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

Vyxal, 5 4 bytes

ɾḂYṅ

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Explained

ɾḂYṅ
ɾ     # range of [1,input+1]
 Ḃ    # dupe and reverse
   Y  # interleave
    ṅ # join with no spaces

Answer 2

tinylisp 2, 46 32 bytes

(\(N)(} ,(m u({ ,(z(1to N)(to1 N

An anonymous function that takes an integer and returns a string. Try it at Replit. Example session:

tl2> (\(N)(} ,(m u({ ,(z(1to N)(to1 N
(() (N) (} , (m u ({ , (z (1to N) (to1 N))))))
tl2> (def interleave-numbers _)
interleave-numbers
tl2> (interleave-numbers 4)
"14233241"

Explanation

(\(N)(} ,(m u({ ,(z(1to N)(to1 N))))))
(\                                   ) ; Lambda function
  (N)                                  ; that takes an integer N:
                 (z              )     ;  Zip together these two lists:
                   (1to N)             ;   Numbers from 1 to N
                          (to1 N)      ;   Numbers from N to 1
             ({ ,                 )    ;  Right fold on list/string concatenation
         (m u                      )   ;  Map unparse (convert to string) to each number
     (} ,                           )  ;  Left fold on list/string concatenation

The reason for the different folds is that right fold is more efficient, but by default it starts with an empty list, which when concatenating strings would make the result a list of charcodes instead of a string. Left-folding strings, by contrast, results in a string.

Answer 3

Actually, 9 bytes

R;R@Z♂iεj

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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 4

Mathematica, 36 32 bytes

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

thanks Martin

Answer 5

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 6

Perl 5, 28 bytes

27 bytes code + 1 byte for -a.

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

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Answer 7

AWK, 34 30 bytes

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

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

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Answer 8

Jelly, 6 bytes

RżU$FV

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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 9

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 10

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 11

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 12

><>, 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+$:n1-:?!;$!

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Answer 13

Powershell, 37 bytes

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

Answer 14

C (gcc), 43 bytes

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

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Answer 15

Python 3, 51 bytes

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

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Port of Mego's Python 2 answer.

Answer 16

C# (.NET Core), 53 bytes

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

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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 17

Japt -mP, 8 6 bytes

°Us+N´

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           :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 18

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 19

Kotlin, 52 50 bytes

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

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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}"}

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Answer 20

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 21

Zsh, 32 bytes

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

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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}

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Answer 22

Braingolf v0.7, 18 bytes

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 23

Pyt, 5 bytes

řĐ↔Žǰ

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ř          implicit input (n); push [1,2,...,n]
 Đ         duplicate top of stack
  ↔        flip the array at the top of the stack
   Ž       interleave the two arrays
    ǰ      join the contents of the array to create a string; implicit print

Answer 24

Ly, 12 bytes

Rsrp[ul,s`u]

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This one generates the 1..N series on the stack, and it generates the N..1 numbers by manipulating an accumulator on the backup cell.

R             - Generate range of 0..N (where N is from STDIN)
 s            - Save the upper limit to the backup cell
  rp          - Reverse the stack, delete the "0"
    [      ]  - While the stack isn't empty
     u        - Print next number in the range
      l,s     - Load backup cell, decrement and save
         `u   - Increment and print

Answer 25

Go, 90 bytes

import."fmt"
func f(n int)(o string){for i:=1;i<=n;i++{o+=Sprintf("%d%d",i,n-i+1)};return}

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Answer 26

J, 18 bytes

[:;],.&<&":&>:&i.-

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Answer 27

Julia 1.0, 33 22 bytes

~x=join([1:x x:-1:1]')

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1. The forward and reverse sequences are collected from ranges.
2. This method of interleaving arrays in Julia can be adapted for code golf.
3. The print command is vectorized with the "dot" operator so that no delimiters are shown.

• -9 bytes thanks to MarcMush: avoid reshaping the array
• -2 bytes thanks to MarcMush: replace print. with join

Answer 28

Pip, 10 bytes

-2 bytes thanks to DLosc!

Y\,ayWVRVy

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---

Pip, 12 bytes

(Y++,a)WVRVy

How?

(Y++,a)WVRVy  : One arg(integer n)
     a        : First arg
    ,         : Inclusive range from 1
  ++          : Increment
 Y            : Yank
         RV   : Alias for unary reverse
           y  : Yanked range
       WV     : Weave two iterables together, alternating their items

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Answer 29

Clojure, 52 bytes

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

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

Answer 30

Python 2.7, 39 bytes

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