New answers tagged sorting
0
05AB1E, 14 bytes
DāR<IÖ≠©Ï{®ƶ<ǝ
I/O of the string as a list of characters.
Try it online or verify all test cases (feel free to remove the S and J in the test suite to see the actual I/O with character lists).
And I agree with @mbomb007's comment, although it's still pretty short this way. :)
Explanation:
# Example inputs: s=["a",...
0
Python 3.8 (pre-release), 111 109 bytes
lambda n,s:(l:=len(s),r:=sorted(s[~i]for i in range(l)if i%n),[r.insert(i,s[i])for i in range(~-l%n,l,n)])[1]
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1
Perl 6, 53 bytes
{(my@a=$^a.comb)[grep (@a-*-1)%$^n,^@a].=sort;[~] @a}
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It feels like there should be a shorter way of doing this
1
Burlesque, 36 bytes
pe<-jJ-.hdcoJ)-]j)[-++<>#acoz[\[\[<-
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Takes input as N "String"
pe # Parse and push
<- # Reverse the string
j # Put N at the top of the stack
J-.hd # Store N-1 for later
co # Split string into chunks of N
J)-] # Duplicate and take the heads
j)[- # Take the tails of the ...
0
Burlesque, 6 bytes
rasgtp
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ra # Read input as array
sg # Sort and group like values
tp # Transpose
0
05AB1E, 4 bytes
{γζþ
Try it online or verify all test cases.
Explanation:
{ # Sort the (implicit) input-list
# i.e. [5,9,12,5,2,71,23,4,7,2,6,8,2,4,8,9,0,65,4,5,3,2]
# → [0,2,2,2,2,3,4,4,4,5,5,5,6,7,8,8,9,9,12,23,65,71]
γ # Group it into chunks of the same subsequent value
# → [[0],[2,2,2,2],[3],[4,4,4],[5,5,5],[6],[7],[8,8],...
3
Wolfram Language (Mathematica), 219 188 184 bytes
F@A_:=MinimalBy[##&@@@f/@#[[1]]&@*(FindPermutation@A~PermutationProduct~#&)/@GroupElements@PermutationGroup[Cycles@*List/@Join@@((f=Partition[#,2,1]&)/@Values@PositionIndex@A)],Length]
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-31 bytes thanks to @Chris
-4 bytes by making the code much faster (no more ...
3
Jelly, 51 43 bytes
ṪµṢ,$Ḣ©€E¬a="ʋṚƊ×\ẸÞṪTḢ,®ṛ¦Ẹ}¡@¥
WÇẸпFĖẠƇÄ
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A pair of links which when called as a monad returns a list of pairs of 1-indexed indices to swap. Based on the clever J algorithm developed by @Jonah, so be sure to upvote that one too!
Explanation
Helper link
Takes a list containing a list of the remaining items to be ...
5
J, 126 bytes
[:>[:{:"1@}./:~(({.@I.@:~:>@{.)({:@](|.@:{`[`]}~;])[,[:(0{*/,&I.{.)(<0{i.@$)*"1{"0 1=|.@])(,:>@{.))^:(-.@-:>@{.)^:a:;&(0 2$'')
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Here is the same code in a more procedural form, if you're looking for an explanation of algorithm.
f=. 3 :0
sorting=.y NB. copy of input we'll mutate
sorted=./:~y
swaps=.0 2$''
...
0
R, 41 bytes
split(sort(s<-scan()),sequence(table(s)))
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Like Robin Ryder's answer, this makes use of table to get counts of the unique elements in the input.
A fuller explanation to follow.
0
Zsh, 58 bytes
for x (${(u)@})<<<$x&&argv[$@[(i)$x]]=
<<<"
${*:+`$0 $@`}"
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Yay, recursion!
For each element $x of (u)nique ${@}rguments, print it, and set the first (i)nstance of it to the empty string.
Finally, print a newline as a list delimiter, and if the remaining arguments are nonempty, substitute $0 $@, which ...
0
C (gcc), 114 bytes
W;*P;S(D,E)int*D,*E;{do for(P=D;P<E;*P==W?printf("%d ", W),*P=*D,*D++=W,P=E:++P);while(++W);puts("");D-E&&S(D,E);}
This solution takes a while to run because it iterates over all possible integer values from -2^31 to 2^31-1.
Here is a TIO link to a slightly longer solution that runs a lot faster, the only difference is, that ...
1
Perl 6, 25 bytes
*.classify({%.{$_}++}){*}
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Groups each element of the input by how many times it has appeared in the sequence so far, then take only the values
3
R, 54 bytes
for(i in 1:max(t<-table(scan())))print(names(t)[t>=i])
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Builds a table of the counts of the values. For input 1 3 4 2 6 5 1 8 3 8, this gives
t = 1 2 3 4 5 6 8
2 1 2 1 1 1 2
where the first row is names(t) (the different values in the input) and the second is the counts in t.
Then in the for loop, at iteration i, ...
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