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Longest palindromic substrings

From Rosetta Code
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.



F#[edit]

Manacher Function[edit]

 
// 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)
 

The Task[edit]

 
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))
 
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[edit]

Translation of: Wren
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)
}
}
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[edit]

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

-------------- 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)
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[edit]

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
 
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[edit]

 
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
 
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"

Phix[edit]

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 e=length(s) to i+longest-1 by -1 do
if s[e]=s[i] then
string p = s[i..e]
integer lp = length(p)
if lp>=longest and p=reverse(p) 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 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"}
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"}

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

faster[edit]

function Manacher(string text) 
-- Manacher's algorithm (linear time)
-- based on https://www.geeksforgeeks.org/manachers-algorithm-linear-time-longest-palindromic-substring-part-4
-- but with a few tweaks, renames, and bugfixes (in particular the < (positions-1), which I later found LIJIE already said)
sequence res = {}
integer positions = length(text)*2+1
if positions>1 then
sequence LPS = repeat(0,positions)
LPS[2] = 1
integer centerPosition = 1,
centerRightPosition = 2,
maxLPSLength = 0
 
for currentRightPosition=2 to positions-1 do
integer lcp = LPS[currentRightPosition+1],
diff = centerRightPosition - currentRightPosition
-- If currentRightPosition is within centerRightPosition
if diff >= 0 then
-- get currentLeftPosition iMirror for currentRightPosition
integer iMirror = 2*centerPosition-currentRightPosition + 1
lcp = min(LPS[iMirror], diff)
end if
 
-- Attempt to expand palindrome centered at currentRightPosition
-- Here for odd positions, we compare characters and
-- if match then increment LPS Length by ONE
-- If even position, we just increment LPS by ONE without
-- any character comparison
while ((currentRightPosition + lcp) < (positions-1) and (currentRightPosition - lcp) > 0) and
((remainder(currentRightPosition+lcp+1, 2) == 0) or
(text[floor((currentRightPosition+lcp+1)/2)+1] == text[floor((currentRightPosition-lcp-1)/2)+1] )) do
lcp += 1
end while
LPS[currentRightPosition+1] = lcp
maxLPSLength = max(lcp,maxLPSLength)
 
// If palindrome centered at currentRightPosition
// expand beyond centerRightPosition,
// adjust centerPosition based on expanded palindrome.
if (currentRightPosition + lcp) > centerRightPosition then
centerPosition = currentRightPosition
centerRightPosition = currentRightPosition + lcp
end if
end for
for p=1 to positions do
if LPS[p] = maxLPSLength then
integer start = floor((p-1 - maxLPSLength)/2) + 1,
finish = start + maxLPSLength - 1
string r = text[start..finish]
if not find(r,res) then
res = append(res,r)
end if
end if
end for
end if
return res
end function
 
include mpfr.e
mpfr pi = mpfr_init(0,-10001) -- (set precision to 10,000 dp, plus the "3.")
mpfr_const_pi(pi)
string piStr = mpfr_sprintf("%.10000Rf", pi),
s = shorten(piStr)
printf(1,"%s: %v\n",{s,Manacher(piStr)})
Output:

(Same as above if given the same inputs.)
However, while Manacher finishes 10,000 digits in 0s, longest_palindromes takes 1s for 2,000 digits, 15s for 5,000 digits, and 2 mins for 10,000 digits,
which goes to prove that longest_palindromes() above is O(n2), whereas Manacher() is O(n).

3.141592653589793238...05600101655256375679 (10,002 digits): {"398989893","020141020"}

Then again, this is also pretty fast (same output):

Translation of: Raku
function longest_palindromes_raku(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
 
printf(1,"%s: %v\n",{s,longest_palindromes_raku(piStr)})
s = "abbbc"
printf(1,"%s: %v\n",{s,longest_palindromes_raku(s)})
Output:

(first line matches the above, the second was a initially a bug)

3.141592653589793238...05600101655256375679 (10,002 digits): {"398989893","020141020"}
abbbc: {"bbb"}

Python[edit]

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

(This version ignores case but allows non-alphanumerics).

'''Longest palindromic substrings'''
 
 
# 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)
 
 
# 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
]
 
 
# 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
 
 
# ------------------------- TEST -------------------------
# 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',
''
])
)
 
 
# ----------------------- GENERIC ------------------------
 
# 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
 
 
# ---------------------- FORMATTING ----------------------
 
# fTable :: String -> (a -> String) ->
# (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
 
 
# MAIN ---
if __name__ == '__main__':
main()
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[edit]

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.

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
#"

 
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);
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[edit]

/*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); say; say /*search each of the (S) strings. */
x= word(s, i); L= length(x) /*get a string to be examined & length.*/
m= 0
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 .
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)
end /*n*/
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 /*obtain length of possible palindromes*/
$= 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==1 then return 1 /*have found one palindrome. */
end /*j*/; return 0 /* " " some palindrome(s). */
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[edit]

 
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
 
Output:
Input: babaccd
Longest palindromic substrings:
bab
aba

Wren[edit]

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.

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)
}
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]