Longest common suffix

Longest common suffix 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.

Task

The goal is to write a function to find the longest common suffix string amongst an array of strings.

Factor

Works with: Factor version 0.99 2020-07-03

<lang factor>USING: accessors grouping kernel prettyprint sequences sequences.extras ;

! Like take-while, but for matrices and works from the rear.

take-col-while-last ( ... matrix quot: ( ... col -- ... ? ) -- ... new-matrix )

   [ [ <reversed> ] map flip ] dip take-while ; inline

lcs ( seq -- lcs )

   dup first swap [ all-equal? ] take-col-while-last to>> tail* ;

{ "baabababc" "baabc" "bbbabc" } lcs . { "baabababc" "baabc" "bbbazc" } lcs . { "" } lcs .</lang>

Output:

"abc"
"c"
""

Julia

<lang julia>function longestcommonsuffix(strings)

   n, nmax = 0, minimum(length, strings)
   nmax == 0 && return ""
   while n <= nmax && all(s -> s[end-n] == strings[end][end-n], strings)
       n += 1
   end
   return strings[1][end-n+1:end]

end

println(longestcommonsuffix(["baabababc","baabc","bbbabc"])) println(longestcommonsuffix(["baabababc","baabc","bbbazc"]))

println(longestcommonsuffix([""]))</lang>

Output:

abc
c

Perl

Based on Longest_common_prefix Perl entry. <lang perl>use strict; use warnings; use feature 'say';

sub lcs {

   for (0..$#_) { $_[$_] = join , reverse split , $_[$_] }
   join , reverse split , (join("\0", @_) =~ /^ ([^\0]*) [^\0]* (?:\0 \1 [^\0]*)* $/sx)[0];

}

for my $words (

 [ <Sunday Monday Tuesday Wednesday Thursday Friday Saturday> ],
 [ <Sondag Maandag Dinsdag Woensdag Donderdag Vrydag Saterdag dag> ],
 [ 2347, 9312347, 'acx5g2347', 12.02347 ],
 [ <longest common suffix> ],
 [ ('one, Hey!', 'three, Hey!', 'ale, Hey!', 'me, Hey!') ],
 [ 'suffix' ],
 [  ]) {
   say qq{'@$words' ==> '@{[lcs(@$words)]}';

}</lang>

Output:

'Sunday Monday Tuesday Wednesday Thursday Friday Saturday' ==> 'day'
'Sondag Maandag Dinsdag Woensdag Donderdag Vrydag Saterdag dag' ==> 'dag'
'2347 9312347 acx5g2347 12.02347' ==> '2347'
'longest common suffix' ==> ''
'one, Hey! three, Hey! ale, Hey! me, Hey!' ==> 'e, Hey!'
'suffix' ==> 'suffix'
'' ==> ''

Raku

Works with: Rakudo version 2020.07

<lang perl6>sub longest-common-suffix ( *@words ) {

   return  unless +@words;
   my $min = @words».chars.min;
   for 1 .. * {
       return @words[0].substr(* - $min) if $_ > $min;
       next if @words».substr(* - $_).Set == 1;
       return @words[0].substr(* - $_ + 1);
   }

}

say "{$_.raku} - LCS: >{longest-common-suffix $_}<" for

 <Sunday Monday Tuesday Wednesday Thursday Friday Saturday>,
 <Sondag Maandag Dinsdag Woensdag Donderdag Vrydag Saterdag dag>,
 ( 2347, 9312347, 'acx5g2347', 12.02347 ),
 <longest common suffix>,
 ('one, Hey!', 'three, Hey!', 'ale, Hey!', 'me, Hey!'),
 'suffix',
 </lang>

Output:

("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday") - LCS: >day<
("Sondag", "Maandag", "Dinsdag", "Woensdag", "Donderdag", "Vrydag", "Saterdag", "dag") - LCS: >dag<
(2347, 9312347, "acx5g2347", 12.02347) - LCS: >2347<
("longest", "common", "suffix") - LCS: ><
("one, Hey!", "three, Hey!", "ale, Hey!", "me, Hey!") - LCS: >e, Hey!<
"suffix" - LCS: >suffix<
"" - LCS: ><

REXX

Essentially,   this REXX version simply reversed the strings,   and then finds the longest common  prefix. <lang rexx>/*REXX program finds the longest common suffix contained in an array of strings. */ parse arg z; z= space(z) /*obtain optional arguments from the CL*/ if z==|z=="," then z='baabababc baabc bbbabc' /*Not specified? Then use the default.*/ z= space(z); #= words(z) /*#: the number of words in the list. */ say 'There are ' # " words in the list: " z zr= reverse(z) /*reverse Z, find longest common prefix*/ @= word(zr, 1); m= length(@) /*get 1st word in reversed string; len.*/

    do j=2  to #;    x= word(zr, j)             /*obtain a word (string) from the list.*/
    t= compare(@, x)                            /*compare to the "base" word/string.   */
    if t==1          then do;  @=;  leave       /*A mismatch of strings?   Then leave, */
                          end                   /*no sense in comparing anymore strings*/
    if t==0 & @==x   then t= length(@) + 1      /*Both strings equal?  Compute length. */
    if t>=m  then iterate                       /*T ≥ M?  Then it's not longest string.*/
    m= t - 1;        @= left(@, max(0, m) )     /*redefine max length & the base string*/
    end   /*j*/

say /*stick a fork in it, we're all done. */ if m==0 then say 'There is no common suffix.'

        else say 'The longest common suffix is: '   right( word(z, 1), m)</lang>

output   when using the default input:

There are  3  words in the list:  baabababc baabc bbbabc

The longest common suffix is:  abc

Ring

<lang ring> load "stdlib.ring"

pre = ["baabababc","baabc","bbbabc"] len = len(pre) lenList = list(len) sub = list(len)

see "Input:" + nl see pre

for n = 1 to len

   temp = pre[n]
   pre[n] = rever(temp)

next

for n = 1 to len

   lenList[n] = len(pre[n])

next

lenList = sort(lenList) lenMax = lenList[1]

for m = 1 to lenMax

   check = 0 
   sub1 = substr(pre[1],1,m)
   sub2 = substr(pre[2],1,m) 
   sub3 = substr(pre[3],1,m)
   if sub1 = sub2 and sub2 = sub3
      check = 1
   ok
   if check = 1
      longest = m
   ok

next

longPrefix = substr(pre[1],1,longest) longPrefix = rever(longPrefix)

see "Longest common suffix = " + longPrefix + nl

func rever(cstr)

    cStr2 = ""
    for x = len(cStr) to 1 step -1 
        cStr2 = cStr2 + cStr[x]
    next
    return cStr2

</lang>

Output:

Input:
baabababc
baabc
bbbabc
Longest common suffix = abc