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.

Delphi

Library: System.SysUtils

Library: Types

Translation of: Ring

<lang Delphi> program Longest_common_suffix;

{$APPTYPE CONSOLE}

uses

 System.SysUtils,
 Types;

type

 TStringDynArrayHelper = record helper for TStringDynArray
 private
   class function Compare(const s: string; a: TStringDynArray; subSize: integer):
     Boolean;
 public
   function Reverse(value: string): string;
   function LongestSuffix: string;
   function Join(const sep: char): string;
 end;

{ TStringDynArrayHelper }

class function TStringDynArrayHelper.Compare(const s: string; a: TStringDynArray;

 subSize: integer): Boolean;

var

 i: Integer;

begin

 for i := 0 to High(a) do
   if s <> a[i].Substring(0, subSize) then
     exit(False);
 Result := True;

end;

function TStringDynArrayHelper.Join(const sep: char): string; begin

 Result := string.Join(sep, self);

end;

function TStringDynArrayHelper.LongestSuffix: string; var

 ALength: Integer;
 i, lenMin, longest: Integer;
 ref: string;

begin

 ALength := Length(self);

 // Empty list
 if ALength = 0 then
   exit();

 lenMin := MaxInt;
 for i := 0 to ALength - 1 do
 begin
   // One string is empty
   if self[i].IsEmpty then
     exit();

   self[i] := Reverse(self[i]);

   // Get the minimum length of string
   if lenMin > self[i].Length then
     lenMin := self[i].Length;
 end;

 longest := -1;
 repeat
   inc(longest);
   ref := self[0].Substring(0, longest + 1);
 until not compare(ref, Self, longest + 1) or (longest >= lenMin);

 Result := self[0].Substring(0, longest);
 Result := reverse(Result);

end;

function TStringDynArrayHelper.Reverse(value: string): string; var

 ALength: Integer;
 i: Integer;
 c: Char;

begin

 ALength := value.Length;
 Result := value;

 if ALength < 2 then
   exit;

 for i := 1 to ALength div 2 do
 begin
   c := Result[i];
   Result[i] := Result[ALength - i + 1];
   Result[ALength - i + 1] := c;
 end;

end;

var

 List: TStringDynArray;

begin

 List := ['baabababc', 'baabc', 'bbbabc'];

 Writeln('Input:');
 Writeln(List.Join(#10), #10);
 Writeln('Longest common suffix = ', List.LongestSuffix);

 Readln;

end.

</lang>

Output:

Input:
baabababc
baabc
bbbabc

Longest common suffix = abc

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'
'' ==> ''

Phix

Phix allows negative indexes, with -1 as the last element [same as $], and -length(s) the first element of s, so we can just do this: <lang Phix>function longestcommonsuffix(sequence strings)

   string res = ""
   if length(strings) then
       res = strings[1]
       for i=2 to length(strings) do
           string si = strings[i]
           if length(si)<length(res) then
               res = res[-length(si)..$]
           end if
           for j=-1 to -length(res) by -1 do
               if res[j]!=si[j] then
                   res = res[j+1..$]
                   exit
               end if
           end for
           if length(res)=0 then exit end if
       end for
   end if
   return res

end function

sequence tests = {{"baabababc","baabc","bbbabc"},

                 {"baabababc","baabc","bbbazc"},
                 {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"},
                 {"longest", "common", "suffix"},
                 {"suffix"},
                 {""}}

for i=1 to length(tests) do

   printf(1,"%v ==> \"%s\"\n",{tests[i],longestcommonsuffix(tests[i])})

end for</lang>

Output:

{"baabababc","baabc","bbbabc"} ==> "abc"
{"baabababc","baabc","bbbazc"} ==> "c"
{"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"} ==> "day"
{"longest","common","suffix"} ==> ""
{"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