 -7738 6458 373 -73 3120 123 0 or 8491 -904 62778 478 -7738 6458 -73 373 3120 123 0. This test case is also used in the example, and to explain the rules, so I would take a look at that to understand it better. It seems you are only sorting by the single largest digit once, without any of the other rules? \$\endgroup\$
    – Kevin Cruijssen
    Nov 16, 2018 at 7:37
  • \$\begingroup\$ @KevinCruijssen: Yes, my apologies. I mis-read the problem statement. I've adjusted the program to handle the stated requirements. This program is accepting the input integers as quoted strings. That's usually acceptable, but if not I might need to add another byte. \$\endgroup\$
    – recursive
    Nov 16, 2018 at 17:23
  • \$\begingroup\$ Looks good now, +1 from me. And yes, inputting as strings is completely fine. \$\endgroup\$
    – Kevin Cruijssen
    Nov 16, 2018 at 17:54
4
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APL (Dyalog Extended), 19 bytes

{⍵[⍒∪¨(∨'¯'~⍨⍕)¨⍵]}

Try it online!

Fixed at a cost of +2 bytes thanks to OP.

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2
  • \$\begingroup\$ I think you are missing an 'uniquify' somewhere? If I try the example test case in your TIO for example, ¯7738 is placed before 478, but it should be after it: digits [8,7,4] come before digits [8,7,3]. \$\endgroup\$
    – Kevin Cruijssen
    Nov 16, 2018 at 17:52
  • \$\begingroup\$ Thanks, @KevinCruijssen \$\endgroup\$
    – Adalynn
    Nov 16, 2018 at 21:06
4
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C (gcc), 114 111 109 bytes

a;v(n){n=n<0?-n:n;for(a=0;n;n/=10)a|=1<<n%10;n=a;}c(int*a,int*b){a=v(*a)<v(*b);}f(a,n)int*a;{qsort(a,n,4,c);}

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Explanation:

f() uses qsort() to sort provided array in place. Using comparison function c() to compare numbers which evaluates numbers using v(). v() calculates a higher number if bigger digits are present in parameter.

[Edit 1] Improved by 3 bytes. 2 byte credits to Kevin. Thanks

[Edit 2] 2 more bytes improved. Credits to gastropner. Thanks

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3
  • 1
    \$\begingroup\$ You can golf n>0 to n I think in your loop of your method v. \$\endgroup\$
    – Kevin Cruijssen
    Nov 15, 2018 at 11:55
  • \$\begingroup\$ f()'s argument list int*a,n can be shortened to int*a. \$\endgroup\$
    – gastropner
    Nov 19, 2018 at 6:58
  • 1
    \$\begingroup\$ Suggest for(a=0;n=abs(n); instead of n=n<0?-n:n;for(a=0;n; \$\endgroup\$
    – ceilingcat
    Nov 20, 2018 at 8:31
4
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Julia 1.0, 50 47 bytes

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

Try it online!

-3 bytes by Ashlin Harris.

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2
3
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J, 17 bytes

{~[:\:~.@\:~@":@|

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Explanation:

                @|    - find the absolute value and
             @":      - convert to string and
         @\:~         - sort down and
       ~.             - keep only the unique symbols
    \:                - grade down the entire list of strings   
  [:                  - function composition
{~                    - use the graded-down list to index into the input
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3
\$\begingroup\$

JavaScript (SpiderMonkey), 68 bytes

Thanks for @Arnauld for reminding me again that SpiderMonkey uses stable sort, so -4 bytes for removing ||-1.

A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y))

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JavaScript (Node.js), 72 bytes

A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y)||-1)

Try it online!

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4
  • \$\begingroup\$ Or 68 bytes with SpiderMonkey. \$\endgroup\$
    – Arnauld
    Nov 15, 2018 at 11:04
  • 1
    \$\begingroup\$ @Arnauld oh stable sort again ;P \$\endgroup\$
    – Shieru Asakoto
    Nov 15, 2018 at 12:17
  • \$\begingroup\$ Actually, I think V8 is using at least 2 different sort algorithms. It seems like it's stable if the size of the array is less than or equal to \$10\$. \$\endgroup\$
    – Arnauld
    Nov 15, 2018 at 12:37
  • 1
    \$\begingroup\$ @Arnauld V8 use quick sort before Chrome 70. The quick sort algorithm perform an insertion sort when array size is small enough. And latest Chrome had changed to stable sort for matching other browsers' (IE/Firefox/Safari) behavior. \$\endgroup\$
    – tsh
    Nov 16, 2018 at 8:42
3
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Java (JDK), 98 bytes

l->l.sort((a,b)->{int r=0,i=58;for(;r==0&i-->48;)r=(b.indexOf(i)>>9)-(a.indexOf(i)>>9);return r;})

Try it online!

Explanation

l->                           // Consumer<List<String>>
 l.sort(                      //  Use the incorporated sort method which uses a...
  (a,b)->{                    //   Comparator of Strings
   int r=0,                   //    define r as the result, initiated to 0
       i=58;                  //           i as the codepoint to test for.
   for(;r==0&i-->48;)         //    for each digit codepoint from '9' to '0',
                              //     and while no difference was found.
    r=                        //     set r as the difference between
     (b.indexOf(i)>>9)-       //      was the digit found in b? then 0 else -1 using the bit-shift operator
     (a.indexOf(i)>>9);       //      and was the digit found in a? then 0 else -1.
   return r;                  //    return the comparison result.
  }
 )

Note:

I needed a way to map numbers to either 0/1 or 0/-1.

indexOf has the nice property that it's consistently returning -1 for characters not found. -1 right-shifted by any number is always -1. Any positive number right-shifted by a big enough number will always produce 0.

So here we are:

input        input.indexOf('9')      input.indexOf('9')>>9
"999"        0                       0
"111119"     5                       0
"123456"     -1                      -1
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\$\endgroup\$
1
  • 1
    \$\begingroup\$ Ah, yeah, that's what I mean. ;p Nice golf of using >>9 instead of >>32 due to the limited range of the numbers. \$\endgroup\$
    – Kevin Cruijssen
    Nov 15, 2018 at 12:53
3
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Ruby, 55 bytes

->a{a.sort_by{|x|x.abs.digits.sort.reverse|[]}.reverse}

Try it online!

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2
\$\begingroup\$

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

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

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.

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

Common Lisp, 88 bytes

(sort(read)'string> :key(lambda(x)(sort(remove-duplicates(format()"~d"(abs x)))'char>)))

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

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.

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  • \$\begingroup\$ Looks like you'll be able to get away with not adding -, looks like order of those doesn't really matter? \$\endgroup\$
    – ASCII-only
    Nov 18, 2018 at 9:47
  • \$\begingroup\$ Without the -, test case #2 returns ... 321 2132 ... which seems incorrect? \$\endgroup\$
    – dana
    Nov 18, 2018 at 16:11
  • \$\begingroup\$ nah, read the example more carefully \$\endgroup\$
    – ASCII-only
    Nov 18, 2018 at 21:52
  • \$\begingroup\$ OK - I think your right. Thanks for the tip! \$\endgroup\$
    – dana
    Nov 18, 2018 at 22:24
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Perl 5 -nl, 68 bytes

$a{eval"9876543210=~y/$_//dcr"}=$_}{say$a{$_}for reverse sort keys%a

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Husk, 7 bytes

↔Öö↔Oda

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