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[edit]

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![edit]

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[edit]

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[edit]

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[edit]

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[edit]

Procedural[edit]

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[edit]

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[edit]

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[edit]

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[edit]

# 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[edit]

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[edit]

#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++[edit]

#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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

(() => {
    '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[edit]

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[edit]

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[edit]

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[edit]

#!/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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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

Standard ML[edit]

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

Vlang[edit]

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[edit]

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"
[] -> ""