Longest palindromic substrings: Difference between revisions

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for i, line in StrSplit(db, nl, rt){
for i, line in StrSplit(db, "`n", "`r"){
result := "[""", i := 0
result := "[""", i := 0
for i, str in LPS(line)
for i, str in LPS(line)

Revision as of 17:09, 1 September 2021

Longest palindromic substrings is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Let given a string s. The goal is to find the longest palindromic substring in s.

Related tasks


Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences



11l

<lang 11l>F longest_palindrome(s) 

  V t = Array(‘^’s‘$’).join(‘#’)
  V n = t.len
  V p = [0] * n
  V c = 0
  V r = 0
  L(i) 1 .< n - 1
     p[i] = (r > i) & min(r - i, p[2 * c - i]) != 0

     L t[i + 1 + p[i]] == t[i - 1 - p[i]]
        p[i]++

     I i + p[i] > r
        (c, r) = (i, i + p[i])

  V (max_len, center_index) = max(enumerate(p).map((i, n) -> (n, i)))
  R s[(center_index - max_len) I/ 2 .< (center_index + max_len) I/ 2]

L(s) [‘three old rotators’, 

     ‘never reverse’,
     ‘stable was I ere I saw elbatrosses’,
     ‘abracadabra’,
     ‘drome’,
     ‘the abbatial palace’]
  print(‘'’s‘' -> '’longest_palindrome(s)‘'’)</lang>
Output:
'three old rotators' -> 'rotator'
'never reverse' -> 'ever reve'
'stable was I ere I saw elbatrosses' -> 'table was I ere I saw elbat'
'abracadabra' -> 'ada'
'drome' -> 'e'
'the abbatial palace' -> 'abba'

AutoHotkey

<lang AutoHotkey>LPS(str){ found := [], result := [], maxL := 0 while (StrLen(str) >= 2 && StrLen(str) >= maxL){ s := str loop { while (SubStr(s, 1, 1) <> SubStr(s, 0)) ; while 1st chr <> last chr s := SubStr(s, 1, StrLen(s)-1) ; trim last chr if (StrLen(s) < 2 || StrLen(s) < maxL ) break if (s = reverse(s)){ found.Push(s) maxL := maxL < StrLen(s) ? StrLen(s) : maxL break } s := SubStr(s, 1, StrLen(s)-1) ; trim last chr } str := SubStr(str, 2) ; trim 1st chr and try again } maxL := 0 for i, str in found maxL := maxL < StrLen(str) ? StrLen(str) : maxL for i, str in found if (StrLen(str) = maxL) result.Push(str) return result }</lang> Examples:<lang AutoHotkey>db = ( three old rotators never reverse stable was I ere I saw elbatrosses abracadabra drome x the abbatial palace )

for i, line in StrSplit(db, "`n", "`r"){ result := "[""", i := 0 for i, str in LPS(line) result .= str """, """ output .= line "`t> " Trim(result, """, """) (i?"""":"") "]`n" } MsgBox % output return</lang>

Output:
three old rotators			> ["rotator"]
never reverse				> ["ever reve"]
stable was I ere I saw elbatrosses	> ["table was I ere I saw elbat"]
abracadabra				> ["aca", "ada"]
drome					> []
x					> []
the abbatial palace			> ["abba"]

F#

Manacher Function

<lang fsharp> // Manacher Function. Nigel Galloway: October 1st., 2020 let Manacher(s:string) = let oddP,evenP=Array.zeroCreate s.Length,Array.zeroCreate s.Length 

                        let rec fN i g e (l:int[])=match g>=0 && e<s.Length && s.[g]=s.[e] with true->l.[i]<-l.[i]+1; fN i (g-1) (e+1) l |_->()
                        let rec fGo n g Ʃ=match Ʃ<s.Length with
                                           false->oddP
                                          |_->if Ʃ<=g then oddP.[Ʃ]<-min (oddP.[n+g-Ʃ]) (g-Ʃ)
                                              fN Ʃ (Ʃ-oddP.[Ʃ]-1) (Ʃ+oddP.[Ʃ]+1) oddP
                                              match (Ʃ+oddP.[Ʃ])>g with true->fGo (Ʃ-oddP.[Ʃ]) (Ʃ+oddP.[Ʃ]) (Ʃ+1) |_->fGo n g (Ʃ+1)
                        let rec fGe n g Ʃ=match Ʃ<s.Length with
                                           false->evenP
                                          |_->if Ʃ<=g then evenP.[Ʃ]<-min (evenP.[n+g-Ʃ]) (g-Ʃ)
                                              fN Ʃ (Ʃ-evenP.[Ʃ]) (Ʃ+evenP.[Ʃ]+1) evenP
                                              match (Ʃ+evenP.[Ʃ])>g with true->fGe (Ʃ-evenP.[Ʃ]+1) (Ʃ+evenP.[Ʃ]) (Ʃ+1) |_->fGe n g (Ʃ+1)
                        (fGo 0 -1 0,fGe 0 -1 0)

</lang>

The Task

<lang fsharp> let fN g=if g=[||] then (0,0) else g|>Array.mapi(fun n g->(n,g))|>Array.maxBy snd let lpss s=let n,g=Manacher s in let n,g=fN n,fN g in if (snd n)*2+1>(snd g)*2 then s.[(fst n)-(snd n)..(fst n)+(snd n)] else s.[(fst g)-(snd g)+1..(fst g)+(snd g)] let test = ["three old rotators"; "never reverse"; "stable was I ere I saw elbatrosses"; "abracadabra"; "drome"; "the abbatial palace"; ""] test|>List.iter(fun n->printfn "A longest palindromic substring of \"%s\" is \"%s\"" n (lpss n)) </lang>

Output:
A longest palindromic substring of "three old rotators" is "rotator"
A longest palindromic substring of "never reverse" is "ever reve"
A longest palindromic substring of "stable was I ere I saw elbatrosses" is "table was I ere I saw elbat"
A longest palindromic substring of "abracadabra" is "aca"
A longest palindromic substring of "drome" is "d"
A longest palindromic substring of "the abbatial palace" is "abba"
A longest palindromic substring of "" is ""

Go

Translation of: Wren

<lang go>package main

import ( 

   "fmt"
   "sort"

)

func reverse(s string) string { 

   var r = []rune(s)
   for i, j := 0, len(r)-1; i < j; i, j = i+1, j-1 {
       r[i], r[j] = r[j], r[i]
   }
   return string(r)

}

func longestPalSubstring(s string) []string { 

   var le = len(s)
   if le <= 1 {
       return []string{s}
   }
   targetLen := le
   var longest []string
   i := 0
   for {
       j := i + targetLen - 1
       if j < le {
           ss := s[i : j+1]
           if reverse(ss) == ss {
               longest = append(longest, ss)
           }
           i++
       } else {
           if len(longest) > 0 {
               return longest
           }
           i = 0
           targetLen--
       }
   }
   return longest

}

func distinct(sa []string) []string { 

   sort.Strings(sa)
   duplicated := make([]bool, len(sa))
   for i := 1; i < len(sa); i++ {
       if sa[i] == sa[i-1] {
           duplicated[i] = true
       }
   }
   var res []string
   for i := 0; i < len(sa); i++ {
       if !duplicated[i] {
           res = append(res, sa[i])
       }
   }
   return res

}

func main() { 

   strings := []string{"babaccd", "rotator", "reverse", "forever", "several", "palindrome", "abaracadaraba"}
   fmt.Println("The palindromic substrings having the longest length are:")
   for _, s := range strings {
       longest := distinct(longestPalSubstring(s))
       fmt.Printf("  %-13s Length %d -> %v\n", s, len(longest[0]), longest)
   }

}</lang>

Output:
The palindromic substrings having the longest length are:
  babaccd       Length 3 -> [aba bab]
  rotator       Length 7 -> [rotator]
  reverse       Length 5 -> [rever]
  forever       Length 5 -> [rever]
  several       Length 3 -> [eve]
  palindrome    Length 1 -> [a d e i l m n o p r]
  abaracadaraba Length 3 -> [aba aca ada ara]

Haskell

A list version, written out of curiosity. A faster approach could be made with an indexed datatype.

<lang haskell>-------------- LONGEST PALINDROMIC SUBSTRINGS ------------

longestPalindromes :: String -> ([String], Int) longestPalindromes [] = ([], 0) longestPalindromes s = go $ palindromes s 

 where
   go xs
     | null xs = (return <$> s, 1)
     | otherwise = (filter ((w ==) . length) xs, w)
     where
       w = maximum $ length <$> xs

palindromes :: String -> [String] palindromes = fmap go . palindromicNuclei 

 where
   go (pivot, (xs, ys)) =
     let suffix = fmap fst (takeWhile (uncurry (==)) (zip xs ys))
     in reverse suffix <> pivot <> suffix

palindromicNuclei :: String -> [(String, (String, String))] palindromicNuclei = 

 concatMap go .
 init . tail . ((zip . scanl (flip ((<>) . return)) []) <*> scanr (:) [])
 where
   go (a@(x:_), b@(h:y:ys))
     | x == h = [("", (a, b))]
     | otherwise =
       [ ([h], (a, y : ys))
       | x == y ]
   go _ = []


TEST -------------------------

main :: IO () main = 

 putStrLn $
 fTable
   "Longest palindromic substrings:\n"
   show
   show
   longestPalindromes
   [ "three old rotators"
   , "never reverse"
   , "stable was I ere I saw elbatrosses"
   , "abracadabra"
   , "drome"
   , "the abbatial palace"
   , ""
   ]

FORMATTING ----------------------

fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String fTable s xShow fxShow f xs = 

 unlines $
 s : fmap (((++) . rjust w ' ' . xShow) <*> ((" -> " ++) . fxShow . f)) xs
 where
   rjust n c = drop . length <*> (replicate n c ++)
   w = maximum (length . xShow <$> xs)</lang>
Output:
Longest palindromic substrings:

                "three old rotators" -> (["rotator"],7)
                     "never reverse" -> (["ever reve"],9)
"stable was I ere I saw elbatrosses" -> (["table was I ere I saw elbat"],27)
                       "abracadabra" -> (["aca","ada"],3)
                             "drome" -> (["d","r","o","m","e"],1)
               "the abbatial palace" -> (["abba"],4)
                                  "" -> ([],0)

Julia

<lang julia>function allpalindromics(s) 

   list, len = String[], length(s)
   for i in 1:len-1, j in i+1:len
       substr = s[i:j]
       if substr == reverse(substr)
           push!(list, substr)
       end
   end
   return list

end

for teststring in ["babaccd", "rotator", "reverse", "forever", "several", "palindrome"] 

   list = sort!(allpalindromics(teststring), lt = (x, y) -> length(x) < length(y))
   println(isempty(list) ? "No palindromes of 2 or more letters found in \"$teststring." :
       "The longest palindromic substring of $teststring is: \"",
       join(list[findall(x -> length(x) == length(list[end]), list)], "\" or \""), "\"")

end

</lang>

Output:
The longest palindromic substring of babaccd is: "bab" or "aba"
The longest palindromic substring of rotator is: "rotator"
The longest palindromic substring of reverse is: "rever"
The longest palindromic substring of forever is: "rever"
The longest palindromic substring of several is: "eve"
No palindromes of 2 or more letters found in "palindrome."

Manacher algorithm

<lang julia> function manacher(str) 

   s =  "^" * join(split(str, ""), "#") * "\$"
   len = length(s)
   pals = fill(0, len)
   center, right = 1, 1
   for i in 2:len-1
       pals[i] = right > i && right - i > 0 && pals[2 * center - i] > 0
       while s[i + pals[i] + 1] == s[i - pals[i] - 1]
           pals[i] += 1
       end
       if i + pals[i] > right
           center, right = i, i + pals[i]
       end
   end
   maxlen, centerindex = findmax(pals)
   start = isodd(maxlen) ? (centerindex-maxlen) ÷ 2 + 1 : (centerindex-maxlen) ÷ 2
   return str[start:(centerindex+maxlen)÷2]

end

for teststring in ["babaccd", "rotator", "reverse", "forever", "several", "palindrome", "abaracadabra"] 

   pal = manacher(teststring)
   println(length(pal) < 2 ? "No palindromes of 2 or more letters found in \"$teststring.\"" :
       "The longest palindromic substring of $teststring is: \"$pal\"")

end

</lang>

Output:
The longest palindromic substring of babaccd is: "aba"
The longest palindromic substring of rotator is: "rotator"
The longest palindromic substring of reverse is: "rever"
The longest palindromic substring of forever is: "rever"
The longest palindromic substring of several is: "eve"
No palindromes of 2 or more letters found in "palindrome."
The longest palindromic substring of abaracadabra is: "ara"

Nim

Simple algorithm but working on Unicode code points. <lang Nim>import sequtils, strutils, unicode

func isPalindrome(s: seq[Rune]): bool = 

 ## Return true if a sequence of runes is a palindrome.
 for i in 1..(s.len shr 1):
   if s[i - 1] != s[^i]:
     return false
 result = true

func lps(s: string): seq[string] = 

 var maxLength = 0
 var list: seq[seq[Rune]]
 let r = s.toRunes
 for first in 0..r.high:
   for last in first..r.high:
     let candidate = r[first..last]
     if candidate.isPalindrome():
       if candidate.len > maxLength:
         list = @[candidate]
         maxLength = candidate.len
       elif candidate.len == maxLength:
         list.add candidate
 if maxLength > 1:
   result = list.mapIt($it)

for str in ["babaccd", "rotator", "several", "palindrome", "piété", "tantôt", "étêté"]: 

 let result = lps(str)
 if result.len == 0:
   echo str, " → ", "<no palindromic substring of two of more letters found>"
 else:
   echo str, " → ", result.join(", ")</lang>
Output:
babaccd → bab, aba
rotator → rotator
several → eve
palindrome → <no palindromic substring of two of more letters found>
piété → été
tantôt → tôt
étêté → étêté

Perl

The short one - find all palindromes with one regex. <lang perl>#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Longest_palindromic_substrings use warnings;

print "Longest Palindrome For $_ = @{[ longestpalindrome($_) ]}\n" 

 for qw(babaccd rotator reverse forever several palindrome abaracadabra);

sub longestpalindrome 

 {
 my @best = {"" => 0};
 pop =~ /(.+) .? (??{reverse $1}) (?{ $best[length $&]{$&}++ }) (*FAIL)/x;
 keys %{pop @best};
 }</lang>
Output:
Longest Palindrome For babaccd = aba bab
Longest Palindrome For rotator = rotator
Longest Palindrome For reverse = rever
Longest Palindrome For forever = rever
Longest Palindrome For several = eve
Longest Palindrome For palindrome = ''
Longest Palindrome For abaracadabra = aba ara aca ada

The faster one - does the million digits of Pi in under half a second. <lang perl>#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Longest_palindromic_substrings use warnings;

  1. @ARGV = 'pi.dat'; # uncomment to use this file or add filename to command line

my $forward = lc do { local $/; @ARGV ? <> : }; $forward =~ s/\W+//g;

my $range = 10; my $backward = reverse $forward; my $length = length $forward; my @best = {"" => 0}; my $len; for my $i ( 1 .. $length - 2 ) 

 {
 do
   {
   my $right = substr $forward, $i, $range;
   my $left = substr $backward, $length - $i, $range;
   ( $right ^ $left ) =~ /^\0\0+/ and                                # evens
     ($len = 2 * length $&) >= $#best and
     $best[ $len ]{substr $forward, $i - length $&, $len}++;
   ( $right ^ "\0" . $left ) =~ /^.(\0+)/ and                        # odds
     ($len = 1 + 2 * length $1) >= $#best and
     $best[ $len ]{substr $forward, $i - length $1, $len}++;
   } while $range < $#best and $range = $#best;
 }

print "Longest Palindrome ($#best) : @{[ keys %{ $best[-1] } ]}\n";

__DATA__ this data borrowed from raku...

Never odd or even Was it a car or a cat I saw? Too bad I hid a boot I, man, am regal - a German am I toot Warsaw was raw</lang>

Output:
Longest Palindrome (27) : ootimanamregalagermanamitoo

Phix

Translation of: Raku
-- demo/rosetta/Longest_palindromic_substrings.exw (plus two older versions)
with javascript_semantics
function longest_palindromes(string s)
--  s = lower/strip_spaces_and_punctuation/utf8_to_utf32, if rqd
    integer longest = 2 -- (do not treat length 1 as palindromic)
--  integer longest = 1 -- (do not treat length 0 as palindromic) [works just fine too]
    sequence res = {}
    for i=1 to length(s) do
        for j=0 to iff(i>1 and s[i-1]=s[i]?2:1) do
            integer rev = j,
                    fwd = 1
            while rev<i and i+fwd<=length(s) and s[i-rev]=s[i+fwd] do
                rev += 1
                fwd += 1
            end while
            string p = s[i-rev+1..i+fwd-1]
            integer lp = length(p)
            if lp>=longest then
                if lp>longest then
                    longest = lp
                    res = {p}
                elsif not find(p,res) then -- (or just "else")
                    res = append(res,p)
                end if
            end if
        end for
    end for
    return res -- (or "sort(res)" or "unique(res)", as needed)
end function
 
constant tests = {"babaccd","rotator","reverse","forever","several","palindrome","abaracadaraba","abbbc"}
for i=1 to length(tests) do
    printf(1,"%s: %v\n",{tests[i],longest_palindromes(tests[i])})
end for
Output:
babaccd: {"bab","aba"}
rotator: {"rotator"}
reverse: {"rever"}
forever: {"rever"}
several: {"eve"}
palindrome: {}
abaracadaraba: {"aba","ara","aca","ada"}
abbbc: {"bbb"}

with longest initialised to 1, you get the same except for palindrome: {"p","a","l","i","n","d","r","o","m","e"}

Python

Defines maximal expansions of any two or three character palindromic nuclei in the string.

(This version ignores case but allows non-alphanumerics). <lang python>Longest palindromic substrings


  1. longestPalindrome :: String -> ([String], Int)

def longestPalindromes(s): 

   All palindromes of the maximal length
      drawn from a case-flattened copy of
      the given string, tupled with the
      maximal length.
      Non-alphanumerics are included here.
   
   k = s.lower()
   palindromes = [
       palExpansion(k)(ab) for ab
       in palindromicNuclei(k)
   ]
   maxLength = max([
       len(x) for x in palindromes
   ]) if palindromes else 1
   return (
       [
           x for x in palindromes if maxLength == len(x)
       ] if palindromes else list(s),
       maxLength
   ) if s else ([], 0)


  1. palindromicNuclei :: String -> [(Int, Int)]

def palindromicNuclei(s): 

   Ranges of all the 2 or 3 character
      palindromic nuclei in s.
   
   cs = list(s)
   return [
       # Two-character nuclei.
       (i, 1 + i) for (i, (a, b))
       in enumerate(zip(cs, cs[1:]))
       if a == b
   ] + [
       # Three-character nuclei.
       (i, 2 + i) for (i, (a, b, c))
       in enumerate(zip(cs, cs[1:], cs[2:]))
       if a == c
   ]


  1. palExpansion :: String -> (Int, Int) -> String

def palExpansion(s): 

   Full expansion of the palindromic
      nucleus with the given range in s.
   
   iEnd = len(s) - 1

   def limit(ij):
       i, j = ij
       return 0 == i or iEnd == j or s[i-1] != s[j+1]

   def expansion(ij):
       i, j = ij
       return (i - 1, 1 + j)

   def go(ij):
       ab = until(limit)(expansion)(ij)
       return s[ab[0]:ab[1] + 1]
   return go


  1. ------------------------- TEST -------------------------
  2. main :: IO ()

def main(): 

   Longest palindromic substrings
   print(
       fTable(main.__doc__ + ':\n')(repr)(repr)(
           longestPalindromes
       )([
           'three old rotators',
           'never reverse',
           'stable was I ere I saw elbatrosses',
           'abracadabra',
           'drome',
           'the abbatial palace',
           
       ])
   )


  1. ----------------------- GENERIC ------------------------
  1. until :: (a -> Bool) -> (a -> a) -> a -> a

def until(p): 

   The result of repeatedly applying f until p holds.
      The initial seed value is x.
   
   def go(f):
       def g(x):
           v = x
           while not p(v):
               v = f(v)
           return v
       return g
   return go


  1. ---------------------- FORMATTING ----------------------
  1. fTable :: String -> (a -> String) ->
  2. (b -> String) -> (a -> b) -> [a] -> String

def fTable(s): 

   Heading -> x display function -> fx display function ->
      f -> xs -> tabular string.
   
   def gox(xShow):
       def gofx(fxShow):
           def gof(f):
               def goxs(xs):
                   ys = [xShow(x) for x in xs]
                   w = max(map(len, ys))

                   def arrowed(x, y):
                       return y.rjust(w, ' ') + ' -> ' + (
                           fxShow(f(x))
                       )
                   return s + '\n' + '\n'.join(
                       map(arrowed, xs, ys)
                   )
               return goxs
           return gof
       return gofx
   return gox


  1. MAIN ---

if __name__ == '__main__': 

   main()</lang>
Output:
Longest palindromic substrings:

                'three old rotators' -> (['rotator'], 7)
                     'never reverse' -> (['ever reve'], 9)
'stable was I ere I saw elbatrosses' -> (['table was i ere i saw elbat'], 27)
                       'abracadabra' -> (['aca', 'ada'], 3)
                             'drome' -> (['d', 'r', 'o', 'm', 'e'], 1)
               'the abbatial palace' -> (['abba'], 4)
                                  '' -> ([], 0)

Raku

Works with: Rakudo version 2020.09

This version regularizes (ignores) case and ignores non alphanumeric characters. It is only concerned with finding the longest palindromic substrings so does not exhaustively find all possible palindromes. If a palindromic substring is found to be part of a longer palindrome, it is not captured separately. Showing the longest 5 palindromic substring groups. Run it with no parameters to operate on the default; pass in a file name to run it against that instead.

<lang perl6>my @chars = ( @*ARGS[0] ?? @*ARGS[0].IO.slurp !! q:to/BOB/ ) .lc.comb: /\w/; 

   Lyrics to "Bob" copyright Weird Al Yankovic
   https://www.youtube.com/watch?v=JUQDzj6R3p4

   I, man, am regal - a German am I
   Never odd or even
   If I had a hi-fi
   Madam, I'm Adam
   Too hot to hoot
   No lemons, no melon
   Too bad I hid a boot
   Lisa Bonet ate no basil
   Warsaw was raw
   Was it a car or a cat I saw?

   Rise to vote, sir
   Do geese see God?
   "Do nine men interpret?" "Nine men," I nod
   Rats live on no evil star
   Won't lovers revolt now?
   Race fast, safe car
   Pa's a sap
   Ma is as selfless as I am
   May a moody baby doom a yam?

   Ah, Satan sees Natasha
   No devil lived on
   Lonely Tylenol
   Not a banana baton
   No "x" in "Nixon"
   O, stone, be not so
   O Geronimo, no minor ego
   "Naomi," I moan
   "A Toyota's a Toyota"
   A dog, a panic in a pagoda

   Oh no! Don Ho!
   Nurse, I spy gypsies - run!
   Senile felines
   Now I see bees I won
   UFO tofu
   We panic in a pew
   Oozy rat in a sanitary zoo
   God! A red nugget! A fat egg under a dog!
   Go hang a salami, I'm a lasagna hog!
   BOB
  1. "

my @cpfoa = flat (1 ..^ @chars).race(:1000batch).map: -> \idx { 

   my @s;
   for 1, 2 {
      my int ($rev, $fwd) = $_, 1;
      loop {
           quietly last if ($rev > idx) || (@chars[idx - $rev] ne @chars[idx + $fwd]);
           $rev = $rev + 1;
           $fwd = $fwd + 1;
       }
       @s.push: @chars[idx - $rev ^..^ idx + $fwd].join if $rev + $fwd > 2;
       last if @chars[idx - 1] ne @chars[idx];
   }
   next unless +@s;
   @s

}

"{.key} ({+.value})\t{.value.unique.sort}".put for @cpfoa.classify( *.chars ).sort( -*.key ).head(5);</lang>

Output:

Returns the length, (the count) and the list: 

29 (2)	doninemeninterpretninemeninod godarednuggetafateggunderadog
26 (1)	gohangasalamiimalasagnahog
23 (1)	arwontloversrevoltnowra
21 (4)	imanamregalagermanami mayamoodybabydoomayam ootnolemonsnomelontoo oozyratinasanitaryzoo
20 (1)	ratsliveonnoevilstar

This isn't intensively optimised but isn't too shabby either. When run against the first million digits of pi: 1000000 digits of pi text file (Pass in the file path/name at the command line) we get: 

13 (1)	9475082805749
12 (1)	450197791054
11 (8)	04778787740 09577577590 21348884312 28112721182 41428782414 49612121694 53850405835 84995859948
10 (9)	0045445400 0136776310 1112552111 3517997153 5783993875 6282662826 7046006407 7264994627 8890770988
9 (98)	019161910 020141020 023181320 036646630 037101730 037585730 065363560 068363860 087191780 091747190 100353001 104848401 111262111 131838131 132161231 156393651 160929061 166717661 182232281 193131391 193505391 207060702 211878112 222737222 223404322 242424242 250171052 258232852 267919762 272636272 302474203 313989313 314151413 314424413 318272813 323212323 330626033 332525233 336474633 355575553 357979753 365949563 398989893 407959704 408616804 448767844 450909054 463202364 469797964 479797974 480363084 489696984 490797094 532121235 546000645 549161945 557040755 559555955 563040365 563828365 598292895 621969126 623707326 636414636 636888636 641949146 650272056 662292266 667252766 681565186 684777486 712383217 720565027 726868627 762727267 769646967 777474777 807161708 819686918 833303338 834363438 858838858 866292668 886181688 895505598 896848698 909565909 918888819 926676629 927202729 929373929 944525449 944848449 953252359 972464279 975595579 979202979 992868299

in right around 7 seconds on my system.

REXX

<lang rexx>/*REXX program finds and displays the longest palindromic string(s) in a given string. */ parse arg s /*obtain optional argument from the CL.*/ if s==|s=="," then s='babaccd rotator reverse forever several palindrome abaracadaraba' 

                                                /* [↑] the case of strings is respected*/
   do i=1  for words(s);          x= word(s, i) /*obtain a string to be examined.      */
   L= length(x);                  m= 0          /*get the string's length; Set max len.*/
                 do LL=2  for L-1               /*start with palindromes of length two.*/
                 if find(1)  then m= max(m, LL) /*Found a palindrome?  Set M=new length*/
                 end   /*LL*/
   LL= max(1, m)
   call find 0                                  /*find all palindromes with length  LL.*/
   say ' longest palindromic substrings for string: '        x
   say '────────────────────────────────────────────'copies('─', 2 + L)
         do n=1  for words(@)                   /*show longest palindromic substrings. */
         say '    (length='LL")  "  word(@, n)  /*display a         "      substring.  */
         end   /*n*/;       say;    say         /*display a two─blank separation fence.*/
   end         /*i*/

exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ find: parse arg short /*if SHORT==1, only find 1 palindrome.*/ 

     @=                                         /*initialize palindrome list to a null.*/
       do j=1  for L-LL+1;  $= substr(x, j, LL) /*obtain a possible palindromic substr.*/
       if $\==reverse($)  then iterate          /*Not a palindrome?       Then skip it.*/
       @= @ $                                   /*add a palindromic substring to a list*/
       if short  then return 1                  /*we have found  one   palindrome.     */
       end   /*j*/;   return 0                  /* "   "    "    some  palindrome(s).  */</lang>
output   when using the default input:
 longest palindromic substrings for string:  babaccd
─────────────────────────────────────────────────────
    (length=3)   bab
    (length=3)   aba


 longest palindromic substrings for string:  rotator
─────────────────────────────────────────────────────
    (length=7)   rotator


 longest palindromic substrings for string:  reverse
─────────────────────────────────────────────────────
    (length=5)   rever


 longest palindromic substrings for string:  forever
─────────────────────────────────────────────────────
    (length=5)   rever


 longest palindromic substrings for string:  several
─────────────────────────────────────────────────────
    (length=3)   eve


 longest palindromic substrings for string:  palindrome
────────────────────────────────────────────────────────
    (length=1)   p
    (length=1)   a
    (length=1)   l
    (length=1)   i
    (length=1)   n
    (length=1)   d
    (length=1)   r
    (length=1)   o
    (length=1)   m
    (length=1)   e


 longest palindromic substrings for string:  abaracadaraba
───────────────────────────────────────────────────────────
    (length=3)   aba
    (length=3)   ara
    (length=3)   aca
    (length=3)   ada
    (length=3)   ara
    (length=3)   aba

Ring

<lang ring> load "stdlib.ring"

st = "babaccd" palList = []

for n = 1 to len(st)-1 

   for m = n+1 to len(st)
       sub = substr(st,n,m-n)
       if ispalindrome(sub) and len(sub) > 1
          add(palList,[sub,len(sub)])
       ok
   next

next

palList = sort(palList,2) palList = reverse(palList) resList = [] add(resList,palList[1][1])

for n = 2 to len(palList) 

   if palList[1][2] = palList[n][2]
      add(resList,palList[n][1])
   ok

next

see "Input: " + st + nl see "Longest palindromic substrings:" + nl see resList </lang>

Output:
Input: babaccd
Longest palindromic substrings:
bab
aba

Wren

Library: Wren-seq
Library: Wren-fmt

I've assumed that the expression 'substring' includes the string itself and that substrings of length 1 are considered to be palindromic. Also that if there is more than one palindromic substring of the longest length, then all such distinct ones should be returned.

The Phix entry examples have been used. <lang ecmascript>import "/seq" for Lst import "/fmt" for Fmt

var longestPalSubstring = Fn.new { |s| 

   var len = s.count
   if (len <= 1) return [s]
   var targetLen = len
   var longest = []
   var i = 0
   while (true) {
       var j = i + targetLen - 1
       if (j < len) {
           var ss = s[i..j]
           if (ss == ss[-1..0]) longest.add(ss)
           i = i + 1
       } else {
           if (longest.count > 0) return longest
           i = 0
           targetLen = targetLen - 1
       }
   }

}

var strings = ["babaccd", "rotator", "reverse", "forever", "several", "palindrome", "abaracadaraba"] System.print("The palindromic substrings having the longest length are:") for (s in strings) { 

   var longest = Lst.distinct(longestPalSubstring.call(s))
   Fmt.print("  $-13s Length $d -> $n", s, longest[0].count, longest)

}</lang>

Output:
The palindromic substrings having the longest length are:
  babaccd       Length 3 -> [bab, aba]
  rotator       Length 7 -> [rotator]
  reverse       Length 5 -> [rever]
  forever       Length 5 -> [rever]
  several       Length 3 -> [eve]
  palindrome    Length 1 -> [p, a, l, i, n, d, r, o, m, e]
  abaracadaraba Length 3 -> [aba, ara, aca, ada]
Retrieved from "https://rosettacode.org/wiki/Longest_palindromic_substrings?oldid=302642"