Stitch Together a Palindrome from Palindromic Substrings

Question

Given a string l, find all palindromic substrings p of l (including duplicates and single character strings). Next, rearrange all sub-strings in p into a valid palindrome (there may be multiple correct answers). If it is not possible to rearrange p into a single palindrome, your program may have undefined behavior (error, stack-overflow, exiting, hanging/untimely murder of John Dvorak, etc...)

---

Examples

Valid Test Cases

l = anaa
p = ['a', 'n', 'a', 'a', 'aa', 'ana']
result = anaaaaana or aanaaanaa or aaananaaa
l = 1213235
p = ['1', '2', '1', '3', '2', '3', '5', '121', '323']
result = 1213235323121
l = racecar
p = ['r', 'a', 'c', 'e', 'c', 'a', 'r', 'cec', 'aceca', 'racecar']
result = racecarcecaacecracecar (there are others)
l = 11233
p = ['1', '11', '1', '2', '3', '33', '3']
result = 113323311 or 331121133
l = abbccdd
p = ['a', 'b', 'bb', 'b', 'c', 'cc', 'c', 'd', 'dd', 'd']
result = bbccddaddccbb or ccbbddaddbbcc or (etc...)
l = a
p = ['a']
result = a

Invalid Test Cases (Not Possible)

l = 123456789
p = ['1', '2', '3', '4', '5', '6', '7', '8', '9']
result = <not possible, behavior undefined>
l = hjjkl
p = ['h', 'j', 'jj', 'j', 'k', 'l']
result = <not possible, behavior undefined>
l = xjmjj
p = ['x', 'j', 'jmj', 'm', 'j', 'jj', 'j']
result = <not possible, behavior undefined>

---

Rules

• If the input word is a palindrome itself, it will always be valid as input.
• Only one substring should be returned, which one you choose is arbitrary as long as it's valid.
• If the input has no viable output, your code may have undefined behavior.
• Inputs will only contain ASCII-Printable characters between 0x20-0x7E.
• This is code-golf, lowest byte-count is the winner.

• Fatalize: The first proposed result for "abbccdd" is wrong: the last two letters should be "bb", not "dd".
• Shaggy: Can we return an array of substrings, rather than a single string?
• alephalpha: Can I take a list of characters as input?
• John Dvorak: By hanging being acceptable behavior, do you mean hanging the person who gave it input?
• Magic Octopus Urn: @JohnDvorak clarified.

Answer 1

Brachylog, 10 bytes

{s.↔}fpc.↔

Try it online!

Fails (i.e. prints false.) if not possible.

Explanation

{ }f Find all...
 s. ...substrings of the input...
 .↔ ...which are their own reverse
 p Take a permutation of this list of palindromes
 c. The output is the concatenation of this permutation
 .↔ The output is its own reverse

Answer 2

Coconut, 140 bytes

s->p(map(''.join,permutations(p(v for k in n(s)for v in n(k[::-1])))))[0]
from itertools import*
n=scan$((+))
p=list..filter$(x->x==x[::-1])

Try it online!

Answer 3

JavaScript (ES6), 193 bytes

"Look Ma, no permutation built-in!" (So yes ... it's long ...)

Returns an empty array if there's no solution.

f=(s,a=[].concat(...[...s].map((_,i,a)=>a.map((_,j)=>s.slice(i,j+1)))).filter(P=s=>[...s].reverse().join``==s&&s),m=S=[])=>S=a.map((_,i)=>f(s,b=[...a],[...m,b.splice(i,1)]))>''?S:P(m.join``)||S

Demo

f=(s,a=[].concat(...[...s].map((_,i,a)=>a.map((_,j)=>s.slice(i,j+1)))).filter(P=s=>[...s].reverse().join``==s&&s),m=S=[])=>S=a.map((_,i)=>f(s,b=[...a],[...m,b.splice(i,1)]))>''?S:P(m.join``)||S
console.log(f('anaa'))
console.log(f('1213235'))
console.log(f('hjjkl'))
console.log(f('a'))

How?

Let's split the code into smaller parts.

We define P(), a function that returns s if s is a palindrome, or false otherwise.

P = s => [...s].reverse().join`` == s && s

We compute all substrings of the input string s. Using P(), we isolate the non-empty palindromes and store them in the array a.

a = [].concat(...[...s].map((_, i, a) => a.map((_, j) => s.slice(i, j + 1)))).filter(P)

The main recursive function f() takes a as input and compute all its permutations. It updates S whenever the permutation itself is a palindrome (once joined), and eventually returns the final value of S.

f = ( // given:
 a, // a[] = input array
 m = S = [] // m[] = current permutation of a[]
) => // and S initialized to []
 S = a.map((_, i) => // for each element at position i in a[]:
 f( // do a recursive call with:
 b = [...a], // b[] = copy of a[] without the i-th element
 [...m, b.splice(i, 1)] // the element extracted from a[] added to m[]
 ) // end of recursive call
 ) > '' ? // if a[] was not empty:
 S // let S unchanged
 : // else:
 P(m.join``) || S // update S to m.join('') if it's a palindrome

Answer 4

Jelly, 13 bytes

ŒḂÐf
ẆÇŒ!F€ÇḢ

Try it online!

Prints 0 in the invalid case.

Answer 5

Stax, 13 bytes

绬►Ö∞j∞:Æ╘τδ

Run test cases (It takes about 10 seconds on my current machine)

This is the corresponding ascii representation of the same program.

:e{cr=fw|Nc$cr=!

It's not quite pure brute-force, but it's just as small as the brute-force implementation I wrote. That one crashed my browser after about 10 minutes. Anyway, here's how it works.

:e Get all contiguous substrings
 {cr=f Keep only those that are palindromes
 w Run the rest of the program repeatedly while a truth value is produced.
 |N Get the next permutation.
 c$ Copy and flatten the permutation.
 cr=! Test if it's palindrome. If not, repeat.
 The last permutation produced will be implicitly printed.

Answer 6

Ruby, (削除) 131 (削除ここまで) (削除) 123 (削除ここまで) 120 bytes

->s{m=->t{t==t.reverse}
(1..z=s.size).flat_map{|l|(0..z-l).map{|i|s[i,l]}}.select(&m).permutation.map(&:join).detect &m}

Try it online!

A lambda accepting a string and returning a string. Returns nil when no solution exists.

-5 bytes: Replace select{|t|l[t]} with select(&l)

-3 bytes: Replace map{..}.flatten with flat_map{...}

-1 bytes: Loop over substring length and substring start, instead of over substring start and substring end

-2 bytes: Declare z at first use instead of beforehand

->s{
 l=->t{t==t.reverse} # Lambda to test for palindromes
 (1..z=s.size).flat_map{|l| # For each substring length
 (0..z-l).map{|i| # For each substring start index
 s[i,l] # Take the substring
 }
 } # flat_map flattens the list of lists of substrings
 .select(&l) # Filter to include only palindromic substrings
 .permutation # Take all orderings of substrings
 .map(&:join) # Flatten each substring ordering into a string
 .detect &l # Find the first palindrome
}

Answer 7

05AB1E, (削除) 13 (削除ここまで) 12 bytes

ŒʒÂQ}œJʒÂQ}¤

Try it online!

-1 byte thanks to Magic Octopus Urn and Emigna.

• Magic Octopus Urn: J automatically factorizes so you don't need €J just J; also, you're supposed to return one of the palindromes, not all. Try it online! is valid for the same byte-count.
• user2956892: @MagicOctopusUrn Fixed, thanks !
• Emigna: Ùć could be ¤ (or a number of other options)
• Magic Octopus Urn: @Emigna not sure why I didn't see the Ù wasn't needed.
• user2956892: Enigma My bad, for an unknown reason I thought we were supposed to display all of the unique palindromes, hence the original Ù. Thanks for the tip, fixed !

Answer 8

Pyth, 13 bytes

h_I#sM.p_I#.:

Try it online!

-1 byte thanks to Mr. Xcoder

• Mr. Xcoder: Lol I was so sure nobody else uses Pyth that I submitted my own separate answer (now deleted) prior to seeing yours. You can use h_I#sM.p_I#.: or e_IDsM.p_I#.: for 13 bytes.
• hyperneutrino: @Mr.Xcoder Oh haha :P yeah I hardly ever use Pyth, don't know why I decided to use it. Thanks!

Answer 9

Python 3, 167 bytes

lambda a:g(sum(k,[])for k in permutations(g(a[i:j+1]for i in range(len(a))for j in range(i,len(a)))))[0]
g=lambda k:[e for e in k if e==e[::-1]]
from itertools import*

Try it online!

-2 bytes thanks to Mr. Xcoder

• Mr. Xcoder: You can use a[i:j+1] if you then use for j in range(i,len(a)) instead, for -2 bytes.

Answer 10

Husk, 12 bytes

ḟS=↔mΣPfS=↔Q

Try it online!

Explanation

ḟS=↔mΣPfS=↔Q Implicit input, a string.
 Q List of substrings.
 f Keep those
 S=↔ that are palindromic (equal to their reversal).
 P Permutations of this list.
 mΣ Flatten each.
ḟ Find an element
 S=↔ that is palindromic.

Answer 11

Japt, (削除) 19 (削除ここまで) 13 bytes

No longer hampered by Japt not being able to get all substrings of a string (nor, too much, by my current levels of exhaustion!).

Outputs undefined if there's no solution.

ã fêQ á m¬æêQ

Try it

ã fêQ á m¬æêQ :Implicit input of string
ã :Substrings
 f :Filter by
 êQ : Is palindrome
 á :Permutations
 m :Map
 ¬ : Join
 æ :First element that
 êQ : Is palindrome

• Magic Octopus Urn: Is your question about a list of substring simply to remove ¬ from your answer :P?
• Shaggy: Thought I could remove m¬ but then I would have needed æ_¬êQ so it wouldn't have saved any bytes anyway!
• Magic Octopus Urn: Hahaha, I'll ensure to be wary of your byte-saving ways from now on ;). I tried removing it myself to check, but realized japt commands don't work like I think they work lol.

Answer 12

J, 53 bytes

[:{:@(#~b"1)@(i.@!@#;@A.])@(#~(0<#*b=.]-:|.)&>)@,<\\.

Try it online!