31
\$\begingroup\$

Given an array of positive integers and two distinct valid indices, return the array with the two elements corresponding to the two indices swapped.

You may choose to use 0-indexing or 1-indexing, but the testcases below will be 0-indexed.

array        m n output
[1,2,3,4]    0 1 [2,1,3,4]
[5,8,9]      0 2 [9,8,5]
[11,13,15,3] 1 2 [11,15,13,3]
[11,13,15,3] 2 1 [11,15,13,3]
[11,15,15,3] 2 1 [11,15,15,3]

This is . Shortest answer in bytes wins. Standard loopholes apply.

\$\endgroup\$
  • \$\begingroup\$ Related, related. \$\endgroup\$ – Leaky Nun Jun 22 '17 at 10:28
  • 1
    \$\begingroup\$ Huh, this may well be a task that many golfing languages have a hard time with but most practical languages find easy. (Lists with mutable elements aren't a common thing for golfing languages to have.) If that is the case, it'll be quite interesting. (The golfing languages will probably still win, though, because they're so much terser they can get away with a more complex algorithm.) \$\endgroup\$ – user62131 Jun 22 '17 at 10:36
  • 7
    \$\begingroup\$ Surprised this probably isn't a dupe, but this challenge is actually creative, since it's a real challenge for many golfing languages out there. \$\endgroup\$ – Erik the Outgolfer Jun 22 '17 at 10:45
  • \$\begingroup\$ @LeakyNun I've got downvotes (and even delete votes) like that in the past, don't worry too much about it... \$\endgroup\$ – Erik the Outgolfer Jun 22 '17 at 10:46
  • \$\begingroup\$ Can m and n be taken as an array? \$\endgroup\$ – Okx Jun 22 '17 at 11:23

35 Answers 35

17
\$\begingroup\$

C/C++, 53 50 39 bytes

f(a,m,n)int*a;{a[m]^=a[n]^=a[m]^=a[n];}

Try it online

Saved 11 bytes thanks to @Dennis

\$\endgroup\$
10
\$\begingroup\$

Operation Flashpoint scripting language, 98 95 bytes

f={t=_this;a=t select 0;b=+a;m=t select 1;n=t select 2;a set[m,b select n];a set[n,b select m]}

Modifies the array directly.

Explanation:

t=_this;                   // Give a shorter name for the array of arguments.

a=t select 0;              // Let 'a' be a pointer to the array that we modify.
                           // (The language doesn't have a concept of pointers really,
                           // yet its array variables are pointers to the actual array.)

b=+a;                      // Make a copy of the original array and save a pointer to it
                           // in the variable 'b'. This saves a few bytes later.

m=t select 1;              // Read the index arguments from the input array and save them
n=t select 2;              // to their respective variables.

a set[m,b select n];       // Do the swapping by reading the values from the copy and
a set[n,b select m]        // writing them to the original array. The last semicolon can
                           // be omitted because there are no more statements following 
                           // the last statement.

Call with:

array = [1,2,3,4];
str = format["%1", array];
[array, 0, 1] call f;
hint format["%1\n%2", str, array];

Output:

enter image description here

\$\endgroup\$
7
\$\begingroup\$

JavaScript ES6, 36 32 bytes

Look, Ma, no temporary variable!

(a,m,n)=>[a[m],a[n]]=[a[n],a[m]]

Try it

Enter a comma separated list of elements for a and 2 integers for m & n.

f=
(a,m,n)=>[a[m],a[n]]=[a[n],a[m]]
oninput=_=>o.innerText=(f(b=i.value.split`,`,+j.value,+k.value),b);o.innerText=(f(b=(i.value="5,8,9").split`,`,j.value=0,k.value=2),b)
*{font-family:sans-serif}input{margin:0 5px 0 0;width:100px;}#j,#k{width:50px;}
<label for=i>a: </label><input id=i><label for=j>m: </label><input id=j type=number><label for=k>n: </label><input id=k type=number><pre id=o>

\$\endgroup\$
  • 2
    \$\begingroup\$ Those instructions modify the array, which means you're allowed to not return the array, which will save you a few bytes. \$\endgroup\$ – Neil Jun 22 '17 at 12:14
  • \$\begingroup\$ @Neil: You're saying to just use (a,m,n)=>[a[m],a[n]]=[a[n],a[m]]? That would only output the 2 swapped elements without the rest of the array (e.g., [5,8,9],0,2 -> [9,5]). \$\endgroup\$ – Shaggy Jun 22 '17 at 13:33
  • \$\begingroup\$ @Neil: Right, which is why we need the a at the end to give us the complete, modified array. Or am I completely missing what you're trying to say? \$\endgroup\$ – Shaggy Jun 22 '17 at 14:16
  • \$\begingroup\$ @Neil: Hmm ... OK, I think I see what you'e getting at now (Sorry, trying to do too many things at the same time today). Thanks for the tip. Is there a consensus on that and, if so, would you have a link handy before I go searching for it myself? \$\endgroup\$ – Shaggy Jun 22 '17 at 14:50
  • 1
    \$\begingroup\$ codegolf.meta.stackexchange.com/a/4942/48934 \$\endgroup\$ – Neil Jun 22 '17 at 14:59
5
\$\begingroup\$

Python 3, 41 32 bytes

-9 bytes thanks to @notjagan

def f(a,m,n):a[m],a[n]=a[n],a[m]

Try it online!

Modifies its argument, which is a valid output format.

\$\endgroup\$
  • 3
    \$\begingroup\$ It's funny how it's not even golfed that much comparing to idiomatic python code. \$\endgroup\$ – Łukasz Rogalski Jun 22 '17 at 13:59
5
\$\begingroup\$

Jelly, 7 bytes

Ṛ,ḷyJ}ị

Try it online!

How it works

Ṛ,ḷyJ}ị  Main link. Left argument: [i, j]. Right argument: A (array)

Ṛ        Reverse; yield [j, i].
  ḷ      Left; yield [i, j].
 ,       Pair; yield [[j, i], [i, j]].
    J}   Indices right; yield all indices of A.
   y     Transliterate; replace j with i and i with j.
      ị  Index into A.
\$\endgroup\$
  • \$\begingroup\$ tfw the wrapper is almost as long as the program... \$\endgroup\$ – Leaky Nun Jun 22 '17 at 13:20
  • \$\begingroup\$ I never knew of the existence of y \$\endgroup\$ – Leaky Nun Jun 22 '17 at 13:23
  • \$\begingroup\$ I knew of y, but didn't think of using it here. That's a pretty clever answer. \$\endgroup\$ – user62131 Jun 22 '17 at 22:40
  • \$\begingroup\$ This got me thinking... is Jelly valid Jelly code? \$\endgroup\$ – M.Herzkamp Jun 23 '17 at 14:16
  • \$\begingroup\$ @M.Herzkamp It is. I doubt its exceptionally useful though. \$\endgroup\$ – Dennis Jun 23 '17 at 15:21
4
\$\begingroup\$

Japt, 17 16 bytes

hV(A=UgV UgW¹hWA

Try it online!

Saved a byte thanks to ETHproductions

\$\endgroup\$
  • 2
    \$\begingroup\$ Nice. Don't think you need the comma though. \$\endgroup\$ – ETHproductions Jun 22 '17 at 11:23
  • \$\begingroup\$ @ETHproductions Thanks, you're right. \$\endgroup\$ – Tom Jun 22 '17 at 11:24
  • \$\begingroup\$ Alternative 16 byte implementation but I'm still convinced there's a shorter solution. \$\endgroup\$ – Shaggy Jun 22 '17 at 12:02
  • 2
    \$\begingroup\$ 15 bytes \$\endgroup\$ – ETHproductions Jun 22 '17 at 16:10
4
\$\begingroup\$

MATL, 7 6 bytes

yyP)w(

Indices are 1-based.

Try it online!

Explanation

Consider inputs [11 13 15 3], [2 3].

yy   % Take two inputs implicitly. Duplicate them
     % STACK: [11 13 15 3], [2 3], [11 13 15 3], [2 3]
P    % Flip
     % STACK: [11 13 15 3], [2 3], [11 13 15 3], [3 2]
)    % Reference indexing (pick indexed entries)
     % STACK: [11 13 15 3], [2 3], [15 13]
w    % Swap
     % STACK: [11 13 15 3], [15 13], [2 3]
(    % Assignment indexing (write values into indexed entries). Implicitly display
     % STACK: [11 15 13 3]
\$\endgroup\$
4
\$\begingroup\$

C# (.NET Core), 48 43 31 bytes

(a,m,n)=>a[m]+=a[n]-(a[n]=a[m])

Try it online!

Swaps the numbers in the original array, no temporary variables used. Nonetheless I can't take credit for this answer as it has been Neil's idea.

\$\endgroup\$
  • \$\begingroup\$ @LeakyNun it does not seem to work, as that leaves a[m] with a value of 0. Try it yourself! \$\endgroup\$ – Charlie Jun 22 '17 at 11:14
  • \$\begingroup\$ (a,m,n)=>a[m]+=a[n]-(a[n]=a[m]) seems to work though. \$\endgroup\$ – Neil Jun 22 '17 at 11:38
  • \$\begingroup\$ (These answers are all also valid in ES6 JavaScript, no?) \$\endgroup\$ – Neil Jun 22 '17 at 11:41
4
\$\begingroup\$

Common Lisp, 42 bytes

-2 bytes thanks to @coredump.

(lambda(a i j)(rotatef(elt a i)(elt a j)))

Try it online!

Quite straight forward, since there is a Common Lisp macro to swap: rotatef.

\$\endgroup\$
  • \$\begingroup\$ You could use ELT instead of AREF \$\endgroup\$ – coredump Jun 23 '17 at 11:10
  • 1
    \$\begingroup\$ @coredump Right, thanks! \$\endgroup\$ – Dada Jun 23 '17 at 11:30
3
\$\begingroup\$

Javascript ES6, 36 34 bytes

(a,m,n)=>(x=a[m],a[m]=a[n],a[n]=x)
  • -2 Bytes because the function is altering the array. No need to return the array. Thanks to @Neil

Demo

f=

(a,m,n)=>(x=a[m],a[m]=a[n],a[n]=x)

let a1=[1,2,3,4], a2=[5,8,9], a3=[11,13,15,3]

f(a1,0,1);
f(a2,0,2);
f(a3,1,2);

console.log(JSON.stringify(a1)); //[2,1,3,4]
console.log(JSON.stringify(a2)); //[9,8,5]
console.log(JSON.stringify(a3)); //[11,15,13,3]

\$\endgroup\$
  • 1
    \$\begingroup\$ Those instructions modify the array, which means you're allowed to not return the array, which will save you a few bytes. \$\endgroup\$ – Neil Jun 22 '17 at 12:14
3
\$\begingroup\$

CJam, 4 bytes

{e\}

Try it online!

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

Java 8, 48 bytes

(a,b,c)->{int t=a[b];a[b]=a[c];a[c]=t;return a;}

Input:

int[] a
int b
int c
\$\endgroup\$
  • \$\begingroup\$ How do you do lambdas with three arguments in Java? \$\endgroup\$ – Leaky Nun Jun 22 '17 at 10:48
  • 1
    \$\begingroup\$ Those instructions modify the array, which means you're allowed to not return the array, which will save you a few bytes. \$\endgroup\$ – Neil Jun 22 '17 at 12:13
  • 1
    \$\begingroup\$ @LeakyNun I'm not Okx, but here is a Try it now example with Okx's current answer and custom interface. \$\endgroup\$ – Kevin Cruijssen Jun 22 '17 at 13:25
  • 1
    \$\begingroup\$ And based on Carlos Alejo's amazing C# answer (with @Neil's help), you can make it even shorter by getting rid of the temporary variable: (a,b,c)->a[b]+=a[c]-(a[c]=a[b]) (31 bytes) \$\endgroup\$ – Kevin Cruijssen Jun 22 '17 at 13:32
  • 1
    \$\begingroup\$ cough cough Collections::swap is 17 bytes... at least assuming this holds for this challenge... \$\endgroup\$ – Socratic Phoenix Jun 23 '17 at 0:57
2
\$\begingroup\$

05AB1E, 10 bytes

D²è³ǝ¹³è²ǝ

Try it online!

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

Octave, 28 bytes

@(a,n){a(n)=a(flip(n)),a}{2}

Try it online!

Quite satisfied with this one actually :)

Takes input on the form: f([1,2,3,4],[1,2]), 1-indexed.

Explanation:

@(a,n)                         % Anonymous function that takes two 1-dimensional
                               % arrays as input
      {               , }      % Create a cell with two elements
       a(n)=a(flip(n))         % One element are the two number at indices given by
                               % the second input array. This will be a 1x2 array
      {a(n)=a(flip(n)),a}      % Place those two in a cell together with the entire array a
                               % a is now updated, thanks to Octave's inline assignment
      {a(n)=a(flip(n)),a}{2}   % Return the second element
\$\endgroup\$
2
\$\begingroup\$

Jellyfish, 7 bytes

p
ZRi
i

Takes a list and a pair of indices. Try it online!

Explanation

Jellyfish happens to have a "modify items at indices" function, Z, which does exactly what we need. The two is grab the inputs from STDIN. Z takes as arguments the second input, the reversal function R, and the list. Then Z performs the modification, and p prints the result.

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

R, 38 bytes

function(x,a,b){x[c(a,b)]=x[c(b,a)];x}

Feels rather long, but I can't get it much shorter. Sadly it requires the explicit returning through x, requiring {} around the function body. pryr::f() doesn't recognise the need for x as function argument so doesn't work :/.

\$\endgroup\$
  • \$\begingroup\$ I think function(x,i)replace(x,i,rev(i)) would work, even with pryr syntax. \$\endgroup\$ – Giuseppe Jun 22 '17 at 17:09
  • \$\begingroup\$ @Giuseppe Ah, I was looking for a convenient function to do the swap, but was searching with the wrong terms. Feel free to post that as an own answer. \$\endgroup\$ – JAD Jun 22 '17 at 19:41
  • \$\begingroup\$ @Giuseppe I think you need to do replace(x,i,x[rev(i)]), else you'll place the indices instead of their values. \$\endgroup\$ – JAD Jun 22 '17 at 20:18
2
\$\begingroup\$

Shenzhen I/O, 735 Bytes

23¥, 810 Power, 48 Lines of Code

[traces] 
......................
......................
......................
......................
......................
......................
.14.14.14.............
.94.14.14.............
.A........1C..........
.3554..95556..........
.9554.16..............
.A....................
.2....................
......................

[chip] 
[type] UC6
[x] 4
[y] 2
[code] 
  slx x0
  mov x1 acc
  mov x1 dat
  mov acc x3
  mov dat x3
  mov acc x3
  mov dat x3

[chip] 
[type] UC6
[x] 8
[y] 5
[code] 
  slx x2
  mov x2 x1
  mov x0 dat
  mov x2 x1
  mov x0 acc
  mov x2 x1
  mov dat 

[chip] 
[type] UC4X
[x] 2
[y] 6
[code] 
  slx x0
  mov 0 x3
j:  mov x0 acc
  mov acc x2
  teq acc 0
- jmp j
  mov -999 x1

[chip] 
[type] RAM
[x] 5
[y] 6

SIO

DISCLAIMER: Arrays are 0-terminated in this. Arrays are a pain in the ass to work with in Shenzhen I/O otherwise.

I actually made a steam level for this game. You can play it here.

EDIT: Aaand I just realized I said that the array in was ordered. Heck.

\$\endgroup\$
  • \$\begingroup\$ Welcome to the site This is really cool! do you think that you might be able to remove some of the whitespace in the file and still have Shenzhen IO accept the file? I don't know how much you have fiddled with it, but you should try and see how flexible the format is. \$\endgroup\$ – Wheat Wizard Jun 25 '17 at 2:26
  • \$\begingroup\$ I haven't played around with it! On the other hand, I am axing the header for the puzzle which contains the puzzle name and solution name, so I dunno if I should bother. \$\endgroup\$ – junkmail Jun 25 '17 at 2:33
1
\$\begingroup\$

Swift, 111 65 bytes (0-indexed)

Swift is already notorious for being one of the worst code-golf languages, but here is a function that makes use of ternary expressions:

func t(l:[Int],m:Int,n:Int){var r=l;r[m]=l[n];r[n]=l[m];print(r)}

Check it out! - Usage: t(l:[1,2,3],m:0,n:1).

\$\endgroup\$
  • \$\begingroup\$ Using a default param for r would save you bytes and you can also just mutate the passed array (AFAIK swift array are pass by value) \$\endgroup\$ – Downgoat Jun 22 '17 at 16:55
  • \$\begingroup\$ Default parameter in Swift? How can I do that? \$\endgroup\$ – Mr. Xcoder Jun 22 '17 at 17:40
  • \$\begingroup\$ And parameters are constants in Swift @Downgoat \$\endgroup\$ – Mr. Xcoder Jun 22 '17 at 17:41
1
\$\begingroup\$

k (kona), 13 bytes

{x[y]:x@|y;x}

Pretty basic, but it works. Ex:

k){x[y]:x@|y;x}[1 2 3 4; 0 1]
2 1 3 4
\$\endgroup\$
1
\$\begingroup\$

Perl 5, 32 bytes

-3 bytes thanks to @Dom Hastings!

30 bytes of code + -pa flags.

@F[pop@p,@p]=@F[@p=<>];$_="@F"

Try it online!

Quite straight forward, using array slices.

\$\endgroup\$
  • \$\begingroup\$ Hey hey, tinkered with this a little and managed to save 3 bytes! @F[pop@p,@p]=@F[@p=<>];$_="@F". \$\endgroup\$ – Dom Hastings Jun 22 '17 at 13:27
  • \$\begingroup\$ @DomHastings Hmm, nice, as always! Thanks :) \$\endgroup\$ – Dada Jun 22 '17 at 13:35
1
\$\begingroup\$

Mathematica, 32 bytes

(a=#;a[[{##2}]]=a[[{#3,#2}]];a)&
\$\endgroup\$
  • 3
    \$\begingroup\$ a[[{##2}]]==a[[{#3,#2}]] should be a[[{##2}]]=a[[{#3,#2}]] (using Set, not Equals) \$\endgroup\$ – JungHwan Min Jun 22 '17 at 13:12
1
\$\begingroup\$

C, 42 bytes

Modify the array in place with a temp value.

f(r,m,n){int*a=r;r=a[m];a[m]=a[n];a[n]=r;}

C, 60 58 bytes

A little more interesting, not using any temp value...

f(a,m,n)int*a;{a[m]+=a[n];a[n]-=a[m];a[n]*=-1;a[m]-=a[n];}

C, 49 bytes

Using XOR

f(a,m,n)int*a;{a[m]^=a[n];a[n]^=a[m];a[m]^=a[n];}
\$\endgroup\$
  • \$\begingroup\$ Heh, I was just about to post f(x,i,j,t)int*x;{t=x[i];x[i]=x[j];x[j]=t;}. \$\endgroup\$ – Dennis Jun 22 '17 at 13:47
  • \$\begingroup\$ @Dennis you saved me two bytes on the other solution, thanks! \$\endgroup\$ – cleblanc Jun 22 '17 at 14:00
  • \$\begingroup\$ Wouldn't the second solution be shorter (and safer) with ^? \$\endgroup\$ – Dennis Jun 22 '17 at 14:10
  • \$\begingroup\$ -1 for the XOR version using a define instead of a function #define X(x,y,z)x[y]^=x[z],x[z]^=x[y],x[y]^=x[z] \$\endgroup\$ – Giacomo Garabello Jun 22 '17 at 14:55
  • \$\begingroup\$ f(r,m,n){int*a=r;r=a[m];a[m]=a[n];a[n]=r;} is broken: SIGSEGV. \$\endgroup\$ – Bodo Thiesen Jun 22 '17 at 20:25
1
\$\begingroup\$

Pyth, 17 8 bytes

Saved 9 bytes thanks to Leaky Num.

@LQ.rUQE

Test it online!

This is 0-indexed, and the indices are provided as a tuple: (n, m).

Explanations

@LQ.rUQE

     UQ     # Generate [0, 1, 2, ..., len(input)]
       E    # Get the indices as the tuple (1, 2)
   .r       # Translate each element of UQ to its cyclic successor in E
            # Now the indices are permuted (e.g. [0, 2, 1, ..., len(input)]
@LQ         # For each index, get it's value. Implicit print
\$\endgroup\$
  • \$\begingroup\$ 8 bytes: @LQ.rUQE \$\endgroup\$ – Leaky Nun Jun 22 '17 at 16:49
  • \$\begingroup\$ @LeakyNun It's so different I think you can post it for yourself! \$\endgroup\$ – Jim Jun 22 '17 at 16:50
  • \$\begingroup\$ I'm the OP; I don't post on my own challenge. \$\endgroup\$ – Leaky Nun Jun 22 '17 at 16:50
1
\$\begingroup\$

Mathematica, 20 bytes

#~Permute~Cycles@#2&

Pure function taking two arguments in the following 1-indexed (and possibly abusive) format: the second test case [5,8,9]; 0 2; [9,8,5] would be called as

#~Permute~Cycles@#2& [ {5,8,9} , {{1,3}} ]

(spaces are extraneous and just for visible parsing). Permute is the builtin function that applies a permutation to a list, and Cycles[{{a,b}}] represents the permutation that exchanges the ath and bth elements of a list and ignores the rest.

\$\endgroup\$
  • \$\begingroup\$ What do the ~ do? \$\endgroup\$ – Cyoce Jun 23 '17 at 3:54
  • \$\begingroup\$ ~ is Mathematica's infix notation for a binary function: x~f~y means the same thing as f[x,y]. \$\endgroup\$ – Greg Martin Jun 23 '17 at 7:35
1
\$\begingroup\$

x86 Machine Code, 10 bytes

8B 04 8B 87 04 93 89 04 8B C3

This is a function written in 32-bit x86 machine code that swaps the values at the specified indices in a given array. The array is modified in-place, and the function does not return a value.

A custom calling convention is used, requiring the function's parameters to be passed in registers:

  • The address of the array (pointer to its first element) is passed in the EBX register.
  • The zero-based index of element A is passed in the ECX register.
    (Assumed to be a valid index.)
  • The zero-based index of element B is passed in the EDX register.
    (Assumed to be a valid index.)

This keeps the size down and complies with all formal requirements, but does mean that the function cannot be easily called from other languages like C. You'd need to call it from another assembly-language program. (You could rewrite it to use any input registers, though, without affecting the byte count; there's nothing magical about the ones I chose.)

Ungolfed:

8B 04 8B     mov  eax, DWORD PTR [ebx+ecx*4]   ; get value of element A
87 04 93     xchg eax, DWORD PTR [ebx+edx*4]   ; swap element A and element B
89 04 8B     mov  DWORD PTR [ebx+ecx*4], eax   ; store new value for element A
C3           ret                               ; return, with array modified in-place
\$\endgroup\$
1
\$\begingroup\$

R, 34 bytes

pryr::f(`[<-`(a,c(m,n),a[c(n,m)]))
\$\endgroup\$
1
\$\begingroup\$

Java 8 + InverseY, 27 bytes

java.util.Collections::swap

Just calls the swap function... this is a method reference of the type Consumer3<List, Integer, Integer>.

Try it online! (header and footer for boilerplate & copy of Consumer3 interface)

\$\endgroup\$
  • \$\begingroup\$ You don't need to add " + InverseY". It's valid in vanilla Java 8. \$\endgroup\$ – Olivier Grégoire Jun 23 '17 at 15:57
1
\$\begingroup\$

JavaScript (ES2015), 66 57 49 bytes

A different (alas, longer) approach than previous JavaScript answers

(s,h,o,w=s.splice.bind(s))=>w(h,1,...w(o,1,s[h]))

Source

const swap = (arr, a, b, splice) => {
  splice(a, 1, ...splice(arr[b], 1, arr[a]))
}
\$\endgroup\$
  • 1
    \$\begingroup\$ (s,h,o,w=s.splice.bind(s))=>w(h,1,...w(o,1,s[h])) 49 bytes \$\endgroup\$ – Patrick Roberts Jun 24 '17 at 3:05
  • \$\begingroup\$ Forgot about them default args. Thanks! \$\endgroup\$ – sshow Jun 24 '17 at 23:58
0
\$\begingroup\$

awk, 31 bytes

{c=$a;$a=$b;$b=c;a=$1;b=$2}NR>1

Try it online!

Takes input in the format

1 2
1 2 3 4

and outputs as

2 1 3 4

(1-indexed).

Explanation

The entire program is a missing pattern with an action followed by a pattern with a missing action.

Since a missing pattern runs on each line, the code inside the braces runs for both input lines. The c=$a;$a=$b;$b=c; part swaps the two values at indices a and b (through the temporary variable c). This only has an effect on the second line, since on the first line a and b are not yet defined. The a=$1;b=$2 part defines a to be the first field and b to be the second field, which sets the appropriate values for the first part to run on the second line.

Since a missing action is equivalent to {print}, the pattern prints every line it matches. This pattern in particular is NR>1: that is, print whenever the line number is greater than 1, which happens to be line 2. This runs after the swapping of values has taken place, thus completing the task.

\$\endgroup\$
0
\$\begingroup\$

q/kdb+, 17 bytes

Solution:

{@[x;(|)y;:;x y]}

Example:

q){@[x;(|)y;:;x y]}[1 2 3 4;0 1]
2 1 3 4

Explanation:

A q version of the k answer by Simon. Apply the assign : function to x at indices reverse-y with value of x indexed at y. Broken down you can see more clearly:

q)x:1 2 3 4
q)y:0 1
q)x y
1 2
q)(|)y
1 0
q)x(|)y
2 1
q)@[x;(|)y;:;x y]
2 1 3 4
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.