Longest common suffix

From Rosetta Code
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

Write a function to find the longest common suffix string amongst an array of strings.


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

F lcs(sa)
   I sa.empty
      R ‘’
   I sa.len == 1
      R sa[0]

   V min_len = min(sa.map(s -> s.len))

   L(i) 1 .. min_len
      V p = sa[0][(len)-i]
      L(j) 1 .< sa.len
         I sa[j][(len)-i] != p
            R sa[0][(len)-i+1..]

   R sa[0][(len)-min_len..]

print(lcs([‘11Sunday’, ‘2Sunday’]))
print(lcs([‘Sunday’, ‘Monday’, ‘Tuesday’]))
print(lcs([‘Sunday’, ‘Monday’, ‘Tuesday’, ‘day’]))
print(lcs([‘Sondag’, ‘Maandag’, ‘Dinsdag’, ‘Woensdag’]))
Output:
Sunday
day
day
dag

Action!

DEFINE PTR="CARD"

BYTE Func Equals(CHAR ARRAY a,b)
  BYTE i

  IF a(0)#b(0) THEN
    RETURN (0)
  FI

  FOR i=1 TO a(0)
  DO
    IF a(i)#b(i) THEN
      RETURN (0)
    FI
  OD
RETURN (1)

BYTE FUNC CommonLength(PTR ARRAY texts BYTE count)
  CHAR ARRAY t
  BYTE i,len

  IF count=0 THEN
    RETURN (0)
  FI

  len=255
  FOR i=0 TO count-1
  DO
    t=texts(i)
    IF t(0)<len THEN
      len=t(0)
    FI
  OD
RETURN (len)

PROC Suffix(PTR ARRAY texts BYTE count CHAR ARRAY res)
  CHAR ARRAY t(100)
  CHAR ARRAY tmp
  BYTE i,len,found

  IF count=1 THEN
    SCopy(res,texts(0))
    RETURN
  FI

  len=CommonLength(texts,count)
  WHILE len>0
  DO
    tmp=texts(0)
    SCopyS(res,tmp,tmp(0)-len+1,tmp(0))
    found=1
    FOR i=1 TO count-1
    DO
      tmp=texts(i)
      SCopyS(t,texts(i),tmp(0)-len+1,tmp(0))
      IF Equals(res,t)#1 THEN
        found=0 EXIT
      FI
    OD
    IF found THEN
      RETURN
    FI
    len==-1
  OD
  res(0)=0
RETURN

PROC Test(PTR ARRAY texts BYTE count)
  BYTE i
  CHAR ARRAY res(100)

  Suffix(texts,count,res)
  Print("lcs(")
  IF count>0 THEN
    FOR i=0 TO count-1
    DO
      PrintF("""%S""",texts(i))
      IF i<count-1 THEN
        Print(",")
      FI
    OD
  FI
  PrintF(")=""%S""%E",res)
RETURN

PROC Main()
  CHAR ARRAY
    t1="Sunday", t2="Monday", t3="Tuesday", t4="Wednesday",
    t5="Thursday", t6="Friday", t7="Saturday",
    t8="throne", t9="throne", t10="dungeon", t11="",
    t12="prefix", t13="suffix"
  PTR ARRAY texts(20)

  texts(0)=t1 texts(1)=t2 texts(2)=t3
  texts(3)=t4 texts(4)=t5 texts(5)=t6
  texts(6)=t7
  Test(texts,7)

  texts(0)=t8 texts(1)=t9
  Test(texts,2)

  texts(0)=t8 texts(1)=t10
  Test(texts,2)

  texts(0)=t8 texts(1)=t11 texts(2)=t9
  Test(texts,3)

  texts(0)=t10
  Test(texts,1)

  texts(0)=t11
  Test(texts,1)

  Test(texts,0)

  texts(0)=t12 texts(1)=t13
  Test(texts,2)
RETURN
Output:

Screenshot from Atari 8-bit computer 

lcs("Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday")="day"
lcs("throne","throne")="throne"
lcs("throne","dungeon")=""
lcs("throne","","throne")=""
lcs("dungeon")="dungeon"
lcs("")=""
lcs()=""
lcs("prefix","suffix")="fix"

Ada

with Ada.Strings.Unbounded;
with Ada.Text_Io.Unbounded_IO;

procedure Longest_Common_Suffix is
   use Ada.Text_Io;
   use Ada.Text_Io.Unbounded_Io;
   use Ada.Strings.Unbounded;

   subtype Ustring is Unbounded_String;

   function "+"(S : String) return Ustring
     renames To_Unbounded_String;

   type String_List is array (Positive range <>) of Ustring;

   function Longest_Suffix (List : String_List) return Ustring is
      Suffix : Ustring := List (List'First);
   begin
      for A in List'First + 1 .. List'Last loop
         declare
            Word  : Ustring renames List (A);
            Found : Boolean := False;
            Len   : constant Natural :=
              Natural'Min (Length (Suffix), Length (Word));
         begin
            for P in reverse 1 .. Len loop
               if Tail (Suffix, P) = Tail (Word, P) then
                  Suffix := Tail (Word, P);
                  Found := True;
                  exit;
               end if;
            end loop;
            if not Found then
               Suffix := +"";
            end if;
         end;
      end loop;
      return Suffix;
   end Longest_Suffix;

   procedure Put (List : String_List) is
   begin
      Put ("[");
      for S of List loop
         Put ("'"); Put (S); Put ("' ");
      end loop;
      Put ("]");
   end Put;

   procedure Test (List : String_List) is
   begin
      Put (List); Put (" -> '");
      Put (Longest_Suffix (List));
      Put ("'");
      New_Line;
   end Test;

   Case_1 : constant String_List := (+"baabababc", +"baabc", +"bbbabc");
   Case_2 : constant String_List := (+"baabababc", +"baabc", +"bbbazc");
   Case_3 : Constant String_List := (+"Sunday", +"Monday", +"Tuesday",
                                     +"Wednesday", +"Thursday", +"Friday",
                                     +"Saturday");
   Case_4 : constant String_List := (+"longest", +"common", +"suffix");
   Case_5 : constant String_List := (1 => +"suffix");
   Case_6 : Constant String_List := (1 => +"");
begin
   Test (Case_1);
   Test (Case_2);
   Test (Case_3);
   Test (Case_4);
   Test (Case_5);
   Test (Case_6);
end Longest_Common_Suffix;
Output:
['baabababc' 'baabc' 'bbbabc' ] -> 'abc'
['baabababc' 'baabc' 'bbbazc' ] -> 'c'
['Sunday' 'Monday' 'Tuesday' 'Wednesday' 'Thursday' 'Friday' 'Saturday' ] -> 'day'
['longest' 'common' 'suffix' ] -> ''
['suffix' ] -> 'suffix'
['' ] -> ''

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

Based on the Algol 68 sample for the Longest Common Prefix task. 

# find the longest common suffix of two strings #
PRIO COMMONSUFFIX = 1;
OP   COMMONSUFFIX = ( STRING a, b )STRING:
    BEGIN
        INT a pos := UPB a; INT a min = LWB a;
        INT b pos := UPB b; INT b min = LWB b;
        WHILE
            IF a pos < a min OR b pos < b min THEN FALSE
            ELSE a[ a pos ] = b[ b pos ]
            FI
        DO
            a pos -:= 1; b pos -:= 1
        OD;
        a[ a pos + 1 : UPB a ]
    END # COMMONSUFFIX # ;
# get the length of a string #
OP  LEN = ( STRING a )INT: ( UPB a + 1 ) - LWB a;
# find the longest common suffix of an array of STRINGs #
OP  LONGESTSUFFIX = ( []STRING list )STRING:
    IF  UPB list < LWB list
    THEN
        # no elements #
        ""
    ELIF UPB list = LWB list
    THEN
        # only one element #
        list[ LWB list ]
    ELSE
        # more than one element #
        STRING suffix := list[ LWB list ] COMMONSUFFIX list[ 1 + LWB list ];
        FOR pos FROM 2 + LWB list TO UPB list WHILE suffix /= "" DO
            STRING next suffix := list[ pos ] COMMONSUFFIX suffix;
            IF LEN next suffix < LEN suffix
            THEN
                # this element has a smaller common suffix #
                suffix := next suffix
            FI
        OD;
        suffix
    FI ;
# test the LONGESTSUFFIX operator #
PROC test suffix = ( []STRING list, STRING expected result )VOID:
    BEGIN
        STRING suffix = LONGESTSUFFIX list;
        print( ( "longest common suffix of (" ) );
        FOR pos FROM LWB list TO UPB list DO print( ( " """, list[ pos ], """" ) ) OD;
        print( ( " ) is: """, suffix, """ "
               , IF suffix = expected result THEN "as expected" ELSE "NOT AS EXPECTED" FI 
               , newline
               )
             )
    END # test suffix # ;
[ 1 : 0 ]STRING empty list; # for recent versions of Algol 68G, can't just put "()" for an empty list #
BEGIN
    test suffix( ( "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday" ), "day" );
    test suffix( ( "throne", "throne" ), "throne" );
    test suffix( ( "throne", "dungeon" ), "" );
    test suffix( ( "throne", "", "throne" ), "" );
    test suffix( ( "cheese" ), "cheese" );
    test suffix( ( "" ), "" );
    test suffix( empty list, "" );
    test suffix( ( "prefix", "suffix" ), "fix" );
    test suffix( ( "send", "lend" ), "end" )
END
Output:
longest common suffix of ( "Sunday" "Monday" "Tuesday" "Wednesday" "Thursday" "Friday" "Saturday" ) is: "day" as expected
longest common suffix of ( "throne" "throne" ) is: "throne" as expected
longest common suffix of ( "throne" "dungeon" ) is: "" as expected
longest common suffix of ( "throne" "" "throne" ) is: "" as expected
longest common suffix of ( "cheese" ) is: "cheese" as expected
longest common suffix of ( "" ) is: "" as expected
longest common suffix of ( ) is: "" as expected
longest common suffix of ( "prefix" "suffix" ) is: "fix" as expected
longest common suffix of ( "send" "lend" ) is: "end" as expected

APL

Works with: Dyalog APL
lcs ← ⌽+/∘(∧\)∘⊃∘(∧/2=/(⌊/≢¨)↑¨⌽¨)↑⌽∘⊃
Output:
      lcs 'baabababc' 'baabc' 'bbbabc'
abc
      lcs 'baabababc' 'baabc' 'bbbazc'
c
      lcs 'Sunday' 'Monday' 'Tuesday' 'Wednesday' 'Thursday' 'Friday' 'Saturday'
day
      lcs 'longest' 'common' 'suffix'

      lcs ,⊂''

AppleScript

Procedural

The simplest solution in AppleScript seems to be to reverse the strings, apply the AppleScriptObjC solution for the Longest common prefix task, and reverse the result. 

use AppleScript version "2.4" -- OS X 10.10 (Yosemite) or later
use framework "Foundation"

on longestCommonSuffix(textList)
    -- Eliminate any non-texts from the input.
    if (textList's class is record) then return ""
    set textList to (textList as list)'s text
    if (textList is {}) then return ""
    
    set astid to AppleScript's text item delimiters
    set AppleScript's text item delimiters to ""
    repeat with i from 1 to (count textList)
        set item i of textList to (reverse of characters of item i of textList) as text
    end repeat
    set lcs to (reverse of characters of longestCommonPrefix(textList)) as text
    set AppleScript's text item delimiters to astid
    
    return lcs
end longestCommonSuffix

on longestCommonPrefix(textList)
    -- Eliminate any non-texts from the input.
    if (textList's class is record) then return ""
    set textList to (textList as list)'s text
    if (textList is {}) then return ""
    
    -- Convert the AppleScript list to an NSArray of NSStrings.
    set stringArray to current application's class "NSArray"'s arrayWithArray:(textList)
    
    -- Compare the strings case-insensitively using a built-in NSString method.
    set lcp to stringArray's firstObject()
    repeat with i from 2 to (count stringArray)
        set lcp to (lcp's commonPrefixWithString:(item i of stringArray) options:(current application's NSCaseInsensitiveSearch))
        if (lcp's |length|() is 0) then exit repeat
    end repeat
    
    -- Return the NSString result as AppleScript text.
    return lcp as text
end longestCommonPrefix

-- Tests and results:
longestCommonSuffix({"throne", "sousaphone"}) --> "one"
longestCommonSuffix({"prefix", "suffix"}) --> "fix"
longestCommonSuffix({"remark", "spark", "aardvark"}) --> "ark"
longestCommonSuffix({"ectoplasm", "banana"}) --> ""

Functional

and for more productivity, and higher re-use of library functions, we can write a functional definition (rather than a procedure): 

------------------- LONGEST COMMON SUFFIX ------------------


-- longestCommonSuffix :: [String] -> String
on longestCommonSuffix(xs)
    if 1 < length of xs then
        reverse of map(my fst, ¬
            takeWhile(my allSame, ¬
                transpose(map(my |reverse|, xs)))) as text
    else
        xs as text
    end if
end longestCommonSuffix


---------------------------- TESTS --------------------------
on run
    script test
        on |λ|(s)
            set xs to words of s
            showList(xs) & " -> '" & longestCommonSuffix(xs) & "'"
        end |λ|
    end script
    
    unlines(map(test, {¬
        "throne sousaphone tone", ¬
        "prefix suffix infix", ¬
        "remark spark aardvark lark", ¬
        "ectoplasm banana brick"}))
end run


--------------------- GENERIC FUNCTIONS --------------------

-- all :: (a -> Bool) -> [a] -> Bool
on all(p, xs)
    -- True if p holds for every value in xs
    tell mReturn(p)
        set lng to length of xs
        repeat with i from 1 to lng
            if not |λ|(item i of xs, i, xs) then return false
        end repeat
        true
    end tell
end all


-- allSame :: [a] -> Bool
on allSame(xs)
    if 2 > length of xs then
        true
    else
        script p
            property h : item 1 of xs
            on |λ|(x)
                h = x
            end |λ|
        end script
        all(p, rest of xs)
    end if
end allSame


-- comparing :: (a -> b) -> (a -> a -> Ordering)
on comparing(f)
    script
        on |λ|(a, b)
            tell mReturn(f)
                set fa to |λ|(a)
                set fb to |λ|(b)
                if fa < fb then
                    -1
                else if fa > fb then
                    1
                else
                    0
                end if
            end tell
        end |λ|
    end script
end comparing


-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
    set lng to length of xs
    set acc to {}
    tell mReturn(f)
        repeat with i from 1 to lng
            set acc to acc & (|λ|(item i of xs, i, xs))
        end repeat
    end tell
    return acc
end concatMap


-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
    tell mReturn(f)
        set v to startValue
        set lng to length of xs
        repeat with i from 1 to lng
            set v to |λ|(v, item i of xs, i, xs)
        end repeat
        return v
    end tell
end foldl


-- fst :: (a, b) -> a
on fst(tpl)
    if class of tpl is record then
        |1| of tpl
    else
        item 1 of tpl
    end if
end fst


-- intercalate :: String -> [String] -> String
on intercalate(delim, xs)
    set {dlm, my text item delimiters} to ¬
        {my text item delimiters, delim}
    set str to xs as text
    set my text item delimiters to dlm
    str
end intercalate


-- justifyLeft :: Int -> Char -> String -> String
on justifyLeft(n, cFiller)
    script
        on |λ|(strText)
            if n > length of strText then
                text 1 thru n of (strText & replicate(n, cFiller))
            else
                strText
            end if
        end |λ|
    end script
end justifyLeft


-- length :: [a] -> Int
on |length|(xs)
    set c to class of xs
    if list is c or string is c then
        length of xs
    else
        (2 ^ 29 - 1) -- (maxInt - simple proxy for non-finite)
    end if
end |length|


-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
    -- The list obtained by applying f
    -- to each element of xs.
    tell mReturn(f)
        set lng to length of xs
        set lst to {}
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs, i, xs)
        end repeat
        return lst
    end tell
end map


-- maximumBy :: (a -> a -> Ordering) -> [a] -> a
on maximumBy(f, xs)
    set cmp to mReturn(f)
    script max
        on |λ|(a, b)
            if a is missing value or cmp's |λ|(a, b) < 0 then
                b
            else
                a
            end if
        end |λ|
    end script
    
    foldl(max, missing value, xs)
end maximumBy


-- min :: Ord a => a -> a -> a
on min(x, y)
    if y < x then
        y
    else
        x
    end if
end min


-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
    -- 2nd class handler function lifted into 1st class script wrapper. 
    if script is class of f then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn


-- Egyptian multiplication - progressively doubling a list, appending
-- stages of doubling to an accumulator where needed for binary 
-- assembly of a target length
-- replicate :: Int -> a -> [a]
on replicate(n, a)
    set out to {}
    if 1 > n then return out
    set dbl to {a}
    
    repeat while (1 < n)
        if 0 < (n mod 2) then set out to out & dbl
        set n to (n div 2)
        set dbl to (dbl & dbl)
    end repeat
    return out & dbl
end replicate


-- reverse :: [a] -> [a]
on |reverse|(xs)
    if class of xs is text then
        (reverse of characters of xs) as text
    else
        reverse of xs
    end if
end |reverse|


-- showList :: [a] -> String
on showList(xs)
    script show
        on |λ|(x)
            if text is class of x then
                "'" & x & "'"
            else
                x as text
            end if
        end |λ|
    end script
    if {} ≠ xs then
        "[" & intercalate(", ", map(show, xs)) & "]"
    else
        "[]"
    end if
end showList


-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
    set c to class of xs
    if list is c then
        if 0 < n then
            items 1 thru min(n, length of xs) of xs
        else
            {}
        end if
    else if string is c then
        if 0 < n then
            text 1 thru min(n, length of xs) of xs
        else
            ""
        end if
    else if script is c then
        set ys to {}
        repeat with i from 1 to n
            set v to |λ|() of xs
            if missing value is v then
                return ys
            else
                set end of ys to v
            end if
        end repeat
        return ys
    else
        missing value
    end if
end take


-- takeWhile :: (a -> Bool) -> [a] -> [a]
-- takeWhile :: (Char -> Bool) -> String -> String
on takeWhile(p, xs)
    if script is class of xs then
        takeWhileGen(p, xs)
    else
        tell mReturn(p)
            repeat with i from 1 to length of xs
                if not |λ|(item i of xs) then ¬
                    return take(i - 1, xs)
            end repeat
        end tell
        return xs
    end if
end takeWhile


-- transpose :: [[a]] -> [[a]]
on transpose(rows)
    set w to length of maximumBy(comparing(|length|), rows)
    set paddedRows to map(justifyLeft(w, "x"), rows)
    
    script cols
        on |λ|(_, iCol)
            script cell
                on |λ|(row)
                    item iCol of row
                end |λ|
            end script
            concatMap(cell, paddedRows)
        end |λ|
    end script
    
    map(cols, item 1 of rows)
end transpose


-- unlines :: [String] -> String
on unlines(xs)
    -- A single string formed by the intercalation
    -- of a list of strings with the newline character.
    set {dlm, my text item delimiters} to ¬
        {my text item delimiters, linefeed}
    set str to xs as text
    set my text item delimiters to dlm
    str
end unlines
Output:
['throne', 'sousaphone', 'tone'] -> 'one'
['prefix', 'suffix', 'infix'] -> 'fix'
['remark', 'spark', 'aardvark', 'lark'] -> 'ark'
['ectoplasm', 'banana', 'brick'] -> ''

Arturo

lcs: function [l][
	ret: ""
    idx: 0

    lst: map l => reverse
 
    while [true] [
        thisLetter: ""
        loop lst 'word [
        	if idx=size word -> return reverse ret
        	if thisLetter="" -> thisLetter: get split word idx
        	if thisLetter<>get split word idx -> return reverse ret
        ]
        ret: ret ++ thisLetter
        idx: idx + 1
    ]
]

print lcs ["Sunday" "Monday" "Tuesday" "Wednesday" "Thursday" "Friday" "Saturday"]
print lcs ["throne" "throne"]
print lcs ["throne" "dungeon"]
print lcs ["cheese"]
print lcs ["prefix" "suffix"]
Output:
day
throne

cheese
fix

AutoHotkey

Longest_common_suffix(data){
	for num, v in StrSplit(data.1)
		for i, word in data	
			if (SubStr(word, 1-num) <> SubStr(data.1, 1-num))
				return num=1 ? "" : SubStr(word, 2-num)
	return SubStr(word, 1-num)
}
Examples:
MsgBox % ""
. "`n" Longest_common_suffix(["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"])
. "`n" Longest_common_suffix(["throne", "throne"])
. "`n" Longest_common_suffix(["throne", "dungeon"])
. "`n" Longest_common_suffix(["throne", "", "throne"])
. "`n" Longest_common_suffix(["cheese"])
. "`n" Longest_common_suffix([""])
. "`n" Longest_common_suffix(["prefix", "suffix"])
. "`n" Longest_common_suffix(["bar", "foobar"])
return
Output:
day
throne


cheese

fix
bar

AWK

# syntax: GAWK -f LONGEST_COMMON_SUFFIX.AWK
BEGIN {
    arr1[++n1] = "AAbcd,Abcd,abcd,bcd"
    arr1[++n1] = "11Sunday,2Sunday"
    arr1[++n1] = "Sunday,Monday"
    arr1[++n1] = "Sunday,Monday,day"
    arr1[++n1] = "Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday"
    arr1[++n1] = "crucifix,infix,prefix,suffix"
    arr1[++n1] = "identical,identical"
    arr1[++n1] = ","
    arr1[++n1] = "this,has,nothing,in,common"
    for (i=1; i<=n1; i++) {
      n2 = split(arr1[i],arr2,",")
      min_wid = 999
      for (j=1; j<=n2; j++) {
        leng = length(arr2[j])
        if (min_wid > leng) {
          min_wid = leng
          min_col = j
        }
      }
      cols = 0
      for (j=1; j<=min_wid; j++) {
        delete arr3
        for (k=1; k<n2; k++) {
          arr3[substr(arr2[k],length(arr2[k])+1-j)] = ""
          arr3[substr(arr2[k+1],length(arr2[k+1])+1-j)] = ""
        }
        if (length(arr3) == 1) {
          cols++
        }
      }
      printf("'%s' : '%s'\n",arr1[i],(cols == 0) ? "" : substr(arr2[min_col],length(arr2[min_col])+1-cols))
    }
    exit(0)
}
Output:
'AAbcd,Abcd,abcd,bcd' : 'bcd'
'11Sunday,2Sunday' : 'Sunday'
'Sunday,Monday' : 'nday'
'Sunday,Monday,day' : 'day'
'Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday' : 'day'
'crucifix,infix,prefix,suffix' : 'fix'
'identical,identical' : 'identical'
',' : ''
'this,has,nothing,in,common' : ''

BaCon

FUNCTION Common_Suffix$(data$)

    LOCAL x, size
    LOCAL delim$

    REPEAT
        delim$ = ""
        INCR size

        FOR x = 1 TO AMOUNT(data$)
            delim$ = APPEND$(delim$, 0, RIGHT$(TOKEN$(data$, x), size))
        NEXT
    UNTIL AMOUNT(UNIQ$(delim$)) <> 1

    RETURN RIGHT$(TOKEN$(data$, 1), size-1)

ENDFUNCTION

PRINT "The common suffix is: '", Common_Suffix$("baabababc baabc bbbabc"), "'"
PRINT "The common suffix is: '", Common_Suffix$("Monday Tuesday Wednesday Thursday Friday Saturday Sunday"), "'"
PRINT "The common suffix is: '", Common_Suffix$("longest common suffix"), "'"
PRINT "The common suffix is: '", Common_Suffix$("prefix suffix"), "'"
PRINT "The common suffix is: '", Common_Suffix$(""), "'"
Output:
The common suffix is: 'abc'
The common suffix is: 'day'
The common suffix is: ''
The common suffix is: 'fix'
The common suffix is: ''

C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct node_t {
    char *elem;
    int length;
    struct node_t *next;
} node;

node *make_node(char *s) {
    node *t = malloc(sizeof(node));
    t->elem = s;
    t->length = strlen(s);
    t->next = NULL;
    return t;
}

void append_node(node *head, node *elem) {
    while (head->next != NULL) {
        head = head->next;
    }
    head->next = elem;
}

void print_node(node *n) {
    putc('[', stdout);
    while (n != NULL) {
        printf("`%s` ", n->elem);
        n = n->next;
    }
    putc(']', stdout);
}

char *lcs(node *list) {
    int minLen = INT_MAX;
    int i;

    char *res;
    node *ptr;

    if (list == NULL) {
        return "";
    }
    if (list->next == NULL) {
        return list->elem;
    }

    for (ptr = list; ptr != NULL; ptr = ptr->next) {
        minLen = min(minLen, ptr->length);
    }
    if (minLen == 0) {
        return "";
    }

    res = "";
    for (i = 1; i < minLen; i++) {
        char *suffix = &list->elem[list->length - i];

        for (ptr = list->next; ptr != NULL; ptr = ptr->next) {
            char *e = &ptr->elem[ptr->length - i];
            if (strcmp(suffix, e) != 0) {
                return res;
            }
        }

        res = suffix;
    }

    return res;
}

void test(node *n) {
    print_node(n);
    printf(" -> `%s`\n", lcs(n));
}

void case1() {
    node *n = make_node("baabababc");
    append_node(n, make_node("baabc"));
    append_node(n, make_node("bbbabc"));
    test(n);
}

void case2() {
    node *n = make_node("baabababc");
    append_node(n, make_node("baabc"));
    append_node(n, make_node("bbbazc"));
    test(n);
}

void case3() {
    node *n = make_node("Sunday");
    append_node(n, make_node("Monday"));
    append_node(n, make_node("Tuesday"));
    append_node(n, make_node("Wednesday"));
    append_node(n, make_node("Thursday"));
    append_node(n, make_node("Friday"));
    append_node(n, make_node("Saturday"));
    test(n);
}

void case4() {
    node *n = make_node("longest");
    append_node(n, make_node("common"));
    append_node(n, make_node("suffix"));
    test(n);
}

void case5() {
    node *n = make_node("suffix");
    test(n);
}

void case6() {
    node *n = make_node("");
    test(n);
}

int main() {
    case1();
    case2();
    case3();
    case4();
    case5();
    case6();
    return 0;
}
Output:
[`baabababc` `baabc` `bbbabc` ] -> `abc`
[`baabababc` `baabc` `bbbazc` ] -> `c`
[`Sunday` `Monday` `Tuesday` `Wednesday` `Thursday` `Friday` `Saturday` ] -> `day`
[`longest` `common` `suffix` ] -> ``
[`suffix` ] -> `suffix`
[`` ] -> ``

C++

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

std::string lcs(const std::vector<std::string>& strs) {
    std::vector<std::string::const_reverse_iterator> backs;
    std::string s;
    
    if (strs.size() == 0) return "";
    if (strs.size() == 1) return strs[0];
    
    for (auto& str : strs) backs.push_back(str.crbegin());
    
    while (backs[0] != strs[0].crend()) {
        char ch = *backs[0]++;
        for (std::size_t i = 1; i<strs.size(); i++) {
            if (backs[i] == strs[i].crend()) goto done;
            if (*backs[i] != ch) goto done;
            backs[i]++;
        }
        s.push_back(ch);
    }
    
done:
    reverse(s.begin(), s.end());
    return s;
}

void test(const std::vector<std::string>& strs) {
    std::cout << "[";
    for (std::size_t i = 0; i<strs.size(); i++) {
        std::cout << '"' << strs[i] << '"';
        if (i != strs.size()-1) std::cout << ", ";
    }
    std::cout << "] -> `" << lcs(strs) << "`\n";
}

int main() {
    std::vector<std::string> t1 = {"baabababc", "baabc", "bbabc"};
    std::vector<std::string> t2 = {"baabababc", "baabc", "bbazc"};
    std::vector<std::string> t3 = 
        {"Sunday", "Monday", "Tuesday", "Wednesday", "Friday", "Saturday"};
    std::vector<std::string> t4 = {"longest", "common", "suffix"};
    std::vector<std::string> t5 = {""};
    std::vector<std::string> t6 = {};
    std::vector<std::string> t7 = {"foo", "foo", "foo", "foo"};

    std::vector<std::vector<std::string>> tests = {t1,t2,t3,t4,t5,t6,t7};
    
    for (auto t : tests) test(t);
    return 0;
}
Output:
["baabababc", "baabc", "bbabc"] -> `abc`
["baabababc", "baabc", "bbazc"] -> `c`
["Sunday", "Monday", "Tuesday", "Wednesday", "Friday", "Saturday"] -> `day`
["longest", "common", "suffix"] -> ``
[""] -> ``
[] -> ``
["foo", "foo", "foo", "foo"] -> `foo`

Cowgol

include "cowgol.coh";
include "strings.coh";

sub lcs(arr: [[uint8]], len: intptr): (s: [uint8]) is
    if len == 0 then
        s := "";
        return;
    elseif len == 1 then
        s := [arr];
        return;
    end if;
    
    s := [arr];
    var slen := StrLen(s);
    s := s + slen;
    
    arr := @next arr;
    len := len - 1;

    while len > 0 and slen > 0 loop
        var c := [arr];
        var clen := StrLen(c);
        c := c + clen;
        
        if clen > slen then
            clen := slen;
        end if;
        
        while clen > 0 and [c] == [s] loop
            c := @prev c;
            s := @prev s;
            clen := clen - 1;
        end loop;
        
        slen := StrLen(s);
        s := s + slen;
        
        arr := @next arr;
        len := len - 1;
    end loop;
    
    s := s - slen + 1;
end sub;

sub test(arr: [[uint8]], len: intptr) is
    var s := arr;
    var l := len;
    print_char('[');
    while l > 0 loop
        print_char('"');
        print([s]);
        print_char('"');
        s := @next s;
        l := l - 1;
        if l > 0 then
            print(", ");
        end if;
    end loop;
    print("] -> `");
    print(lcs(arr, len));
    print("`\n");
end sub;

var test1: [uint8][] := {"baabababc", "baabc", "bbbabc"};
var test2: [uint8][] := {"baabababc", "baabc", "bbbazc"};
var test3: [uint8][] :=
    {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday",
     "Friday", "Saturday"};
var test4: [uint8][] := {"longest", "common", "suffix"};
var test5: [uint8][] := {""};
var test6: [uint8][] := {};

test(&test1[0], @sizeof test1);
test(&test2[0], @sizeof test2);
test(&test3[0], @sizeof test3);
test(&test4[0], @sizeof test4);
test(&test5[0], @sizeof test5);
test(&test6[0], @sizeof test6);
Output:
["baabababc", "baabc", "bbbabc"] -> `abc`
["baabababc", "baabc", "bbbazc"] -> `c`
["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"] -> `day`
["longest", "common", "suffix"] -> ``
[""] -> ``
[] -> ``

D

Translation of: Kotlin
import std.algorithm;
import std.stdio;

string lcs(string[] a) {
    auto le = a.length;
    if (le == 0) {
        return "";
    }
    if (le == 1) {
        return a[0];
    }
    auto le0 = a[0].length;
    auto minLen = a.map!"a.length".reduce!"min(a,b)";
    if (minLen == 0) {
        return "";
    }
    auto res = "";
    foreach (i; 1..minLen) {
        auto suffix = a[0][le0 - i .. $];
        foreach (e; a[1..$]) {
            if (!e.endsWith(suffix)) {
                return res;
            }
        }
        res = suffix;
    }
    return "";
}

void main() {
    auto tests = [
        ["baabababc", "baabc", "bbbabc"],
        ["baabababc", "baabc", "bbbazc"],
        ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"],
        ["longest", "common", "suffix"],
        ["suffix"],
        [""]
    ];
    foreach (test; tests) {
        writeln(test, " -> `", lcs(test), '`');
    }
}
Output:
["baabababc", "baabc", "bbbabc"] -> `abc`
["baabababc", "baabc", "bbbazc"] -> `c`
["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"] -> `day`
["longest", "common", "suffix"] -> ``
["suffix"] -> `suffix`
[""] -> ``

Delphi

Library: System.SysUtils
Library: Types
Translation of: Ring
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.
Output:
Input:
baabababc
baabc
bbbabc

Longest common suffix = abc

Factor

Works with: Factor version 0.99 2020-07-03
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 .
Output:
"abc"
"c"
""


FreeBASIC

Translation of: Ring
Dim As String pre(1 To ...) = {"baabababc","baabc","bbbabc"}
Dim As Integer longitud = Ubound(pre)
Dim As Integer lenList(longitud)
Dim As Integer n, m, longest
Dim As String temp

Function rever(Byval text As String) As String
    Dim As String text2 = text
    Dim As Integer x, lt = Len(text)
    For x = 0 To lt Shr 1 - 1
        Swap text2[x], text2[lt - x - 1]
    Next x
    Return text2
End Function

Print "There are"; longitud; " words in the list:  ";
For n = 1 To longitud
    Print pre(n); "  ";
    temp = pre(n)
    pre(n) = rever(temp)
    lenList(n) = Len(pre(n))
Next n

For m = 1 To lenList(1)
    Dim As String sub1 = Mid(pre(1),1,m)
    Dim As String sub2 = Mid(pre(2),1,m) 
    Dim As String sub3 = Mid(pre(3),1,m)
    If sub1 = sub2 And sub2 = sub3 Then longest = m
Next m

Dim As String longPrefix = Mid(pre(1),1,longest)
longPrefix = rever(longPrefix)

Print !"\n\nThe longest common suffix is:  "; longPrefix
Sleep
Output:
There are 3 words in the list:  baabababc  baabc  bbbabc

The longest common suffix is:  abc


Go

Translation of: Wren
package main

import (
    "fmt"
    "strings"
)

func lcs(a []string) string {
    le := len(a)
    if le == 0 {
        return ""
    }
    if le == 1 {
        return a[0]
    }
    le0 := len(a[0])
    minLen := le0
    for i := 1; i < le; i++ {
        if len(a[i]) < minLen {
            minLen = len(a[i])
        }
    }
    if minLen == 0 {
        return ""
    }
    res := ""
    a1 := a[1:]
    for i := 1; i <= minLen; i++ {
        suffix := a[0][le0-i:]
        for _, e := range a1 {
            if !strings.HasSuffix(e, suffix) {
                return res
            }
        }
        res = suffix
    }
    return res
}

func main() {
    tests := [][]string{
        {"baabababc", "baabc", "bbbabc"},
        {"baabababc", "baabc", "bbbazc"},
        {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"},
        {"longest", "common", "suffix"},
        {"suffix"},
        {""},
    }
    for _, test := range tests {
        fmt.Printf("%v -> \"%s\"\n", test, lcs(test))
    }
}
Output:
[baabababc baabc bbbabc] -> "abc"
[baabababc baabc bbbazc] -> "c"
[Sunday Monday Tuesday Wednesday Thursday Friday Saturday] -> "day"
[longest common suffix] -> ""
[suffix] -> "suffix"
[] -> ""

Haskell

This task clearly needs a little more work to bring it up to the usual standard – it's rather underspecified, and bereft of test samples – but one response, for the moment, might be something like: 

import Data.List (transpose)

longestCommonSuffix :: [String] -> String
longestCommonSuffix =
  foldr (flip (<>) . return . head) [] .
  takeWhile (all =<< (==) . head) . transpose . fmap reverse

main :: IO ()
main =
  mapM_
    (putStrLn . longestCommonSuffix)
    [ [ "Sunday"
      , "Monday"
      , "Tuesday"
      , "Wednesday"
      , "Thursday"
      , "Friday"
      , "Saturday"
      ]
    , [ "Sondag"
      , "Maandag"
      , "Dinsdag"
      , "Woensdag"
      , "Donderdag"
      , "Vrydag"
      , "Saterdag"
      , "dag"
      ]
    ]
Output:
day
dag


J

lcs =: [: |. [: ({. #~ [: *./\ [: *./ 2 =/\ ]) >@(|. each)

test1 =: 'baabababc';'baabc';'bbabc'
test2 =: 'baabababc';'baabc';'bbazc'
test3 =: 'Sunday';'Monday';'Tuesday';'Wednesday';'Friday';'Saturday'
test4 =: 'longest';'common';'suffix'
tests =: test1;test2;test3;<test4
echo@((1{":),' -> ', 1{":@<@lcs) each tests
exit''
Output:
│baabababc│baabc│bbabc│ -> │abc│
│baabababc│baabc│bbazc│ -> │c│
│Sunday│Monday│Tuesday│Wednesday│Friday│Saturday│ -> │day│
│longest│common│suffix│ -> ││

Java

Translation of: Kotlin
import java.util.List;

public class App {
    private static String lcs(List<String> a) {
        var le = a.size();
        if (le == 0) {
            return "";
        }
        if (le == 1) {
            return a.get(0);
        }
        var le0 = a.get(0).length();
        var minLen = le0;
        for (int i = 1; i < le; i++) {
            if (a.get(i).length() < minLen) {
                minLen = a.get(i).length();
            }
        }
        if (minLen == 0) {
            return "";
        }
        var res = "";
        var a1 = a.subList(1, a.size());
        for (int i = 1; i < minLen; i++) {
            var suffix = a.get(0).substring(le0 - i);
            for (String e : a1) {
                if (!e.endsWith(suffix)) {
                    return res;
                }
            }
            res = suffix;
        }
        return "";
    }

    public static void main(String[] args) {
        var tests = List.of(
            List.of("baabababc", "baabc", "bbbabc"),
            List.of("baabababc", "baabc", "bbbazc"),
            List.of("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"),
            List.of("longest", "common", "suffix"),
            List.of("suffix"),
            List.of("")
        );
        for (List<String> test : tests) {
            System.out.printf("%s -> `%s`\n", test, lcs(test));
        }
    }
}
Output:
[baabababc, baabc, bbbabc] -> `abc`
[baabababc, baabc, bbbazc] -> `c`
[Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday] -> `day`
[longest, common, suffix] -> ``
[suffix] -> `suffix`
[] -> ``

JavaScript

(() => {
    'use strict';

    // --------------- LONGEST COMMON SUFFIX ---------------

    // longestCommonSuffix :: [String] -> String
    const longestCommonSuffix = xs => (
            1 < xs.length ? (
                takeWhile(allSame)(
                    transpose(xs.map(reverse))
                ).map(fst).reverse()
            ) : xs
        ).join('');


    // ----------------------- TEST ------------------------
    const main = () =>
        fTable('Longest common suffix:\n')(unwords)(
            quoted("'")
        )(longestCommonSuffix)([
            'throne saxophone tone',
            'prefix suffix infix',
            'remark spark aardvark lark',
            'ectoplasm banana brick'
        ].map(words))


    // ----------------- GENERIC FUNCTIONS -----------------

    // allSame :: [a] -> Bool
    const allSame = xs =>
        2 > xs.length || (
            h => xs.slice(1).every(x => h === x)
        )(xs[0]);


    // fst :: (a, b) -> a
    const fst = tpl =>
        // First member of a pair.
        tpl[0];


    // fTable :: String -> (a -> String) -> 
    // (b -> String) -> (a -> b) -> [a] -> String
    const fTable = s =>
        // Heading -> x display function ->
        //           fx display function ->
        //    f -> values -> tabular string
        xShow => fxShow => f => xs => {
            const
                ys = xs.map(xShow),
                w = Math.max(...ys.map(length));
            return s + '\n' + zipWith(
                a => b => a.padStart(w, ' ') + ' -> ' + b
            )(ys)(
                xs.map(x => fxShow(f(x)))
            ).join('\n');
        };


    // length :: [a] -> Int
    const length = xs =>
        // Returns Infinity over objects without finite
        // length. This enables zip and zipWith to choose
        // the shorter argument when one is non-finite,
        // like cycle, repeat etc
        'GeneratorFunction' !== xs.constructor
        .constructor.name ? xs.length : Infinity;


    // map :: (a -> b) -> [a] -> [b]
    const map = f =>
        // The list obtained by applying f
        // to each element of xs.
        // (The image of xs under f).
        xs => [...xs].map(f);


    // reverse :: [a] -> [a]
    const reverse = xs =>
        'string' !== typeof xs ? (
            xs.slice(0).reverse()
        ) : xs.split('').reverse().join('');


    // str :: a -> String
    const str = x =>
        Array.isArray(x) && x.every(
            v => ('string' === typeof v) && (1 === v.length)
        ) ? (
            x.join('')
        ) : x.toString();


    // take :: Int -> [a] -> [a]
    // take :: Int -> String -> String
    const take = n =>
        // The first n elements of a list,
        // string of characters, or stream.
        xs => 'GeneratorFunction' !== xs
        .constructor.constructor.name ? (
            xs.slice(0, n)
        ) : [].concat.apply([], Array.from({
            length: n
        }, () => {
            const x = xs.next();
            return x.done ? [] : [x.value];
        }));


    // takeWhile :: (a -> Bool) -> [a] -> [a]
    // takeWhile :: (Char -> Bool) -> String -> String
    const takeWhile = p =>
        xs => (
            n => xs.slice(
                0, 0 < n ? until(
                    i => n === i || !p(xs[i])
                )(i => 1 + i)(0) : 0
            )
        )(xs.length);


    // transpose :: [[a]] -> [[a]]
    const transpose = xss => {
        // If some of the rows are shorter than the following rows, 
        // their elements are skipped:
        // > transpose [[10,11],[20],[],[30,31,32]] == [[10,20,30],[11,31],[32]]
        const go = xss =>
            0 < xss.length ? (() => {
                const
                    h = xss[0],
                    t = xss.slice(1);
                return 0 < h.length ? [
                    [h[0]].concat(t.reduce(
                        (a, xs) => a.concat(
                            0 < xs.length ? (
                                [xs[0]]
                            ) : []
                        ),
                        []
                    ))
                ].concat(go([h.slice(1)].concat(
                    t.map(xs => xs.slice(1))
                ))) : go(t);
            })() : [];
        return go(xss);
    };


    // until :: (a -> Bool) -> (a -> a) -> a -> a
    const until = p => f => x => {
        let v = x;
        while (!p(v)) v = f(v);
        return v;
    };

    // unwords :: [String] -> String
    const unwords = xs =>
        // A space-separated string derived
        // from a list of words.
        xs.join(' ');


    // quoted :: Char -> String -> String
    const quoted = c =>
        // A string flanked on both sides
        // by a specified quote character.
        s => c + s + c;

    // words :: String -> [String]
    const words = s =>
        // List of space-delimited sub-strings.
        s.split(/\s+/);


    // zipWithList :: (a -> b -> c) -> [a] -> [b] -> [c]
    const zipWith = f =>
        // A list constructed by zipping with a
        // custom function, rather than with the
        // default tuple constructor.
        xs => ys => ((xs_, ys_) => {
            const lng = Math.min(length(xs_), length(ys_));
            return take(lng)(xs_).map(
                (x, i) => f(x)(ys_[i])
            );
        })([...xs], [...ys]);

    // MAIN ---
    return main();
})();
Output:
Longest common suffix:

     throne saxophone tone -> 'one'
       prefix suffix infix -> 'fix'
remark spark aardvark lark -> 'ark'
    ectoplasm banana brick -> ''

jq

Works with: jq

Works with gojq, the Go implementation of jq

This entry uses `longest_common_prefix` from Longest_common_suffix#jq and so the definition is not repeated here. 

# Input: an array of strings
def longest_common_suffix:
  def r: explode | reverse | implode;
  if length == 0 then ""        # by convention
  elif length == 1 then .[0]    # for speed
  else map(r)
  | longest_common_prefix
  | r
  end;

Test Cases 

def test:
  ["baabababc","baabc","bbbabc"],
  ["baabababc","baabc","bbbazc"],
  [""],
  [],
  ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"],
  ["throne","cone", "bone"]
  | "\(.) => \(longest_common_suffix)";

test
Output:
["baabababc","baabc","bbbabc"] => abc
["baabababc","baabc","bbbazc"] => c
[""] => 
[] => 
["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"] => day
["throne","cone","bone"] => one

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([""]))
Output:
abc
c

Kotlin

Translation of: Go
fun lcs(a: List<String>): String {
    val le = a.size
    if (le == 0) {
        return ""
    }
    if (le == 1) {
        return a[0]
    }
    val le0 = a[0].length
    var minLen = le0
    for (i in 1 until le) {
        if (a[i].length < minLen) {
            minLen = a[i].length
        }
    }
    if (minLen == 0) {
        return ""
    }
    var res = ""
    val a1 = a.subList(1, a.size)
    for (i in 1..minLen) {
        val suffix = a[0].substring(le0 - i)
        for (e in a1) {
            if (!e.endsWith(suffix)) {
                return res
            }
        }
        res = suffix
    }
    return ""
}

fun main() {
    val tests = listOf(
        listOf("baabababc", "baabc", "bbbabc"),
        listOf("baabababc", "baabc", "bbbazc"),
        listOf("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"),
        listOf("longest", "common", "suffix"),
        listOf("suffix"),
        listOf("")
    )
    for (test in tests) {
        println("$test -> `${lcs(test)}`")
    }
}
Output:
[baabababc, baabc, bbbabc] -> `abc`
[baabababc, baabc, bbbazc] -> `c`
[Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday] -> `day`
[longest, common, suffix] -> ``
[suffix] -> `suffix`
[] -> ``

Ksh

#!/bin/ksh

# Longest common suffix

#	# Variables:
#
typeset -a arr1=( "Sunday" "Monday" "Tuesday" "Wednesday" "Thursday" "Friday" "Saturday" )
typeset -a arr2=( "baabababc" "baabc" "bbbabc" )
typeset -a arr3=( "baabababc" "babc" "bbbabc" )
typeset -a arr4=( "longest" "common" "suffix" )
typeset -a arr5=( "suffix" )
typeset -a arr6=( "" )

#	# Functions:
#

#	# Function _minlenele(arr) - return the shortest element in an array
#
function _minlenele {
	typeset _arr ; nameref _arr="$1"
	typeset _min _i ; integer _i

	_min=${_arr[0]}
	for ((_i=1; _i<${#_arr[*]}; _i++)); do
		(( ${#_arr[_i]} < ${#_min} )) && _min=${_arr[_i]}
	done
	echo "${_min}"
}

 ######
# main #
 ######

for array in arr1 arr2 arr3 arr4 arr5 arr6; do
	nameref arr=${array}
	printf "\n( %s ) -> " "${arr[*]}"
	suffix=$(_minlenele arr)
	for ((j=${#suffix}; j>0; j--)); do
		for ((i=0; i<${#arr[*]}; i++)); do
			[[ ${arr[i]%${suffix: -${j}}} == ${arr[i]} ]] && continue 2
		done
		printf "'%s'" ${suffix: -${j}}
		break
	done
	typeset +n arr
done
echo
Output:
( Sunday Monday Tuesday Wednesday Thursday Friday Saturday ) -> 'day'
( baabababc baabc bbbabc ) -> 'abc'
( baabababc babc bbbabc ) -> 'babc'
( longest common suffix ) -> 
( suffix ) -> 'suffix'
(  ) ->

Mathematica/Wolfram Language

ClearAll[LCS]
LCS[x_List] := Module[{l, s},
  If[Length[x] > 0,
   l = Min[StringLength /@ x];
   s = Characters[StringTake[x, -l]];
   s //= Transpose;
   l = LengthWhile[Reverse@s, Apply[SameQ]];
   StringTake[First[x], -l]
   ,
   ""
   ]
  ]
LCS[{"throne", "sousaphone"}]
LCS[{"prefix", "suffix"}]
LCS[{"remark", "spark", "aardvark"}]
LCS[{"throne", "", "throne"}]
LCS[{"throne", "throne"}]
LCS[{"cheese"}]
LCS[{""}]
LCS[{}]
LCS[{"ectoplasm", "banana"}]
Output:
"one"
"fix"
"ark"
""
"throne"
"cheese"
""
""
""

Nim

import sequtils, strformat, strutils

func lcs(list: varargs[string]): string =
  if list.len == 0: return
  result = list[0]
  for i in 1..list.high:
    var newLength = 0
    for j in 1..result.len:
      if j > list[i].len or list[i][^j] != result[^j]:
        break
      inc newLength
    result = result[^newLength..^1]

proc test(list: varargs[string]) =
  let lst = list.mapIt('"' & it & '"').join(", ")
  echo &"lcs({lst}) = \"{lcs(list)}\""


test("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday")
test("throne", "throne")
test("throne", "dungeon")
test("cheese")
test("")
test()
test("prefix", "suffix")
test("send", "lend")
Output:
lcs("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday") = "day"
lcs("throne", "throne") = "throne"
lcs("throne", "dungeon") = ""
lcs("cheese") = "cheese"
lcs("") = ""
lcs() = ""
lcs("prefix", "suffix") = "fix"
lcs("send", "lend") = "end"

Perl

Based on Longest_common_prefix Perl entry. 

use strict;
use warnings;
use feature 'say';

sub lcs {
    for (0..$#_) { $_[$_] = reverse $_[$_] }
    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)]}';
}
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: 

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

Python

Pending a fuller task statement and some test samples:

Works with: Python version 3
'''Longest common suffix'''

from itertools import takewhile
from functools import reduce


# longestCommonSuffix :: [String] -> String
def longestCommonSuffix(xs):
    '''Longest suffix shared by all
       strings in xs.
    '''
    def allSame(cs):
        h = cs[0]
        return all(h == c for c in cs[1:])

    def firstCharPrepended(s, cs):
        return cs[0] + s
    return reduce(
        firstCharPrepended,
        takewhile(
            allSame,
            zip(*(reversed(x) for x in xs))
        ),
        ''
    )


# -------------------------- TEST --------------------------
# main :: IO ()
def main():
    '''Test'''

    samples = [
        [
            "Sunday", "Monday", "Tuesday", "Wednesday",
            "Thursday", "Friday", "Saturday"
        ], [
            "Sondag", "Maandag", "Dinsdag", "Woensdag",
            "Donderdag", "Vrydag", "Saterdag"
        ]
    ]
    for xs in samples:
        print(
            longestCommonSuffix(xs)
        )


# MAIN ---
if __name__ == '__main__':
    main()
Output:
day
dag

Quackery

commonprefix is defined at Longest common prefix#Quackery. 

  [ [] swap
    witheach 
      [ reverse nested join ]
    commonprefix
    reverse ]                 is commonsuffix ( [ --> $ )

  $ "monday tuesday wednesday thursday friday saturday sunday"
  nest$ commonsuffix echo$
Output:
day

Raku

Works with: Rakudo version 2020.07
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',
  ''
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. 

/*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)
output   when using the default input:
There are  3  words in the list:  baabababc baabc bbbabc

The longest common suffix is:  abc

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
Output:
Input:
baabababc
baabc
bbbabc
Longest common suffix = abc

Ruby

Testcases taken from Go. 

tests = [["baabababc", "baabc", "bbbabc"],
 ["baabababc", "baabc", "bbbazc"],
 ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"],
 ["longest", "common", "suffix"],
 ["suffix"],
 [""], 
]
def lcs(ar)
  i = (0..ar.first.size).detect{|s| ar.all?{|word| word.end_with? ar.first[s..-1]} }
  ar.first[i..-1]
end

tests.each{|test| p lcs(test) }
Output:
"abc"
"c"
"day"
""
"suffix"
""

sed

$q
N
s/\(.*\)\(\n\).*\1$/\2\1/
D

$ printf '%s\n' Sunday '' Monday Tuesday | sed -f suffix.sed

$ printf '%s\n' Sunday Monday Tuesday | sed -f suffix.sed
day
$ printf '%s\n' Sunday Monday | sed -f suffix.sed
nday

Standard ML

val lcs =
  let
    val commonSuffix = fn (s0, s1) =>
      let
        val rec pairTakeREq = fn (0, _) => s0 | (_, 0) => s1 | (i, j) =>
          let
            val i' = i - 1 and j' = j - 1
          in
            if String.sub (s0, i') = String.sub (s1, j')
            then pairTakeREq (i', j')
            else String.extract (s0, i, NONE)
          end
      in
        pairTakeREq (size s0, size s1)
      end
  in
    fn [] => "" | x :: xs => foldl commonSuffix x xs
  end

V (Vlang)

Translation of: go
fn lcs(a []string) string {
    // Special cases first
    match a.len {
        0 {
            return ""
        }
        1 {
            return a[0]
        }
        else {}
    }
    le0 := a[0].len
    mut min_len := le0
    for i in 1..a.len {
        if a[i].len < min_len {
            min_len = a[i].len
        }
    }
    if min_len == 0 {
        return ""
    }
    mut res := ""
    a1 := a[1..]
    for i := 1; i <= min_len; i++ {
        suffix := a[0][le0-i..]
        for e in a1 {
            if e.index(suffix) or {0} + suffix.len != e.len {
                return res
            }
        }
        res = suffix
    }
    return res
}
 
// Normally something like this would be a Testlcs function in *_test.go
// and use the testing package to report failures.
fn main() {
    for l in [
        ["baabababc", "baabc", "bbbabc"],
        ["baabababc", "baabc", "bbbazc"],
        ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"],
        ["longest", "common", "suffix"],
        ["suffix"],
        [""],
     ] {
        println("lcs($l) = ${lcs(l)}")
    }
}
Output:
Same as Go entry

Wren

var lcs = Fn.new { |a|
    if (a.count == 0) return ""
    if (a.count == 1) return a[0]
    var minLen = a.reduce(a[0].count) { |min, s| (s.count < min) ? s.count : min }
    if (minLen == 0) return ""
    var res = ""
    for (i in 1..minLen) {
        var suffix = a[0][-i..-1]
        for (e in a.skip(1)) {
            if (!e.endsWith(suffix)) return res
        }
        res = suffix
    }
    return res
}

var tests = [
    ["baabababc","baabc","bbbabc"],
    ["baabababc","baabc","bbbazc"],
    ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"],
    ["longest", "common", "suffix"],
    ["suffix"],
    [""]
]
for (test in tests) System.print("%(test) -> \"%(lcs.call(test))\"")
Output:
[baabababc, baabc, bbbabc] -> "abc"
[baabababc, baabc, bbbazc] -> "c"
[Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday] -> "day"
[longest, common, suffix] -> ""
[suffix] -> "suffix"
[] -> ""
Retrieved from "https://rosettacode.org/wiki/Longest_common_suffix?oldid=315098"