Interleave numbers from 1 to n, with the same numbers reversed

Question

A simple one:

Take a positive integer n less than 1000, and output the integers from 1 to n interleaved with the integers from n to 1. You must concatenate the numbers so that they appear without any delimiters between them.

Test cases:

n = 1
11

n = 4
14233241

n = 26
12622532442352262172081991810171116121513141413151216111710189198207216225234243252261

n = 100
110029939849759669579489399210911190128913881487158616851784188319822081218022792378247725762675277428732972307131703269336834673566366537643863396240614160425943584457455646554754485349525051515052495348544755465645574458435942604161406239633864376536663567346833693270317130722973287427752676257724782379228021812082198318841785168615871488138912901191109299389479569659749839921001

This is code-golf so the shortest submission in bytes in each language wins. Explanations are encouraged.

Answer 1

JavaScript (ES6), 30 bytes

f=(n,k=1)=>n?f(n-1,k+1)+n+k:''

How?

This is pretty straightforward but it's worth noting that the string is built from tail to head. An empty string at the beginning is appended last and allows the coercion of the final result to a string to happen.

Below is the detail of the recursion for f(4):

f(4) =                                            // initial call
f(4, 1) =                                         // applying the default value to k
f(3, 2) + 4 + 1 =                                 // recursive call #1
(f(2, 3) + 3 + 2) + 4 + 1 =                       // recursive call #2
((f(1, 4) + 2 + 3) + 3 + 2) + 4 + 1 =             // recursive call #3
(((f(0, 5) + 1 + 4) + 2 + 3) + 3 + 2) + 4 + 1 =   // recursive call #4
((('' + 1 + 4) + 2 + 3) + 3 + 2) + 4 + 1 =        // n = 0 --> end of recursion
'' + 1 + 4 + 2 + 3 + 3 + 2 + 4 + 1 =              // final sum
'14233241'                                        // final result

Test cases

f=(n,k=1)=>n?f(n-1,k+1)+n+k:''

console.log(f(1))
console.log(f(4))
console.log(f(26))
console.log(f(100))

Answer 2

Python 2, 46 bytes

lambda n:''.join(`x+1`+`n-x`for x in range(n))

Thanks to ovs for 4 bytes

Try it online!

Explanation:

lambda n:''.join(`x+1`+`n-x`for x in range(n))
lambda n:                                      # anonymous lambda taking one parameter n
                 `x+1`+`n-x`                   # `x` is repr(x) which is equivalent to str(x) for integers less than INT_MAX
                            for x in range(n)  # integers x in [0, n)

Answer 3

-5 thanks to Ørjan Johansen

Haskell, 33 bytes

f n=do a<-[1..n];[a,n-a+1]>>=show

Try it online!

Answer 4

R, 35 bytes

n=scan();cat(rbind(1:n,n:1),sep="")

Try it online

rbind(1:n,n:1) creates a 2 row matrix with 1 to n in the first row and n to 1 in the second. The cat function collapses this matrix, reading down each column.

Answer 5

Bash, 25 bytes

printf %s`seq $1 -1 1|nl`

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Prints decreasing sequence, number lines increasing and printf joins lines

Space delimited, 20 bytes : seq $1 -1 1|nl|xargs

Answer 6

05AB1E, 6 5 bytes

Saved a byte using the new interleave built-in as suggested by rev

LÂ.ιJ

Try it online!

Explanation

L        # range [1 ... input]
 Â       # create a reversed copy
  .ι     # interleave the lists
    J    # join

Answer 7

CJam, 10 bytes

ri,:)_W%]z

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Explanation

ri   e# Read input and convert to integer N.
,    e# Turn into range [0 1 ... N-1].
:)   e# Increment to get [1 2 ... N].
_W%  e# Duplicate and reverse the copy.
]    e# Wrap both in an array to get [[1 2 ... N] [N ... 2 1]]
z    e# Transpose to get [[1 N] [2 N-1] ... [N-1 2] [N 1]]
     e# This list is printed implicitly at the end of the program,
     e# but without any of the array structure (so it's essentially flattened,
     e# each number is converted to a string and then all the strings
     e# are joined together and printed).

Answer 8

Ruby, 29 bytes

->n{n.times{|x|$><<x+1<<n-x}}

Explanation:

->n{n.times{|x|                # x in range [0..n-1]
               $><<            # output on console
                   x+1<<n-x}}  # x+1, then n-x

Try it online!

Answer 9

Haskell, 65 48 47 bytes

1 byte saved thanks to Laikoni:

f n=show=<<(\(l,r)->[l,r])=<<zip[1..][n,n-1..1]

---

6 bytes saved thanks to nimi:

f n=show=<<(\(l,r)->[l,r])=<<zip[1..n][n,n-1..1]

---

Previous answer and explanation:

f n=concatMap show$concatMap(\(l,r)->[l,r])(zip[1..n][n,n-1..1])

There's already a better Haskell answer here, but I'm new to both Haskell and code golfing, so I may as well post it :)

This function zips the list [1..n] with its reverse, resulting in a list of tuples.

[(1,n),(2,n-1),(3,n-2)..(n,1)]

Then it uses concatMap to map a lambda to this list of tuples that results in a list of lists...

[[1,n],[2,n-1],[3,n-2]..[n,1]]

...and concatenates it.

[1,n,2,n-1,3,n-2..n,1]

Then a final concatMap maps show to the list and concatenates it into a single string.

f 26 "12622532442352262172081991810171116121513141413151216111710189198207216225234243252261"

Answer 10

Whitespace, 71 bytes

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Explanation

sssn  ; push 0 - seed the stack with 0 (this will be our 1->n counter, a)
sns   ; dup
tntt  ; getnum - read n (stored on the heap)
sns   ; dup
ttt   ; retr - pull n onto the stack (this will be our n->1 counter, b)
nssn  ; label 'loop'
snt   ; swap - bring a to the top
ssstn ; push 1
tsss  ; add - increment a
sns   ; dup
tnst  ; putnum - output a as a number
snt   ; swap - bring b to the top
sns   ; dup
tnst  ; putnum - output b as a number
ssstn ; push 1
tsst  ; sub - decrement b
sns   ; dup
ntstn ; jez 'exit' if b is 0
nsnn  ; jmp 'loop'

The first couple of instructions are needed to set up the stack correctly, Whitespace's input commands write to the heap so we need to copy b (the input value) back onto the stack. We start with a = 0 since it is shorter to declare 0 instead of 1 (saves a byte) and we only need to reorder the increment instruction to cope. After that we just loop and increment a, output a, output b, decrement b, until b reaches 0 (checked after the decrement).

Answer 11

Röda, 21 19 bytes

{seq 1,_<>seq _1,1}

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This is an anonymous function that takes input from the stream.

Explanation

{seq 1,_<>seq _1,1}               Anonymous function, takes integer n from the stream
        <>                        Interleave
 seq 1,_                            the range 1 .. n with
          seq _1,1                  the range n .. 1

Answer 12

Pyth, 7 bytes

jksC_BS

Try it online: Demonstration

Explanation:

jksC_BSQ   implicit Q (=input number) at the end
      SQ   create the range [1, ..., Q]
    _B     bifurcate by inversion, this gives [[1, ..., Q], [Q, ..., 1]]
  sC       zip and flatten result
jk         join to a string

Answer 13

Octave, 29 bytes

@(n)printf("%d",[1:n;n:-1:1])

Try it online!

Answer 14

Perl 6, 20 bytes

{[~] 1..*Z~($_...1)}

Test it

With an input of 100000 this takes roughly 10 seconds, including compilation and printing the output.

Expanded:

{                # bare block lambda with implicit parameter ｢$_｣

  [~]            # reduce using concatenation operator ｢&infix:«~»｣
                 # (shorter than ｢join '',｣)

    1 .. *       # Range from 1 to infinity

    Z~           # zip using concatenation operator

    ( $_ ... 1 ) # deduced sequence starting at the input
                 # going down to 1
}

The Z~ needs the ~ because otherwise it generates a list of lists which will stringify with spaces.

There is no need to limit the Range starting at 1, because Z stops when any of the input lists run out.
This saves two bytes (a space would be needed after $_)

Answer 15

Java 61 bytes

(int n)->{for(int i=0;i<n;System.out.print(i+1+""+(n-i++)));}

Answer 16

Jelly, 5 bytes

RṚĖVV

Try it online!

How it works

RṚĖVV  Main link. Argument: n

R      Range; yield [1, ..., n].
 Ṛ     Reverse; yield [n, ..., 1].
  Ė    Enumerate; yield [[1, n], ..., [n, 1]].
   V   Eval; convert each flat array to a string, interpret it as a Jelly program,
       and yield the output. This concatenates the integers in each pair, yielding
       a flat array of integers
    V  Repeat the previous step, concatenating the intgegers from before.

Answer 17

Clojure, 61 bytes

#(let[a(range 1(+ 1 %))](apply str(interleave a(reverse a))))

Literally does what is asked. I believe it can be outgolfed by a less trivial solution.

See it online

Answer 18

PHP, 36 35 29 bytes

for(;$argn;)echo++$i,$argn--;

Saved one byte thanks to Jörg Hülsermann.
Saved six bytes thanks to Christoph.

Answer 19

Aceto, 25 22 bytes

)&
pX`=
(pl0
id@z
r}Z)

Explanation:

We read an integer and put it on two stacks.

id
r}

On one, we call range_up (Z), on the other range_down (z), then we set a catch mark to be able to return to this place later:

  @z
  Z)

We then check if the current stack is empty and exit if so:

 X`=
  l0

Otherwise, we print from both stacks and jump back to the catch mark:

)&
p
(p

Answer 20

Brachylog, 10 9 bytes

⟦₁g↔ᶻczcc

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Explanation

⟦₁           [1, …, Input]
  g↔ᶻc       [[1, …, Input],[Input, …, 1]]
      z      Zip
       cc    Concatenate twice

Answer 21

R, 41 bytes

pryr::f(for(i in 1:x){cat(i);cat(x-i+1)})

pryr::f() creates a function that takes one input. Loops over 1:x and prints each element of 1:x along with each element of x:1. Prints to STDOUT.

Answer 22

MATL, 13 11 9 bytes

2 bytes saved thanks to @Luis

:tPv1eVXz

Try it at MATL Online

Explanation

        % Implicitly grab input as a number, N
:       % Create an array from 1..N
tP      % Create a reversed copy
v       % Vertically concatenate the two
1e      % Reshape it into a row vector
V       % Convert to a string
Xz      % Remove whitespace and implicitly display

Answer 23

V, 20 bytes

ywo1@"­ñykPjñkògJ

Try it online!

Explain:

yw                    ' Copy the input number (for looping later)
  o1                 ' Insert a 1 under the input (on a newline)
     @"               ' [Copy register] number of times
       ­ñ      ñ       ' Do the thing inside of this loop
        ykP           ' Copy the current line and line above it, and paste above both
           j        ' decrement the current (top) number, and increment the one below
               k      ' Go to the top line
                ògJ   ' recursively join all of the lines

Answer 24

Scala, 43 bytes

It's not the best but it's my first code golf.

n=>1.to(n).foldLeft("")((s,x)=>s+x+(n-x+1))

Answer 25

Cubix, 17 bytes

....1I>sO)su.@?(O

Try it online!

cubified:

    . .
    . .
1 I > s O ) s u
. @ ? ( O . . .
    . .
    . .

Pushes 1, reads in the input (I), then enters the loop which swaps the top of the stack, outputs it, increments, swaps, outputs the top of the stack, decrements, and stops if the top of the stack is 0.

Answer 26

Common Lisp, 63 54 bytes

(lambda(n)(dotimes(i n)(format t"~a~a"(1+ i)(- n i))))

Try it online!

Answer 27

K (ngn/k), 16 bytes

{,//$r,'|r:1+!x}

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Answer 28

MathGolf, 5 bytes

{îkï-

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Explanation:

{      Run a for loop over implicit input
 î     Push 1 based index of loop
  k    Push inputted number
   ï-  Subtract 0 based index of loop
       Implicitly output all this joined together

Answer 29

Mouse-2002, 32 30 bytes

-2 moved conditional to start of loop (z.^ ... ) instead of (... z.0>^)

?n:n.z:(z.^a.1+a:a.!z.!z.1-z:)

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Explanation:

?n:                                 ~ get input and store in n
   n.z:                             ~ copy n into z
       (z.^                         ~ stop if z equals 0
           a.1+a:                   ~ add 1 to a
                 a.!                ~ print a
                    z.!             ~ print z
                       z.1-z:)      ~ substract 1 from z

Answer 30

Actually, 9 bytes

R;R@Z♂iεj

Try it online!, or run all test cases

Explanation:

R;R@Z♂iεj
R          range(1, n+1)
 ;R        duplicate and reverse
   @Z      swap and zip
     ♂i    make 1D
       εj  join with empty string

Answer 31

QBIC, 18 bytes

[:|Z=Z+!a$+!b-a+1$

Explanation

[:|       FOR a = 1 to n
Z=Z+      Add to Z$ 
!b$+      a cast of the loop counter as string
!b-a+1$   and a cast of (n+1) minus the loop counter to string
          Z$ is printed implicitly at the end of QBIC

Answer 32

Mathematica, 36 32 bytes

Row@Riffle[x=Range@#,Reverse@x]&

thanks Martin

Answer 33

CJam, 13 12 bytes

1 byte removed thanks to Martin Ender

ri:X{_)X@-}%

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Explanation

ri               e# Read integer n
  :X             e# Copy n into variable X
     {     }%    e# Map over the array [0 1 2 ... n-1]
      _          e# Duplicate
       )         e# Add 1
        X        e# Push n
         @       e# Rotate
          -      e# Subtract

Answer 34

Perl 5, 28 bytes

27 bytes code + 1 byte for -a.

map{print$_,"@F"+1-$_}1..$_

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Answer 35

Braingolf v0.7, 18 bytes [non-competing]

VR.MUvU&,R{vMR}>&_

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Explanation:

VR.MUvU&,R{vMR}>&_  Implicit input of n to stack
VR                  Create new stack then return to main
  .M                Duplicate n and move duplicate to new stack
    U               Replace stack with 1-n, where n is last item on stack
     vU             Replace 2nd stack with 1-n
       &,           Reverse second stack
         R          Return to main
          {   }     Map loop, runs for each item in the stack
           vMR      Move to next stack, move last item to main stack
               >    Cleanup after loop
                &_  Print all items in stack with no delimiter

Answer 36

Japt -P, 11 8 bytes

Saved 2 byte thanks to @Shaggy

õ í1õU)c

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Explanation:

õ í1õU)c
õ              Range [1...Input]
  í            Pair with:
   1õU           Range [Input...1]  
      )c       Flatten   
-P             Join into a string  

Answer 37

Powershell, 37 bytes

param($n)-join(1..$n|%{$_,($n-$_+1)})

Answer 38

Jelly, 6 bytes

RżU$FV

Try it online!

Explanation:

         Example: n = 6
R        Create a range from 1 to n             [1, 2, 3, 4, 5, 6]
  U$     and a reversed copy of that same range [6, 5, 4, 3, 2, 1]
 ż       and interleave them                    [ [1, 6], [2, 5, ... ]
    F    Flatten the list                       [1, 6, 2, 5, 3, 4 ...]
     V   And Eval: Jelly code consisting of only numbers would simply print those numbers,

Answer 39

Retina, 20 bytes

.+
$*__
\B
$.`$.'
_

Try it online! Includes test cases. Explanation: Each pair of numbers sums to n+1, so we convert n to unary and add 1. Then, we match between each pair of _s, counting the number of _s to the left and right. This generates the pairs of numbers. Finally we delete the _s now that they've served their purpose.

Answer 40

><>, 16 12+3 = 19 15 bytes

l:n}:n1-:?!;

Input is expected on the stack, so +3 bytes for the -v flag.

Thanks to @TealPelican for pointing out a very clever way to save 4 bytes by using the size of the stack itself - on the first iteration it'll be 1, then 2, then 3... That way, the first number in each pair manages itself, no manual incrementing required!

Previous version:

1:n1+$:n1-:?!;$!

Try it online!

Answer 41

Perl 5, 25 21 + 1 = 26 22 bytes

Takes input from stdin, without a trailing newline.

Runs with the -n flag:

print++$x.$_--while$_

4 bytes saved thanks to Dada.

Answer 42

C (gcc), 43 bytes

a;f(n){for(a=1;n;a++)printf("%d%d",a,n--);}

Try it online!

Answer 43

Python 3, 51 bytes

lambda n:''.join("%d%d"%(x+1,n-x)for x in range(n))

Try it online!

Port of Mego's Python 2 answer.

Answer 44

Japt -mP, 8 6 bytes

°Us+N´

Try it

           :For each U in the range [0,input)
°U         : Prefix increment U
  s        : Convert to a string
   +       : Append
    N´     :  Postfix decrement the (singleton) array of inputs
           :Implicitly join and output

Answer 45

C# (.NET Core), 53 bytes

a=>{for(int i=0;++i<=a;)Console.Write(i+""+(a-i+1));}

Try it online!

Uses an Action delegate to pull in the input and not require a return.

Ungolfed:

a => {
    for(int i = 0; ++i <= a;)           // initialize i and increment until a
        Console.Write(i + "" + (a - i + 1));    // output i and "inverse" of i
                                                    // empty string required to set as string
                                                    // parentheses required because C# is dumb and can't figure out what a minus sign is supposed to do without them
}

Answer 46

Forth (gforth), 41 bytes

: f 1+ dup 1 do i 1 .r 1- dup 1 .r loop ;

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Explanation

Loops from 1 to n and outputs the index as well as n-index in a right-aligned space of size 1 (forces no space after output)

Code Explanation

: f               \ start new word definition
  1+ dup 1        \ set up arguments to loop from 1 to n
  do              \ start loop
    i 1 .r        \ output the index in a right-aligned space of size 1
    1- dup        \ subtract 1 from current value of n and duplicate
    1 .r          \ output new n in right-aligned space of size 1
  loop            \ end loop
;                 \ end word definition

Answer 47

Kotlin, 52 50 bytes

fun g(e:Int)=(1..e).forEach{print("$it${e-it+1}")}

Try it online

Older version, the one above also prints it, the one below produces a string.

fun y(n:Int)=(1..n).joinToString(""){"$it${n-it+1}"}

Try it online!

Answer 48

APL (Dyalog Unicode), 10 bytes^SBCS

(∊⍪,⌽)⍕¨∘⍳

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Explanation:

(∊⍪,⌽)⍕¨∘⍳  ⍝ Monadic function train
       ⍕¨∘⍳  ⍝ Generate integers from 1 to input, convert each to string
             ⍝ Call this vector "X"
  ⍪          ⍝ "Table" X, making a single-column matrix
   ,⌽       ⍝ Reverse X and concatenate it with the above table
             ⍝ This results in a two-column matrix with integers from
             ⍝ 1 to N and N to 1 side-by-side
 ∊           ⍝ "Enlist" the above - left-to-right, top-to-bottom,
             ⍝ recursively concatenate all items in the matrix

Answer 49

AWK, 34 30 bytes

{for(i=1;i<=$1;i++)printf i$1-i+1}

{for(;i<$1;i++)printf i+1$1-i}

Try it online!

Answer 50

Clojure, 52 bytes

#(apply str(mapcat(fn[i][(inc i)(- % i)])(range %)))

Concatenates pair-wise integers (head and tail of the seq).

Answer 51

Python 2.7, 39 bytes

x=''.join(`i+1`+`n-i`for i in range(n))

Answer 52

Pyke, 9 bytes

SD_],sm`s

Try it here!

Answer 53

Mathematica, 52 bytes

As a traditional function:

f[x_] := StringJoin[ToString /@ Riffle[Range[x], Reverse[Range[x]]]]

As an anonymous function

(x = Range[#]; StringJoin[ToString /@ Riffle[x, Reverse[x]]]) &

As a code block that accepts the input in the variable n

x = Range[n]; StringJoin[ToString /@ Riffle[x, Reverse[x]]]

The code is pretty simple -- generate the list of integers with Range[], interleave this list (using Riffle[]) with a copy of the list that has been reversed, then convert all the integers to strings and concatenate them with StringJoin[].

Could have done it in 30 bytes if we were allowed to leave the answer in the form of a list (thus, {1,4,2,3,3,2,4,1} rather than "14233241"). Also, in the hypothetical Mthmtca with short command names it would have been something like 21 bytes.

Performance is tolerable for n <= 1,000,000. I didn't test past that.

Answer 54

R, 43 39 37 bytes

This is an anonymous function.

function(n)cat(rbind(1:n,n:1),sep="")

It's 35 bytes if I use pryr::f() like in the other answer:

pryr::f(cat(rbind(1:n,n:1),sep=""))

but needs to be installed (Enter install.packages("pryr") in the R console).

Answer 55

Python 2, 67 bytes

A simple one, just for the collection :)

n=input()+1
i=1
s=""
while i<n:
    s+=`i`+`n-i`
    i+=1
print s

Answer 56

Hexagony, 15 bytes

{?\"!\{$.(.@).!

Embiggened:

   { ? \
  " ! \ {
 $ . ( . @
  ) . ! .
   . . .

Try it online!

Answer 57

LOGO, 48 bytes

to f :n
for[i 1 :n][(type :i :n+1-:i)]show "
end

Define a function f that when invoke with parameter = number n, print the result string.

Answer 58

braingasm, 9 bytes

;[>+:<:-]

Simple: Read a number from stdin to the current cell; While the current cell is not zero, go to next cell (initially 0), increase it, output its value, go back, output that value and decrease it.

Answer 59

Java 8, 53 52 bytes

n->{for(int i=1;n>0;System.out.print(n--+""+i++));};

Try it online!

-1 bytes thanks to Sara J!

Answer 60

Kotlin, 44 bytes

{n->(1..n).joinToString(""){"$it${n-it+1}"}}

{n->                                          // n is Int input
    (1..n)                                    // range from 1 to end
          .joinToString("")                   // join items with no separator
                           {                  // transform by
                            "$it${n-it+1}"    // current number and number from end of range
                                          }}  

Lambda that takes an Int and returns a String.

Try it online!

Answer 61

dc, 17 bytes

[znddn1-d0<M]dsMx

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Answer 62

APL(NARS), 13 chars, 26 bytes

{∊k,¨⌽k←⍕¨⍳⍵}

How to use & test:

  f←{∊k,¨⌽k←⍕¨⍳⍵}
  f 1
11
  f 26
12622532442352262172081991810171116121513141413151216111710189198207216225234243252261

Answer 63

Ink, 47 bytes

=f(n)
~temp i=n
-(l)
~i--
{n-i}{1+i}{i:->l}->->

Try it online!

Answer 64

Stax, 5 bytes

çK^↑Ñ

Run and debug it

Answer 65

Zsh, 32 bytes

repeat $1 printf $[++i]$[$1+1-i]

Try it online!

I like trying to beat Bash+coreutils with pure zsh, but sadly that is not the case here.

---

Cooler, but one byte longer (33 bytes):

a=({$1..1})
<<<${(j::)${(n)a}:^a}

Try it online!