Given a list of positive integers that contains at least 3 distinct entries, output a permutation of that list that isn't sorted in ascending or descending order.
1,2,3 -> 2,1,3 or 3,1,2 or 1,3,2 or 2,3,1
1,2,3,3 -> 2,1,3,3 or 3,1,2,3 or 1,3,2,3 etc..
Thanks @Arnauld and @NoOneIsHere for the title!
a=>[a.sort((x,y)=>x-y).pop(),...a]
Sort the array in ascending order, pop the last element and use it as the first element of a new array. Then destructure the remaining elements of the original array into the new array (In JS, both sort and pop modify the original array).
o.innerText=(f=
a=>[a.sort((x,y)=>x-y).pop(),...a]
)(i.value=[1,2,3]);oninput=_=>o.innerText=f(i.value.split`,`)<input id=i><pre id=o>
a.sort()?
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sort method sorts lexicographically.
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Ṣṙ- also works (just felt like saying that; you probably knew :P)
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Ṣṙ1 only three bytes? In UTF-8, it's 7 bytes.
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Commented
Aug 15, 2017 at 0:46
min at the end saves a byte.
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Prompt A
SortA(LA
max(LA→B
dim(LA)-1→dim(LA
augment({B},LA
Prompts for input in the format {1,2,3,4}.
TI-Basic is a tokenized language, all tokens used here are one-byte.
Explanation:
Prompt A # 3 bytes, store user input in LA
SortA(LA # 4 bytes, sort LA ascending
max(LA→B # 6 bytes, save the last value in the sorted list to B
dim(LA)-1→dim(LA # 11 bytes, remove the last value from LA
augment({B},LA # 7 bytes, prepend B to LA and implicitly print the result
O#`
s`(.*)¶(.*)
$2¶$1
Try it online! Sort and rotate as per usual. At least there's no unary conversion this time.
l->{l.sort(null);l.add(l.remove(0));}
-31 bytes thanks to @Nevay (forgot Java 8 had a List#sort(Comparator) method..)
Modifies the input-ArrayList, instead of returning a new one.
Explanation:
l->{ // Method with ArrayList parameter and no return-type
l.sort(null); // Sort the input-list (no need for a Comparator, thus null)
l.add(l.remove(0)); // Remove the first element, and add it last
} // End of method
l->{l.sort(null);java.util.Collections.rotate(l,1);} to save 16 bytes.
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l->{l.sort(null);l.add(l.remove(0));} to save 31 bytes (requires the usage of a not fixed sized list).
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add and remove must be implemented; nothing is said about fixed-sized list... Kevin Cruijssen, given that there are much better alternatives in the previous comments, I'll wait for an edit before +1ing.
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Commented
Aug 14, 2017 at 13:52
import Data.List
f(a:b)=b++[a];f.sort
Use view patterns to match on the head of a sorted version of the input list, then append the first item of the list to the tail of the remaining list.
View patterns aren't worth it. Sort the list, take the head off, append it to the end. In this case, it turns out that the naive solution typed out compactly is the best.
-XViewPatterns. Counting those the standard way f(a:b)=b++[a];f.sort is shorter.
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{first {&&![>=] $_},.permutations}
*.sort[1..*,0].flat
Note that [1..*,0] would result in ((2,3),1), so .flat is there to turn it into (2,3,1)
RotateLeft@*Sort
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Commented
Aug 14, 2017 at 3:46
&nasprl
Ugh, ruining the sort is so expensive!
Explanation:
&nasprl
&n # take input as a list of numbers
a # sort
sp # save top of stack and pop
r # reverse stack
l # load saved item
Takes input from stdin. Sorts the list and then moves the first element to the end, ensuring that it is no longer sorted. Saved three bytes due to Giuseppe.
c(sort(x<-scan())[-1],min(x))
Another implementation, same byte count:
c((x<-sort(scan()))[-1],x[1])
c(sort(x<-scan())[-1],min(x)) is 29 bytes using essentially the same idea as yours.
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S╜
I think this is dissimilar enough from totallyhuman's post to post a new answer; I hope you don't mind :P EDIT: DAMMIT YOU NINJA'D ME
def f(a):a[1:]=a[a.sort():0:-1]
Yet another Python solution.
Sadly, this one has the same length to HyperNeutrino's answer.
O#`
O^#-2`
O#` Sort the list
O^#-2` Reverse sort the list other than the last element
This leaves the list with the 2nd highest element first and the highest element last which is never correctly sorted
Submitted on mobile. Please don't kill me for problems.
->a{a.sort.rotate}
.>SQ1
SQ - sort input list
.>SQ1 - rotate input list cyclicaly by 1
a=>sorted(a)[1to,0]
-2 bytes indirectly thanks to xnor
Not yet working on TIO; waiting for a pull.
lambda a:a[1:a.sort()]+a[:1]
a.sort() sorts a in place and returns None. None can be used as a slicing index and is the same as omitting that index.
lambda a:a.sort()or[a.pop(),*a]
Try it online! or Verify all test cases.
Inspired by Shaggy's JS answer.
requires PHP 5.4 or later for short array syntax.
sort($a=&$argv);print_r([array_pop($a)]+$a);
sort arguments, replace 0-th argument with removed last argument, print.
Run with -nr or try it online.
The 0-th argument is the script file name, "-" if you call PHP with -r. "-" is compared to the other arguments as a string, and since ord("-")==45, it is smaller than any number. The numbers themselves, although strings, are compared as numbers: "12" > "2".
php -nr '<code>' 3 4 2 5 1 and sort($a=&$argv) lead to $a=["-","1","2","3","4","5"] →
[array_pop($a)]+$a is [0=>"5"]+[0=>"-",1=>"1",2=>"2",3=>"3",4=>"4"],
which results in [0=>"5",1=>"1",2=>"2",3=>"3",4=>"4"].
+ operator does not append, it merges (without reordering the indexes; but that doesn´t matter here). The important point is that $a points to $argv and $argv[0] contains the script´s file name, the arguments start at index 1. I extended the description. Thanks for the question.
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f(x)=sort(x)[[2:end;1]]
Slightly shorter than, but equivalent to f(x)=circshift(sort(x),1).
I wish I could makeamethod based on select that was more, compact but I can not
[2,[1,3]]. \$\endgroup\$