Interleave strings

Question

Inspiration.* I cannot believe we have not had this challenge before:

Task

Given one or more printable ASCII strings, interleave them by taking one character from each string, cyclically until out of characters. If a string runs out of characters before the others, just skip that one from then on.

Examples

SIMPLE gives SIMPLE

POLLS and EPEES gives PEOPLELESS

LYES and APRONS gives LAYPERSONS

ABCDE and a c and 123 567 gives Aa1B 2Cc3D E567

"\n$?* and ​ (empty string) and ,(.)" (trailing space) gives ",\(n.$)?"* (trailing space)

---

^{* There are shorter APL solutions.}

Answer 1

Jelly, 1 byte

Z

Try it online!

The “transpose” built-in will do exactly this to a list of strings.

Answer 2

Python 2, 101 89 86 69 bytes

I'm hoping I can get this into a lambda somehow, shortening it down by making it recursive. It isn't ideal because you would hope transposing is shorter, unfortunately it isn't (from what I have managed to come up with so far).

f=lambda s:' '*any(s)and''.join(x[:1]for x in s)+f([x[1:]for x in s])

Old solutions:

w=input();o=''
while any(w):
 for i in range(len(w)):o+=w[i][:1];w[i]=w[i][1:]
print o

lambda s:''.join(''.join([c,''][c<' ']for c in x)for x in map(None,*[list(y)for y in s]))

w=input();o=''
while any(x>=' 'for x in w):
 for i in range(len(w)):o+=w[i][:1];w[i]=w[i][1:]
print o

^{thanks to mathmandan for making me feel dumb ;) saved me a bunch of bytes! (on an old solution)}

Answer 3

Perl 6, 34 32 bytes

{[~] flat roundrobin |$_».comb}

{roundrobin(|$_».comb).flat.join}

A lambda that takes an array of strings as argument, and returns a string.

(Try it online)

Answer 4

CJam, 4 bytes

qN/z

Try it online!

We can also write an unnamed function for 4 bytes, which expects a list of strings on top of the stack:

{zs}

Try it online!

Answer 5

Pyth - 3 bytes

Very simple, will add expansion later, on mobile.

s.T

Test Suite

s                         Join all the strings together
 .T                       Transpose, without chopping off overhang
  (Q implicit)

Answer 6

J, 13 bytes

({~/:)&;#\&.>

Try it online!

Based on the inspiration for this question.

Another way to do it takes 27 bytes but operates using transpose. Most of the bytes are to handle the automatically added zeroes from padding.

[:u:0<:@-.~[:,@|:(1+3&u:)&>

Explanation

({~/:)&;#\&.>  Input: list of boxed strings S
          &.>  For each boxed string x in S
        #\       Get the length of each prefix from shortest to longest
                 This forms the range [1, 2, ..., len(x)]
                 Rebox it
(    )         Operate on S and the prefix lengths
      &;         Raze both
   /:            Grade up the raze of the prefix lengths
 {~              Index into the raze of S using the grades
               Return

Answer 7

JavaScript (ES6), 52 46 bytes

f=([[c,...s],...a])=>s+a?c+f(s+s?[...a,s]:a):c

Takes input as an array of strings and outputs as a single string.

Test snippet

f=([[c,...s],...a])=>s+a?c+f(s+s?[...a,s]:a):c

g=a=>console.log("Input:",JSON.stringify(a),"Output:",JSON.stringify(f(a)))

g(["SIMPLE"])
g(["POLLS","EPEES"])
g(["LYES","APRONS"])
g(["ABCDE","a c","123 567"])
g(["\"\\n$?*",",(.)\" "]) // Backslash and quote are escaped, but in/output are correct

Answer 8

Haskell, 33 bytes

import Data.List
concat.transpose

Try it on Ideone. Usage:

Prelude Data.List> concat.transpose$["ABCDE","a c","123 567"]
"Aa1B 2Cc3D E567"

---

Without using a build-in: (38 34 bytes)

f[]=[]
f x=[h|h:_<-x]++f[t|_:t<-x]

Try it on Ideone. 4 bytes off thanks to Zgarb! Usage:

Prelude> f["ABCDE","a c","123 567"]
"Aa1B 2Cc3D E567"

Answer 9

C, 114 84 bytes

-20 bytes for not calculating the length.

i,b;f(char**s){b=1;while(b){i=-1;b=0;while(s[++i]>0)if(*s[i])putchar(*s[i]++),++b;}}

Accepts array of char pointers and requires last item to be a null-pointer (see usage).

Ungolfed and usage:

i,b;f(char**s){
 b=1;
 while(b){
  i=-1;
  b=0;
  while(s[++i]>0)
   if(*s[i])
    putchar(*s[i]++),++b;
 }
}


int main(){
 char*a[]={ 
//  "POLLS","EPEES"
//  "LYES","APRONS"
 "ABCDE","a c","123 567"
 ,0};
 f(a);
 puts("");
}

Answer 10

PHP, 68 67 bytes

for(;$f=!$k=$f;$i++)for(;y|$v=$argv[++$k];$f&=""==$c)echo$c=$v[$i];

Loops over command line arguments. Run with -r.

After the inner loop, $f is 1 when all strings are finished, 0 else (bitwise & casts ""==$c to int).
Next iteration of the outer loop: copy $f to $k (saves one byte from $k=0) and toggle $f:
When all strings are done, $f is now false and the loop gets broken.

Answer 11

Retina, 13 bytes

Byte count assumes ISO 8859-1 encoding.

O$#`.
$.%`
¶

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Explanation

O$#`.
$.%`

This is based on the standard transposition technique in Retina. We sort (O) all non-linefeed characters (.), by ($#) the number of characters in front of them on the same line ($.%`), i.e. their horizontal position.

The second stage then simply removes linefeeds from the input.

Answer 12

JavaScript (ES6), 46 bytes

f=([[c,...s],...a])=>c?c+f([...a,s]):a+a&&f(a)

<textarea oninput=o.textContent=f(this.value.split`\n`)></textarea><div id=o>

Answer 13

Java, 19+155=174 160

String f(java.util.Queue<String> q){String s,r="";while(!q.isEmpty()){s=q.poll();r+=s.isEmpty()?"":s.charAt(0);if(s.length()>1)q.add(s.substring(1));}return r;}

Ungolfed:

  String f(java.util.Queue<String> q) {
    String s, r = "";
    while (!q.isEmpty()) {
      s = q.poll();
      r += s.isEmpty() ? "" : s.charAt(0);
      if (s.length() > 1) {
        q.add(s.substring(1));
      }
    }
    return r;
  }

Output:

SIMPLE

PEOPLELESS

LAYPERSONS

Aa1B 2Cc3D E567

",(n.$)?"* ​

First modification: merged string declaration to save some bytes. Removed import, it was used by the main() method (not shown here) that also needed LinkedList. It is fewer bytes to referece Queue directly.

Answer 14

Python 2, 58 bytes

lambda*m:''.join(map(lambda*r:''.join(filter(None,r)),*m))

Try it online!

Answer 15

PHP, 77 bytes

Golfed

function($a){for($d=1;$d!=$s;$i++){$d=$s;foreach($a as$v)$s.=$v[$i];}echo$s;}

Anonymous function that takes an array of strings.

I'm sure this could be golfed more, but it's early. On each iteration, we grab the i-th letter from each given string and append it to our final string, one at a time. PHP just throws warnings if we access bits of strings that don't exist, so that's fine. We only stop when no changes have been made after looping through all the strings once.

I feel like the usage of $d can be golfed more, but it's early. :P

Answer 16

Actually, 7 6 bytes

Golfing suggestions welcome! Try it online!

Edit: -1 byte thanks to Teal pelican.

a Z♂ΣΣ

Ungolfing

          Implicit input each string.
a         Invert the stack so that the strings are in the correct order.
<space>   Get the number of items on the stack, len(stack).
Z         Zip all len(stack) strings into one, transposing them.
♂Σ        sum() every transposed list of chars into strings.
Σ         sum() again to join the strings together.

Answer 17

Python 3, 59 bytes

f=lambda s:s and s[0][:1]+f(s[1:]+[s[0][1:]][:s[0]>""])or""

A recursive function that takes a list of strings and returns a string. Try it online!

Explanation

We're going to build the interleaved string one character at a time. There are several cases to consider:

• Is the first string in the list nonempty? Take its first character, move the rest of it to the end of the list, and recurse.
• Is the first string in the list empty? Remove it from the list and recurse.
• Is the list of strings empty? Then we're done. Return empty string.

The implementation goes as follows:

f = lambda s: s and ... or ""

Define f to be a function of one argument, s, which is a list of strings. If s is nonempty, do the ... part (see below); otherwise, return "".

s[0][:1] + f(...)

Since we now know s is nonempty, we can refer to its first element, s[0]. We don't know yet whether s[0] is empty, though. If it isn't, we want to concatenate its first character to a recursive call; if it is, we can concatenate the empty string to a recursive call. Slicing everything left of index 1 works in both cases.

s[1:] + ...

The argument to the recursive call is going to be all but the first element of s, concatenated to either a 1-element list (the rest of s[0] if it was nonempty) or an empty list (if s[0] was empty).

[s[0][1:]][:s[0] > ""]

We put s[0][1:] (the part of s[0] after the first character, which is the empty string if s[0] is empty) into a singleton list and then take a slice of that list. If s[0] is empty, s[0] > "" is False, which is treated as 0 in a slice; thus, we slice from index 0 to index 0--an empty slice. If s[0] is not empty, s[0] > "" is True, which is treated as 1; thus, we slice from index 0 to index 1--the whole 1-element list.

---

If lists of single-character strings can be used in place of strings, this solution can be 54 bytes:

f=lambda s:s and s[0][:1]+f(s[1:]+(s[0]and[s[0][1:]]))

Try it online!

Answer 18

K (ngn/k), 13 bytes

<!/,/'(!#:)'\

Try it online!

Port of miles's J and DLosc's BQN. For a language with flatten and grade/sort-by but not zip-longest, I guess this is one of the best possible approaches.

More straightforward port would be {(,/x)@<,/!'#'x} because K doesn't have fork or over. But it is possible to use K train by abusing repeat-scan:

<!/,/'(!#:)'\
      (   )'\   Given a list of strings, apply this to each list
                until convergence and collect all intermediate values
                (including the input):
       !#:        get the indices of the list
                This always converges after step 1, giving (x;!#x)
   ,/'          Flatten each
<!/             Sort first by second

Answer 19

Elisp, 242 Bytes

This is literally my first time so it could probably be smaller and it only works with two arguments (assigned to a and b)

(setq a (elt argv 0))(setq b (elt argv 1))(with-temp-buffer (insert a b)(setq i 0)(ignore-errors (while (< i (length b))(progn (goto-char (+(length a) 2 i))(transpose-chars (* -1 (- (length a) i 1)))(setq i (+ i 1)))))(princ (buffer-string)))

Ungolfed version

(setq a (elt argv 0))
(setq b (elt argv 1))
(with-temp-buffer
  (insert a b)
  (setq i 0)
  (ignore-errors
(while (< i (length b))
  (progn
    (goto-char (+(length a) 2 i))
    (transpose-chars (* -1 (- (length a) i 1)))
    (setq i (+ i 1))
    )
  )
)
  (princ (buffer-string))
  )

But if you strip away all the boilerplate code its actually much shorter (134).

(insert a b)
(setq i 0)
(while (< i (length b))
  (goto-char (+(length a) 2 i))
  (transpose-chars (* -1 (- (length a) i 1)))
  (setq i (+ i 1))
  )

Elisp really sucks. Someone needs to modify it for golfing if haven't already.

Answer 20

Bash + GNU utilities, 55

 eval paste `sed "s/.*/<(fold -1<<<'&')/g"`|tr -d \\n\\t

I/O via STDIN (line-separated) and STDOUT.

The sed formats each line to a bash process substitution. These are then evaled into paste to do the actual interleaving. tr then removes unnecessary newlines and tabs.

Ideone.

Answer 21

PHP, 63 bytes

Note: uses IBM-850 encoding

for(;$s^=1;$i++)for(;n|$w=$argv[++$$i];$s&=$x<~■)echo$x=$w[$i];

Run like this:

php -r 'for(;$s^=1;$i++)for(;n|$w=$argv[++$$i];$s&=$x<~■)echo$x=$w[$i];' "\"\n\$?*" "" ",(.)\" " 2>/dev/null;echo
> ",\(n.$)?"* 

Explanation

for(                       # Iterate over string index.
  ;
  $s ^= 1;                 # Continue until $s (stop iterating) is 1.
                           # Flip $s so each iteration starts with $s
                           # being 1.
  $i++                     # Increment string index.
)
  for(
    ;
    "n" | $w=$argv[++$$i]; # Iterate over all input strings. OR with "n"
                           # to allow for empty strings.
    $s &= $x<~■            # If last character printed was greater than
                           # \x0 (all printable chars), set $s to 0,
                           # causing the loop to continue.
  )
    echo $x = $w[$i];      # Print char $i of current string.

Answer 22

Python 3, 75 Bytes

I know the other Python one is shorter, but this is the first time I've used map ever in my life so I'm pretty proud of it

lambda n:''.join(i[k]for k in range(max(map(len,n)))for i in n if len(i)>k)

Answer 23

Python 2, 128 96

I was hoping not to have to use itertools

a=lambda a:"".join([i for i in reduce(lambda: b,c:b+c, map(None,*map(lambda m:list(m),a)) if i])

Ungolfed

 a=lambda a:                              #Start a lambda taking in a
    "".join(                              #Join the result together with empty string
        [i for i in reduce(               #For every item, apply the function and 'keep'
           lambda: b,c:b+c,               #Add lists from...
                map(None,*map(            #None = Identity function, over a map of...
                    lambda m:list(m), a)  #list made for mthe strings m
                   ) if i                 #truthy values only (otherwise the outer map will padd with None.
       ])

Answer 24

R, 73 bytes

for(i in 1:max(nchar(s<-scan(,""))))for(j in seq(s))cat(substr(s[j],i,i))

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Explanation: very simple (but verbose), just loop through printing ith character of the jth string. Fortunately, substr returns an empty string if given an out-of-range input.

Answer 25

Perl 5, + -0n 24 bytes

1while s/^./!print$&/gem

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Explanation

-0n slurps the input into $_ and repeatedly replaces the first char of each line with the result of negating the print of each match (which is the empty string since print returns 1 by default)

Answer 26

Factor, 22 bytes

[ round-robin ""like ]

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Answer 27

Pip, 3 bytes

WVg

Try it online!

The WV (weave) builtin does exactly this. It's commonly used as a binary operator to weave two iterables together, but it can also be used as a unary operator to weave a list of iterables together. Here, g is the list of command-line arguments.

Answer 28

BQN, 23 10 bytes

Massive savings by porting miles's J solution

↕∘≠¨⍋⊸⊏○∾⊢

Anonymous tacit function that takes a list of strings and returns a string. Try it at BQN online!

Explanation

↕∘≠¨⍋⊸⊏○∾⊢
   ¨         For each string in the argument list:
  ≠            Take its length
↕∘             and make a range from 0 to that number (exclusive)
             With that list of lists of indices as the left argument
          ⊢  and the original list of lists of strings as the right argument:
        ○∾     Flatten both lists
      ⊸⊏       Index into the flattened string using
    ⍋          the grade-up of the flattened list of indices

Grade up, for those unfamiliar with APL-related languages, is an operation that takes a list and returns a permutation of its indices, such that if the items in the list were reordered according to the order of the indices, the list would become sorted ascending. For example:

Indices:  0 1 2 3 4 5
List:     3 1 4 1 5 9

Grade up: 1 3 0 2 4 5
Sorted:   1 1 3 4 5 9

This operation is useful for sorting one list according to the values of another list, as seen in the above solution.

Answer 29

Uiua ^SBCS, 9 bytes

⊏⍏∩/◇⊂⍚°⊏

Try it!

Explanation

⊏⍏∩/◇⊂⍚°⊏­⁡​‎‎⁡⁠⁢⁣‏⁠‎⁡⁠⁢⁤‏⁠‎⁡⁠⁣⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁣‏⁠‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁢‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌­
      ⍚°⊏  # ‎⁡get list of length-ranges of input
  ∩/◇⊂     # ‎⁢flatten both ranges and input
 ⍏         # ‎⁣get rise (grade up) of flattened ranges
⊏          # ‎⁤index into input

Port of miles' J solution.

Answer 30

Go, 140 bytes

func(s[]string)(o string){for L,e,i,k:=len(s),0,0,0;e!=L;i++{
r:=s[i%L]
if len(r)<=k{e++;continue}
o+=r[k:k+1]
if i%L==L-1{k++;e=0}}
return}

Attempt This Online!

Explanation

func(s[]string)(o string){
for L,e,i,k:=len(s),0,0,0;e!=L;i++{ // loop while the the whole slice is not "empty"
j:=i%L // get the index of the string to look at
r:=s[j] // the string to try to take from
if len(r)<=k{e++;continue} // if the current k-index is farther along than the string's length, skip it
o+=r[k:k+1] // add the first character of the string to the output
if j==L-1{k++;e=0}} // move to the next character and reset the "empty" counter if at the end of the slice
return}