Longest common prefix

It is often useful to find the common prefix of a set of strings, that is, the longest initial portion of all strings that are identical.

Given a set of strings, R, for a prefix S, it should hold that:

\forall x\ \in \ R:\ S\leq

_pref

\ x

~ "for all members x of set R, it holds true that string S is a prefix of x"

(help here: does not express that S is the longest common prefix of x)

An example use case for this: given a set of phone numbers, identify a common dialing code. This can be accomplished by first determining the common prefix (if any), and then matching it against know dialing codes (iteratively dropping characters from rhs until a match is found, as the lcp function may match more than the dialing code).

Test cases

For a function, lcp, accepting a list of strings, the following should hold true (the empty string, $\varepsilon$ , is considered a prefix of all strings):

lcp("interspecies","interstellar","interstate") = "inters"
lcp("throne","throne") = "throne"
lcp("throne","dungeon") =  $\varepsilon$ 
lcp("throne", $\varepsilon$ ,"throne") =  $\varepsilon$ 
lcp("cheese") = "cheese"
lcp( $\varepsilon$ ) =  $\varepsilon$ 
lcp( $\emptyset$ ) =  $\varepsilon$ 
lcp("prefix","suffix") =  $\varepsilon$ 
lcp("foo","foobar") = "foo"

Task inspired by this stackoverflow question: Find the longest common starting substring in a set 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 lcp(sa)
   I sa.empty
      R ‘’
   I sa.len == 1
      R sa[0]

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

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

   R sa[0][0 .< min_len]

F test(sa)
   print(String(sa)‘ -> ’lcp(sa))

test([‘interspecies’, ‘interstellar’, ‘interstate’])
test([‘throne’, ‘throne’])
test([‘throne’, ‘dungeon’])
test([‘throne’, ‘’, ‘throne’])
test([‘cheese’])
test([‘’])
test([‘prefix’, ‘suffix’])
test([‘foo’, ‘foobar’])

Output:

[interspecies, interstellar, interstate] -> inters
[throne, throne] -> throne
[throne, dungeon] ->
[throne, , throne] ->
[cheese] -> cheese
[] ->
[prefix, suffix] ->
[foo, foobar] -> foo

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 Prefix(PTR ARRAY texts BYTE count CHAR ARRAY res)
  CHAR ARRAY t(100)
  BYTE i,len,found

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

  len=CommonLength(texts,count)
  WHILE len>0
  DO
    SCopyS(res,texts(0),1,len)
    found=1
    FOR i=1 TO count-1
    DO
      SCopyS(t,texts(i),1,len)
      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)

  Prefix(texts,count,res)
  Print("lcp(")
  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="interspecies", t2="interstellar", t3="interstate",
    t4="throne", t5="throne", t6="dungeon", t7="",
    t8="prefix", t9="suffix", t10="foo", t11="foobar"
  PTR ARRAY texts(20)

  texts(0)=t1 texts(1)=t2 texts(2)=t3
  Test(texts,3)

  texts(0)=t4 texts(1)=t5
  Test(texts,2)

  texts(0)=t4 texts(1)=t6
  Test(texts,2)

  texts(0)=t4 texts(1)=t7 texts(2)=t5
  Test(texts,3)

  texts(0)=t7
  Test(texts,1)

  Test(texts,0)

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

  texts(0)=t10 texts(1)=t11
  Test(texts,2)
RETURN

Output:

Screenshot from Atari 8-bit computer

lcp("interspecies","interstellar","interstate")="inters"
lcp("throne","throne")="throne"
lcp("throne","dungeon")=""
lcp("throne","","throne")=""
lcp("")=""
lcp()=""
lcp("prefix","suffix")=""
lcp("foo","foobar")="foo"

Ada

with Ada.Text_IO;
with Ada.Strings.Unbounded;

procedure Longest_Prefix is

   package Common_Prefix is
      procedure Reset;
      procedure Consider (Item : in String);
      function Prefix return String;
   end Common_Prefix;

   package body Common_Prefix
   is
      use Ada.Strings.Unbounded;
      Common : Unbounded_String;
      Active : Boolean := False;

      procedure Reset is
      begin
         Common := Null_Unbounded_String;
         Active := False;
      end Reset;

      procedure Consider (Item : in String) is
         Len : constant Natural := Natural'Min (Length (Common), Item'Length);
      begin
         if not Active then
            Active := True;
            Common := To_Unbounded_String (Item);
         else
            for I in 1 .. Len loop
               if Element (Common, I) /= Item (Item'First + I - 1) then
                  Head (Common, I - 1);
                  return;
               end if;
            end loop;
            Head (Common, Len);
         end if;
      end Consider;

      function Prefix return String is (To_String (Common));

   end Common_Prefix;

   use Common_Prefix;
begin
   Consider ("interspecies");   Consider ("interstellar");   Consider ("interstate");
   Ada.Text_IO.Put_Line (Prefix);

   Reset;   Consider ("throne");   Consider ("throne");
   Ada.Text_IO.Put_Line (Prefix);

   Reset;   Consider ("throne");   Consider ("dungeon");
   Ada.Text_IO.Put_Line (Prefix);

   Reset;   Consider ("prefix");   Consider ("suffix");
   Ada.Text_IO.Put_Line (Prefix);

   Reset;   Consider ("foo");   Consider ("foobar");
   Ada.Text_IO.Put_Line (Prefix);

   Reset;   Consider ("foo");
   Ada.Text_IO.Put_Line (Prefix);

   Reset;   Consider ("");
   Ada.Text_IO.Put_Line (Prefix);

   Reset;
   Ada.Text_IO.Put_Line (Prefix);

end Longest_Prefix;

Aime

lcp(...)
{
    integer n;
    record r;
    text l;

    ucall(r_fix, 1, r, 0);
    n = 0;
    if (~r) {
        l = r.low;
        n = prefix(r.high, l);
    }

    l.cut(0, n);
}

main(void)
{
    o_("\"", lcp("interspecies", "interstellar", "interstate"), "\"\n");
    o_("\"", lcp("throne", "throne"), "\"\n");
    o_("\"", lcp("throne", "dungeon"), "\"\n");
    o_("\"", lcp("throne", "", "throne"), "\"\n");
    o_("\"", lcp("cheese"), "\"\n");
    o_("\"", lcp(""), "\"\n");
    o_("\"", lcp(), "\"\n");
    o_("\"", lcp("prefix", "suffix"), "\"\n");
    o_("\"", lcp("foo", "foobar"), "\"\n");

    0;
}

Output:

"inters"
"throne"
""
""
"cheese"
""
""
""
"foo"

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.win32

# find the longest common prefix of two strings #
PRIO COMMONPREFIX = 1;
OP   COMMONPREFIX = ( STRING a, b )STRING:
    BEGIN
        INT a pos := LWB a; INT a max = UPB a;
        INT b pos := LWB b; INT b max = UPB b;
        WHILE
            IF a pos > a max OR b pos > b max THEN FALSE
            ELSE a[ a pos ] = b[ b pos ]
            FI
        DO
            a pos +:= 1; b pos +:= 1
        OD;
        a[ LWB a : a pos - 1 ]
    END # COMMONPREFIX # ;

# get the length of a string #
OP  LEN = ( STRING a )INT: ( UPB a + 1 ) - LWB a;
            
# find the longest common prefix of an array of STRINGs #
OP  LONGESTPREFIX = ( []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 prefix := list[ LWB list ] COMMONPREFIX list[ 1 + LWB list ];
        FOR pos FROM 2 + LWB list TO UPB list DO
            STRING next prefix := list[ pos ] COMMONPREFIX prefix;
            IF LEN next prefix < LEN prefix
            THEN
                # this element has a smaller common prefix #
                prefix := next prefix
            FI
        OD;
        prefix
    FI ;


# test the LONGESTPREFIX operator #

PROC test prefix = ( []STRING list, STRING expected result )VOID:
    BEGIN
        STRING prefix = LONGESTPREFIX list;
        print( ( "longest common prefix of (" ) );
        FOR pos FROM LWB list TO UPB list DO print( ( " """, list[ pos ], """" ) ) OD;
        print( ( " ) is: """, prefix, """ "
               , IF prefix = expected result THEN "as expected" ELSE "NOT AS EXPECTED" FI 
               , newline
               )
             )
    END # test prefix # ;

[ 1 : 0 ]STRING empty list; # for recent versions of Algol 68G, can't just put "()" for an empty list #

BEGIN
    test prefix( ( "interspecies", "interstellar", "interstate" ), "inters" );
    test prefix( ( "throne", "throne" ), "throne" );
    test prefix( ( "throne", "dungeon" ), "" );
    test prefix( ( "throne", "", "throne" ), "" );
    test prefix( ( "cheese" ), "cheese" );
    test prefix( ( "" ), "" );
    test prefix( empty list, "" );
    test prefix( ( "prefix", "suffix" ), "" );
    test prefix( ( "foo", "foobar" ), "foo" )
END

Output:

longest common prefix of ( "interspecies" "interstellar" "interstate" ) is: "inters" as expected
longest common prefix of ( "throne" "throne" ) is: "throne" as expected
longest common prefix of ( "throne" "dungeon" ) is: "" as expected
longest common prefix of ( "throne" "" "throne" ) is: "" as expected
longest common prefix of ( "cheese" ) is: "cheese" as expected
longest common prefix of ( "" ) is: "" as expected
longest common prefix of ( ) is: "" as expected
longest common prefix of ( "prefix" "suffix" ) is: "" as expected
longest common prefix of ( "foo" "foobar" ) is: "foo" as expected

AppleScript

AppleScriptObjC

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

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:
longestCommonPrefix({"interspecies", "interstellar", "interstate"}) --> "inters"
longestCommonPrefix({"throne", "throne"}) --> "throne"
longestCommonPrefix({"throne", "dungeon"}) --> ""
longestCommonPrefix({"throne", "", "throne"}) --> ""
longestCommonPrefix({""}) --> ""
longestCommonPrefix({}) --> ""
longestCommonPrefix({"prefix", "suffix"}) --> ""
longestCommonPrefix({"foo", "foobar"}) --> "foo"

Functional

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

------------------- LONGEST COMMON PREFIX ------------------


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


---------------------------- TESTS --------------------------
on run
    script test
        on |λ|(xs)
            showList(xs) & " -> '" & longestCommonPrefix(xs) & "'"
        end |λ|
    end script
    
    unlines(map(test, {¬
        {"interspecies", "interstellar", "interstate"}, ¬
        {"throne", "throne"}, ¬
        {"throne", "dungeon"}, ¬
        {"throne", "", "throne"}, ¬
        {"cheese"}, ¬
        {""}, ¬
        {}, ¬
        {"prefix", "suffix"}, ¬
        {"foo", "foobar"}}))
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


-- chars :: String -> [Char]
on chars(s)
    characters of s
end chars


-- 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


-- eq (==) :: Eq a => a -> a -> Bool
on eq(a)
    script
        on |λ|(b)
            a = b
        end |λ|
    end script
end eq


-- 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


-- 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:

['interspecies', 'interstellar', 'interstate'] -> 'inters'
['throne', 'throne'] -> 'throne'
['throne', 'dungeon'] -> ''
['throne', '', 'throne'] -> ''
['cheese'] -> 'cheese'
[''] -> ''
[] -> ''
['prefix', 'suffix'] -> ''
['foo', 'foobar'] -> 'foo'

Arturo

lcp: function [lst][
	ret: ""
    idx: 0
 
    while [true] [
        thisLetter: ""
        loop lst 'word [
        	if idx=size word -> return ret
        	if thisLetter="" -> thisLetter: get split word idx
        	if thisLetter<>get split word idx -> return ret
        ]
        ret: ret ++ thisLetter
        idx: idx + 1
    ]
]
 
print lcp ["interspecies" "interstellar" "interstate"]
print lcp ["throne" "throne"]
print lcp ["throne" "dungeon"]
print lcp ["throne" "" "throne"]
print lcp ["cheese"]
print lcp [""]
print lcp ["prefix" "suffix"]
print lcp ["foo" "foobar"]

Output:

inters
throne


cheese


foo

AutoHotkey

lcp(data){
	for num, v in StrSplit(data.1)
		for i, word in data	
			if (SubStr(word, 1, num) <> SubStr(data.1, 1, num))
				return SubStr(word, 1, num-1)
	return SubStr(word, 1, num)
}

Examples:

MsgBox % ""
. "`n" lcp(["interspecies","interstellar","interstate"])
. "`n" lcp(["throne","throne"])
. "`n" lcp(["throne","dungeon"])
. "`n" lcp(["throne","","throne"])
. "`n" lcp(["cheese"])
. "`n" lcp([""])
. "`n" lcp([])
. "`n" lcp(["prefix","suffix"])
. "`n" lcp(["foo","foobar"])
return

Output:

inters
throne


cheese



foo

AWK

# syntax: GAWK -f LONGEST_COMMON_PREFIX.AWK
BEGIN {
    words_arr[++n] = "interspecies,interstellar,interstate"
    words_arr[++n] = "throne,throne"
    words_arr[++n] = "throne,dungeon"
    words_arr[++n] = "throne,,throne"
    words_arr[++n] = "cheese"
    words_arr[++n] = ""
    words_arr[++n] = "prefix,suffix"
    words_arr[++n] = "foo,foobar"
    for (i=1; i<=n; i++) {
      str = words_arr[i]
      printf("'%s' = '%s'\n",str,lcp(str))
    }
    exit(0)
}
function lcp(str,  arr,hits,i,j,lcp_leng,n,sw_leng) {
    n = split(str,arr,",")
    if (n == 0) { # null string
      return("")
    }
    if (n == 1) { # only 1 word, then it's the longest
      return(str)
    }
    sw_leng = length(arr[1])
    for (i=2; i<=n; i++) { # find shortest word length
      if (length(arr[i]) < sw_leng) {
        sw_leng = length(arr[i])
      }
    }
    for (i=1; i<=sw_leng; i++) { # find longest common prefix
      hits = 0
      for (j=1; j<n; j++) {
        if (substr(arr[j],i,1) == substr(arr[j+1],i,1)) {
          hits++
        }
      }
      if (hits == 0) {
        break
      }
      if (hits + 1 == n) {
        lcp_leng++
      }
    }
    return(substr(str,1,lcp_leng))
}

Output:

'interspecies,interstellar,interstate' = 'inters'
'throne,throne' = 'throne'
'throne,dungeon' = ''
'throne,,throne' = ''
'cheese' = 'cheese'
'' = ''
'prefix,suffix' = ''
'foo,foobar' = 'foo'

BQN

LCP ← (1<≠)◶⟨∾, ((∧`·∧´⊏=1⊸↓)/⊑)(⌊´≠¨)↑¨⊢⟩

LCP¨ ⟨
  ⟨"interspecies", "interstellar", "interstate"⟩
  ⟨"throne", "throne"⟩
  ⟨"throne", "dungeon"⟩
  ⟨"throne", "", "throne"⟩
  ⟨"cheese"⟩
  ⟨""⟩
  ⟨⟩
  ⟨"prefix", "suffix"⟩
  ⟨"foo", "foobar"⟩
⟩

Output:

⟨ "inters" "throne" ⟨⟩ ⟨⟩ "cheese" ⟨⟩ ⟨⟩ ⟨⟩ "foo" ⟩

C

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

char* lcp(int num,...){
	va_list vaList,vaList2;
	int i,j,len,min;
	char* dest;
	char** strings = (char**)malloc(num*sizeof(char*));
	
	va_start(vaList,num);
	va_start(vaList2,num);
	
	for(i=0;i<num;i++){
		len = strlen(va_arg(vaList,char*));
		strings[i] = (char*)malloc((len + 1)*sizeof(char));
		
		strcpy(strings[i],va_arg(vaList2,char*));
		
		if(i==0)
			min = len;
		else if(len<min)
			min = len;
	}
	
	if(min==0)
		return "";
	
	for(i=0;i<min;i++){
		for(j=1;j<num;j++){
			if(strings[j][i]!=strings[0][i]){
				if(i==0)
					return "";
				else{
					dest = (char*)malloc(i*sizeof(char));
					strncpy(dest,strings[0],i-1);
					return dest;
				}
			}
		}
	}
	
	dest = (char*)malloc((min+1)*sizeof(char));
	strncpy(dest,strings[0],min);
	return dest;
}

int main(){

	printf("\nLongest common prefix : %s",lcp(3,"interspecies","interstellar","interstate"));
        printf("\nLongest common prefix : %s",lcp(2,"throne","throne"));
        printf("\nLongest common prefix : %s",lcp(2,"throne","dungeon"));
        printf("\nLongest common prefix : %s",lcp(3,"throne","","throne"));
        printf("\nLongest common prefix : %s",lcp(1,"cheese"));
        printf("\nLongest common prefix : %s",lcp(1,""));
        printf("\nLongest common prefix : %s",lcp(0,NULL));
        printf("\nLongest common prefix : %s",lcp(2,"prefix","suffix"));
        printf("\nLongest common prefix : %s",lcp(2,"foo","foobar"));
	return 0;
}

Output:

Longest common prefix : inter
Longest common prefix : throne
Longest common prefix :
Longest common prefix :
Longest common prefix : cheese
Longest common prefix :
Longest common prefix :
Longest common prefix :
Longest common prefix : foo

C#

Translation of: Java

using System;

namespace LCP {
    class Program {
        public static string LongestCommonPrefix(params string[] sa) {
            if (null == sa) return ""; //special case
            string ret = "";
            int idx = 0;

            while (true) {
                char thisLetter = '\0';
                foreach (var word in sa) {
                    if (idx == word.Length) {
                        // if we reached the end of a word then we are done
                        return ret;
                    }
                    if (thisLetter == '\0') {
                        // if this is the first word then note the letter we are looking for
                        thisLetter = word[idx];
                    }
                    if (thisLetter != word[idx]) {
                        return ret;
                    }
                }

                // if we haven't said we are done then this position passed
                ret += thisLetter;
                idx++;
            }
        }

        static void Main(string[] args) {
            Console.WriteLine(LongestCommonPrefix("interspecies", "interstellar", "interstate"));
            Console.WriteLine(LongestCommonPrefix("throne", "throne"));
            Console.WriteLine(LongestCommonPrefix("throne", "dungeon"));
            Console.WriteLine(LongestCommonPrefix("throne", "", "throne"));
            Console.WriteLine(LongestCommonPrefix("cheese"));
            Console.WriteLine(LongestCommonPrefix(""));
            Console.WriteLine(LongestCommonPrefix(null));
            Console.WriteLine(LongestCommonPrefix("prefix", "suffix"));
            Console.WriteLine(LongestCommonPrefix("foo", "foobar"));
        }
    }
}

Output:

inters
throne


cheese



foo

C++

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

std::set<std::string> createPrefixes ( const std::string & s ) {
   std::set<std::string> result ;
   for ( int i = 1 ; i < s.size( ) + 1 ; i++ )
      result.insert( s.substr( 0 , i )) ;
   return result ;
}

std::set<std::string> findIntersection ( const std::set<std::string> & a ,
      const std::set<std::string> & b ) {
   std::set<std::string> intersection ;
   std::set_intersection( a.begin( ) , a.end( ) , b.begin( ) , b.end( ) ,
	 std::inserter ( intersection , intersection.begin( ) ) ) ;
   return intersection  ;
}

std::set<std::string> findCommonPrefixes( const std::vector<std::string> & theStrings ) {
   std::set<std::string> result ;
   if ( theStrings.size( ) == 1 ) {
      result.insert( *(theStrings.begin( ) ) ) ;
   }
   if ( theStrings.size( ) > 1 ) {
      std::vector<std::set<std::string>> prefixCollector ;
      for ( std::string s : theStrings )
	 prefixCollector.push_back( createPrefixes ( s ) ) ;
      std::set<std::string> neutralElement (createPrefixes( *(theStrings.begin( ) ) )) ;
      result = std::accumulate( prefixCollector.begin( ) , prefixCollector.end( ) ,
	    neutralElement , findIntersection ) ;
   }
   return result ;
}

std::string lcp( const std::vector<std::string> & allStrings ) {
   if ( allStrings.size( ) == 0 ) 
      return "" ;
   if ( allStrings.size( ) == 1 ) {
      return allStrings[ 0 ] ;
   }
   if ( allStrings.size( ) > 1 ) {
      std::set<std::string> prefixes( findCommonPrefixes ( allStrings ) ) ;
      if ( prefixes.empty( ) ) 
	 return "" ;
      else {
	 std::vector<std::string> common ( prefixes.begin( ) , prefixes.end( ) ) ;
	 std::sort( common.begin( ) , common.end( ) , [] ( const std::string & a, 
		  const std::string & b ) { return a.length( ) > b.length( ) ; } ) ;
	 return *(common.begin( ) ) ;
      }
   }
}

int main( ) {
   std::vector<std::string> input { "interspecies" , "interstellar" , "interstate" } ;
   std::cout << "lcp(\"interspecies\",\"interstellar\",\"interstate\") = " << lcp ( input ) << std::endl ;
   input.clear( ) ;
   input.push_back( "throne" ) ;
   input.push_back ( "throne" ) ;
   std::cout << "lcp( \"throne\" , \"throne\"" << ") = " << lcp ( input ) << std::endl ;
   input.clear( ) ;
   input.push_back( "cheese" ) ;
   std::cout << "lcp( \"cheese\" ) = " << lcp ( input ) << std::endl ;
   input.clear( ) ;
   std::cout << "lcp(\"\") = " << lcp ( input ) << std::endl ;
   input.push_back( "prefix" ) ;
   input.push_back( "suffix" ) ;
   std::cout << "lcp( \"prefix\" , \"suffix\" ) = " << lcp ( input ) << std::endl ;
   input.clear( ) ;
   input.push_back( "foo" ) ;
   input.push_back( "foobar" ) ;
   std::cout << "lcp( \"foo\" , \"foobar\" ) = " << lcp ( input ) << std::endl ;
   return 0 ;
}

Another more concise version (C++14 for comparing dissimilar containers):

#include <algorithm>
#include <string>
#include <iostream>
#include <vector>
 
std::string lcp( const std::vector<std::string> & allStrings ) {
	if (allStrings.empty()) return std::string();
	const std::string &s0 = allStrings.front();
	auto end = s0.cend();
	for(auto it=std::next(allStrings.cbegin()); it != allStrings.cend(); it++){
		auto loc = std::mismatch(s0.cbegin(), s0.cend(), it->cbegin(), it->cend());
		if (std::distance(loc.first, end)>0) end = loc.first;
	}
	return std::string(s0.cbegin(), end);
}
 
int main( ) {
   std::vector<std::string> input { "interspecies" , "interstellar" , "interstate" } ;
   std::cout << "lcp(\"interspecies\",\"interstellar\",\"interstate\") = " << lcp ( input ) << std::endl ;
   input.clear( ) ;
   input.push_back( "throne" ) ;
   input.push_back ( "throne" ) ;
   std::cout << "lcp( \"throne\" , \"throne\"" << ") = " << lcp ( input ) << std::endl ;
   input.clear( ) ;
   input.push_back( "cheese" ) ;
   std::cout << "lcp( \"cheese\" ) = " << lcp ( input ) << std::endl ;
   input.clear( ) ;
   std::cout << "lcp(\"\") = " << lcp ( input ) << std::endl ;
   input.push_back( "prefix" ) ;
   input.push_back( "suffix" ) ;
   std::cout << "lcp( \"prefix\" , \"suffix\" ) = " << lcp ( input ) << std::endl ;
   input.clear( ) ;
   input.push_back( "foo" ) ;
   input.push_back( "foobar" ) ;
   std::cout << "lcp( \"foo\" , \"foobar\" ) = " << lcp ( input ) << std::endl ;
   return 0 ;
}

Output:

lcp("interspecies","interstellar","interstate") = inters
lcp( "throne" , "throne") = throne
lcp( "cheese" ) = cheese
lcp("") = 
lcp( "prefix" , "suffix" ) = 
lcp( "foo" , "foobar" ) = foo

CLU

lcp = proc (strs: ss) returns (string)
    ss = sequence[string]
    if ss$empty(strs) then return("") end
    
    pfx: string := ss$bottom(strs)
    for str: string in ss$elements(strs) do
        if string$empty(pfx) then return("") end
        if string$size(str) < string$size(pfx) then
            pfx := string$substr(pfx, 1, string$size(str))
        end
        
        no_match: int := 1
        while no_match <= string$size(pfx) cand
              str[no_match] = pfx[no_match] do
            no_match := no_match + 1 
        end
        
        pfx := string$substr(pfx, 1, no_match-1)
    end
    return(pfx)
end lcp

start_up = proc ()
    ss = sequence[string]
    sss = sequence[ss]
    po: stream := stream$primary_output()
    
    tests: sss := sss$[
        ss$["interspecies","interstellar","interstate"],
        ss$["throne","throne"],
        ss$["throne","dungeon"],
        ss$["throne","","dungeon"],
        ss$["cheese"],
        ss$[""],
        ss$[],
        ss$["prefix","suffix"],
        ss$["foo","foobar"]
    ]
    
    for test: ss in sss$elements(tests) do
        stream$putl(po, "\"" || lcp(test) || "\"")
    end
end start_up

Output:

"inters"
"throne"
""
""
"cheese"
""
""
""
"foo"

D

Translation of: Java

import std.stdio;

string lcp(string[] list ...) {
    string ret = "";
    int idx;

    while(true) {
        char thisLetter = 0;
        foreach (word; list) {
            if (idx == word.length) {
                return ret;
            }
            if(thisLetter == 0) { //if this is the first word then note the letter we are looking for
                thisLetter = word[idx];
            }
            if (thisLetter != word[idx]) { //if this word doesn't match the letter at this position we are done
                return ret;
            }
        }
        ret ~= thisLetter; //if we haven't said we are done then this position passed
        idx++;
    }
}

void main() {
    writeln(lcp("interspecies","interstellar","interstate"));
    writeln(lcp("throne","throne"));
    writeln(lcp("throne","dungeon"));
    writeln(lcp("throne","","throne"));
    writeln(lcp("cheese"));
    writeln(lcp(""));
    writeln(lcp("prefix","suffix"));
    writeln(lcp("foo","foobar"));
}

Output:

inters
throne


cheese


foo

Dyalect

Translation of: C#

func lcp(sa...) {
    if sa.Length() == 0 || !sa[0] {
        return ""
    } 
 
    var ret = ""
    var idx = 0
 
    while true {
        var thisLetter = '\0'
        for word in sa {
            if idx == word.Length() {
                return ret
            }
            if thisLetter == '\0' {
                thisLetter = word[idx]
            }
            if thisLetter != word[idx] {
                return ret
            }
        }
 
        ret += thisLetter
        idx += 1
    }
}
 
print(lcp("interspecies", "interstellar", "interstate"))
print(lcp("throne", "throne"))
print(lcp("throne", "dungeon"))
print(lcp("throne", "", "throne"))
print(lcp("cheese"))
print(lcp(""))
print(lcp(nil))
print(lcp("prefix", "suffix"))
print(lcp("foo", "foobar"))

Output:

inters
throne


cheese



foo

EasyLang

func$ lcp list$[] .
   if len list$[] = 0
      return ""
   .
   shortest$ = list$[1]
   for s$ in list$[]
      if len s$ < len shortest$
         shortest$ = s$
      .
   .
   for i to len shortest$ - 1
      sub$ = substr shortest$ 1 i
      for s$ in list$[]
         if substr s$ 1 i <> sub$
            return substr shortest$ 1 (i - 1)
         .
      .
   .
   return shortest$
.
print lcp [ "interspecies" "interstellar" "interstate" ]
print lcp [ "throne" "throne" ]
print lcp [ "throne" "dungeon" ]
print lcp [ "throne" "" "throne" ]
print lcp [ "cheese" ]
print lcp [ ]
print lcp [ "foo" "foobar" ]

Output:

inters
throne


cheese

foo

EchoLisp

;; find common prefix of two strings
(define (prefix s t ) (for/string ((u s) (v t)) #:break (not (= u v)) u))

(prefix "foo" "foobar") → "foo"

;; fold over a list of strings
(define (lcp strings) 
	(if
	(null? strings) ""
	(foldl prefix (first strings) (rest strings))))

 define lcp-test '(
 ("interspecies" "interstellar" "interstate")
 ("throne" "throne")
 ("throne" "dungeon")
 ("cheese")
 ("") 
 () 
 ("prefix" "suffix")))

;;
(for ((t lcp-test)) (writeln t '→ (lcp t)))
    ("interspecies" "interstellar" "interstate")    →     "inters"    
    ("throne" "throne")     →    "throne"    
    ("throne" "dungeon")     →     ""    
    ("cheese")     →     "cheese"    
    ("")     →     ""    
    null     →     ""    
    ("prefix" "suffix")     →     ""

ed

Uses a backreference to find the longest common prefix for every (vertical bar-separated) string. Inputs should be newline-separated.

# by Artyom Bologov
H
,p
g/.*/s//&|/
,j
g/^([^|]*)[^|]*\|(\1[^|]*\|)*$/s//&: \1/
,p
Q

Output:

$ cat longest-prefix.ed | ed -lEGs longest-prefix.input 
interspecies
interstellar
interstate
interspecies|interstellar|interstate|: inters

Elixir

defmodule LCP do
  @data [
    ["interspecies", "interstellar", "interstate"],
    ["throne", "throne"],
    ["throne", "dungeon"],
    ["throne", "", "throne"],
    ["cheese"],
    [""],
    [],
    ["prefix", "suffix"],
    ["foo", "foobar"]
  ]

  def main do
    Enum.each(@data, fn strs ->
      IO.puts("#{inspect(strs)} -> #{inspect(lcp(strs))}")
    end)
  end

  defp lcp( [] ), do: ""
  defp lcp(strs), do: Enum.reduce(strs, &lcp/2)

  defp lcp(xs, ys), do: lcp(xs, ys, "")

  defp lcp(<<x,xs>>, <<x,ys>>, pre), do: lcp(xs, ys, <<x,pre>>)
  defp lcp(       _,        _, pre), do: String.reverse(pre)
end

Output:

["interspecies", "interstellar", "interstate"] -> "inters"
["throne", "throne"] -> "throne"
["throne", "dungeon"] -> ""
["throne", "", "throne"] -> ""
["cheese"] -> "cheese"
[""] -> ""
[] -> ""
["prefix", "suffix"] -> ""
["foo", "foobar"] -> "foo"

Erlang

-module(lcp). 
-export([ main/0 ]).
 
data() -> [
  ["interspecies", "interstellar", "interstate"],
  ["throne", "throne"],
  ["throne", "dungeon"],
  ["throne", "", "throne"],
  ["cheese"],
  [""],
  [],
  ["prefix", "suffix"],
  ["foo", "foobar"]
].

main() -> [io:format("~p -> \"~s\"~n",[Strs,lcp(Strs)]) || Strs <- data()].

lcp(      []) -> [];
lcp([S|Strs]) -> lists:foldl( fun(X,Y) -> lcp(X,Y,[]) end, S, Strs).

lcp([X|Xs], [X|Ys], Pre) -> lcp(Xs, Ys, [X|Pre]);
lcp(     _,      _, Pre) -> lists:reverse(Pre).

Output:

["interspecies","interstellar","interstate"] -> "inters"
["throne","throne"] -> "throne"
["throne","dungeon"] -> ""
["throne",[],"throne"] -> ""
["cheese"] -> "cheese"
[[]] -> ""
[] -> ""
["prefix","suffix"] -> ""
["foo","foobar"] -> "foo"

Factor

USING: continuations formatting fry kernel sequences strings ;
IN: rosetta-code.lcp

! Find the longest common prefix of two strings.
: binary-lcp ( str1 str2 -- str3 )
    [ SBUF" " clone ] 2dip
    '[ _ _ [ over = [ suffix! ] [ drop return ] if ] 2each ]
    with-return >string ;

! Reduce a sequence of strings using binary-lcp.
: lcp ( seq -- str )
    [ "" ] [ dup first [ binary-lcp ] reduce ] if-empty ;

: lcp-demo ( -- )
    {
        { "interspecies" "interstellar" "interstate" }
        { "throne" "throne" }
        { "throne" "dungeon" }
        { "throne" "" "throne" }
        { "cheese" }
        { "" }
        { }
        { "prefix" "suffix" }
        { "foo" "foobar" }
    } [ dup lcp "%u lcp = %u\n" printf ] each ;

MAIN: lcp-demo

Output:

{ "interspecies" "interstellar" "interstate" } lcp = "inters"
{ "throne" "throne" } lcp = "throne"
{ "throne" "dungeon" } lcp = ""
{ "throne" "" "throne" } lcp = ""
{ "cheese" } lcp = "cheese"
{ "" } lcp = ""
{ } lcp = ""
{ "prefix" "suffix" } lcp = ""
{ "foo" "foobar" } lcp = "foo"

FreeBASIC

' FB 1.05.0 Win64

Function lcp(s() As String) As String
  Dim As Integer lb = LBound(s)
  Dim As Integer ub = UBound(s)
  Dim length As Integer = ub - lb + 1
  If length = 0 Then Return ""    '' empty array
  If length = 1 Then Return s(lb) '' only one element
  ' find length of smallest string
  Dim minLength As Integer = Len(s(lb))
  For i As Integer = lb + 1 To ub
    If Len(s(i)) < minLength Then minLength = Len(s(i))
    If minLength = 0 Then Return ""  '' at least one string is empty
  Next  
  Dim prefix As String
  Dim isCommon As Boolean 
  Do
     prefix = Left(s(lb), minLength)
     isCommon = True 
     For i As Integer = lb + 1 To ub
       If Left(s(i), minLength) <> prefix Then
         isCommon = False
         Exit For
       End If
     Next
     If isCommon Then Return prefix
     minLength -= 1
     If minLength = 0 Then Return ""
  Loop
End Function


Dim s1(1 To 3) As String = {"interspecies","interstellar","interstate"} 
Print "lcp(""interspecies"",""interstellar"",""interstate"") = """; lcp(s1()); """"

Dim s2(1 To 2) As String = {"throne", "throne"}
Print "lcp(""throne"", ""throne"") = """; lcp(s2()); """"

Dim s3(1 To 2) As String = {"throne", "dungeon"}
Print "lcp(""throne"", ""dungeon"") = """; lcp(s3()); """"

Dim s4(1 To 3) As String = {"throne", "", "dungeon"}
Print "lcp(""throne"", """", ""dungeon"") = """; lcp(s4()); """"

Dim s5(1 To 1) As String = {"cheese"}
Print "lcp(""cheese"") = """; lcp(s5()); """"

Dim s6(1 To 1) As String
Print "lcp("""") = """; lcp(s6()); """"
 
Dim s7() As String
Print "lcp() = """; lcp(s7()); """"

Dim s8(1 To 2) As String = {"prefix", "suffix"}
Print "lcp(""prefix"", ""suffix"") = """; lcp(s8()); """"

Dim s9(1 To 2) As String = {"foo", "foobar"}
Print "lcp(""foo"", ""foobar"") = """; lcp(s9()); """"

Print
Print "Press any key to quit"
Sleep

Output:

lcp("interspecies","interstellar","interstate") = "inters"
lcp("throne", "throne") = "throne"
lcp("throne", "dungeon") = ""
lcp("throne", "", "dungeon") = ""
lcp("cheese") = "cheese"
lcp("") = ""
lcp() = ""
lcp("prefix", "suffix") = ""
lcp("foo", "foobar") = "foo"

FutureBasic

include "NSLog.incl"

local fn CommonPrefix( string1 as CFStringRef, string2 as CFStringRef ) as CFStringRef
  NSUInteger              i, minLength
  CFMutableStringRef prefix = fn MutableStringNew
  
  if ( len(string1) == 0 || len(string1) == 0 )
    return @"\"\""
  else
    if len(string1) > len(string2) then minLength = len(string2) else minLength = len(string1)
  end if
  
  for i = 0 to minLength -1
    if ( fn StringCharacterAtIndex( string1, i ) == fn StringCharacterAtIndex( string2, i ) )
      MutableStringAppendFormat( prefix, @"%C", fn StringCharacterAtIndex( string1, i ) )
    else
      break
    end if
  next
  if len(prefix) == 0 then MutableStringAppendString( prefix, @"\"\"" )
end fn = prefix


local fn LongestCommonPrefix( strings as CFArrayRef ) as CFStringRef
  CFStringRef minString, maxString
  CfArrayRef  array
  
  select switch ( len(strings) )
    case 0 : return @"\"\""
    case 1
      if ( len(strings[0]) )
        return fn StringWithFormat( @"%@", strings[0] )
      else
        return  @"\"\""
      end if
    case else
      array = fn ArraySortedArrayUsingSelector( strings, @"compare:" )
      minString = fn ArrayFirstObject( array )
      maxString = fn ArrayLastObject(  array )
      return fn CommonPrefix( minString, maxString )
  end select
end fn = @"\"\""


void local fn PrintLongestCommonPrefix( strings as CFArrayRef )
  CFStringRef tempStr = fn ArrayComponentsJoinedByString( strings, @", " )
  NSLog( @"lcp( %@ ) = %@", tempStr, fn LongestCommonPrefix( strings ) )
end fn

fn PrintLongestCommonPrefix( @[@"interspecies", @"interstellar", @"interstate"] )
fn PrintLongestCommonPrefix( @[@"throne", @"throne"] )
fn PrintLongestCommonPrefix( @[@"throne", @"dungeon"] )
fn PrintLongestCommonPrefix( @[@"throne", @"\"\"", @"throne"] )
fn PrintLongestCommonPrefix( @[@"cheese"] )
fn PrintLongestCommonPrefix( @[@""] )
fn PrintLongestCommonPrefix( @[] )
fn PrintLongestCommonPrefix( @[@"prefix", @"suffix"] )
fn PrintLongestCommonPrefix( @[@"foo", @"foobar"] )

HandleEvents

Output:

lcp( interspecies, interstellar, interstate ) = inters
lcp( throne, throne ) = throne
lcp( throne, dungeon ) = ""
lcp( throne, "", throne ) = ""
lcp( cheese ) = cheese
lcp(  ) = ""
lcp(  ) = ""
lcp( prefix, suffix ) = ""
lcp( foo, foobar ) = foo

Go

package main

import "fmt"

// lcp finds the longest common prefix of the input strings.
// It compares by bytes instead of runes (Unicode code points).
// It's up to the caller to do Unicode normalization if desired
// (e.g. see golang.org/x/text/unicode/norm).
func lcp(l []string) string {
	// Special cases first
	switch len(l) {
	case 0:
		return ""
	case 1:
		return l[0]
	}
	// LCP of min and max (lexigraphically)
	// is the LCP of the whole set.
	min, max := l[0], l[0]
	for _, s := range l[1:] {
		switch {
		case s < min:
			min = s
		case s > max:
			max = s
		}
	}
	for i := 0; i < len(min) && i < len(max); i++ {
		if min[i] != max[i] {
			return min[:i]
		}
	}
	// In the case where lengths are not equal but all bytes
	// are equal, min is the answer ("foo" < "foobar").
	return min
}

// Normally something like this would be a TestLCP function in *_test.go
// and use the testing package to report failures.
func main() {
	for _, l := range [][]string{
		{"interspecies", "interstellar", "interstate"},
		{"throne", "throne"},
		{"throne", "dungeon"},
		{"throne", "", "throne"},
		{"cheese"},
		{""},
		nil,
		{"prefix", "suffix"},
		{"foo", "foobar"},
	} {
		fmt.Printf("lcp(%q) = %q\n", l, lcp(l))
	}
}

Output:

lcp(["interspecies" "interstellar" "interstate"]) = "inters"
lcp(["throne" "throne"]) = "throne"
lcp(["throne" "dungeon"]) = ""
lcp(["throne" "" "throne"]) = ""
lcp(["cheese"]) = "cheese"
lcp([""]) = ""
lcp([]) = ""
lcp(["prefix" "suffix"]) = ""
lcp(["foo" "foobar"]) = "foo"

Haskell

This even works on infinite strings (that have a finite longest common prefix), due to Haskell's laziness.

import Data.List (transpose)

lcp :: (Eq a) => [[a]] -> [a]
lcp = fmap head . takeWhile ((all . (==) . head) <*> tail) . transpose

main :: IO ()
main =
  putStrLn
    ( unlines $
        fmap
          showPrefix
          [ ["interspecies", "interstellar", "interstate"],
            ["throne", "throne"],
            ["throne", "dungeon"],
            ["cheese"],
            [""],
            ["prefix", "suffix"],
            ["foo", "foobar"]
          ]
    )
    >> print (lcp ["abc" <> repeat 'd', "abcde" <> repeat 'f'])

showPrefix :: [String] -> String
showPrefix = ((<>) . (<> " -> ") . show) <*> (show . lcp)

Output:

["interspecies","interstellar","interstate"] -> "inters"
["throne","throne"] -> "throne"
["throne","dungeon"] -> ""
["cheese"] -> "cheese"
[""] -> ""
["prefix","suffix"] -> ""
["foo","foobar"] -> "foo"

"abcd"

J

lcp=: {. {.~ 0 i.~ [: */2 =/\ ]

In other words: compare adjacent strings pair-wise, combine results logically, find first mismatch in any of them, take that many characters from the first of the strings.

Note that we rely on J's handling of edge cases here. In other words: if we have only one string that falls out as the longest prefix, and if we have no strings the result is the empty string.

As the number of adjacent pairs is O(n) where n is the number of strings, this approach could be faster in the limit cases than sorting.

Examples:

   lcp 'interspecies','interstellar',:'interstate'
inters
   lcp 'throne',:'throne'
throne
   lcp 'throne',:'dungeon'

   lcp ,:'cheese'
cheese
   lcp ,:''

   lcp 0 0$''

   lcp 'prefix',:'suffix'

Java

Works with: Java version 1.5+

public class LCP {
    public static String lcp(String... list){
        if(list == null) return "";//special case
        String ret = "";
        int idx = 0;

        while(true){
            char thisLetter = 0;
            for(String word : list){
                if(idx == word.length()){ //if we reached the end of a word then we are done
                    return ret;
                }
                if(thisLetter == 0){ //if this is the first word then note the letter we are looking for
                    thisLetter = word.charAt(idx);
                }
                if(thisLetter != word.charAt(idx)){ //if this word doesn't match the letter at this position we are done
                    return ret;
                }
            }
            ret += thisLetter;//if we haven't said we are done then this position passed
            idx++;
        }
    }
    
    public static void main(String[] args){
        System.out.println(lcp("interspecies","interstellar","interstate"));
        System.out.println(lcp("throne","throne"));
        System.out.println(lcp("throne","dungeon"));
        System.out.println(lcp("throne","","throne"));
        System.out.println(lcp("cheese"));
        System.out.println(lcp(""));
        System.out.println(lcp(null));
        System.out.println(lcp("prefix","suffix"));
        System.out.println(lcp("foo","foobar"));
    }
}

Output:

inters
throne


cheese



foo

JavaScript

ES5

(function () {
    'use strict';

    function lcp() {
        var lst = [].slice.call(arguments),
            n = lst.length ? takewhile(same, zip.apply(null, lst)).length : 0;

        return n ? lst[0].substr(0, n) : '';
    }


    // (a -> Bool) -> [a] -> [a]
    function takewhile(p, lst) {
        var x = lst.length ? lst[0] : null;
        return x !== null && p(x) ? [x].concat(takewhile(p, lst.slice(1))) : [];
    }

    // Zip arbitrary number of lists (an imperative implementation)
    // [[a]] -> [[a]]
    function zip() {
        var lngLists = arguments.length,
            lngMin = Infinity,
            lstZip = [],
            arrTuple = [],
            lngLen, i, j;

        for (i = lngLists; i--;) {
            lngLen = arguments[i].length;
            if (lngLen < lngMin) lngMin = lngLen;
        }

        for (i = 0; i < lngMin; i++) {
            arrTuple = [];
            for (j = 0; j < lngLists; j++) {
                arrTuple.push(arguments[j][i]);
            }
            lstZip.push(arrTuple);
        }
        return lstZip;
    }

    // [a] -> Bool
    function same(lst) {
        return (lst.reduce(function (a, x) {
            return a === x ? a : null;
        }, lst[0])) !== null;
    }


    // TESTS

    return [
        lcp("interspecies", "interstellar", "interstate") === "inters",
        lcp("throne", "throne") === "throne",
        lcp("throne", "dungeon") === "",
        lcp("cheese") === "cheese",
        lcp("") === "",
        lcp("prefix", "suffix") === "",
        lcp("foo", "foobar") == "foo"
    ];

})();

Output:

[true, true, true, true, true, true, true]

We could also, of course, use a functional implementation of a zip for an arbitrary number of arguments (e.g. as below). A good balance is often, however, to functionally compose primitive elements which are themselves iteratively implemented.

The functional composition facilitates refactoring, code reuse, and brisk development, while the imperative implementations can sometimes give significantly better performance in ES5, which does not optimise tail recursion. ( Tail call optimisation is, however, envisaged for ES6 - see https://kangax.github.io/compat-table/es6/ for progress towards its implementation ).

This functionally implemented zip is significantly slower than the iterative version used above:

// Zip arbitrary number of lists (a functional implementation, this time)
// Accepts arrays or strings, and returns [[a]]
function zip() {
    var args = [].slice.call(arguments),
        lngMin = args.reduce(function (a, x) {
            var n = x.length;
            return n < a ? n : a;
        }, Infinity);

    if (lngMin) {
        return args.reduce(function (a, v) {
            return (
                typeof v === 'string' ? v.split('') : v
            ).slice(0, lngMin).map(a ? function (x, i) {
                return a[i].concat(x);
            } : function (x) {
                return [x];
            });
        }, null)
    } else return [];
}

ES6

(() => {
    "use strict";

    // -------------- LONGEST COMMON PREFIX --------------

    // lcp :: (Eq a) => [[a]] -> [a]
    const lcp = xs => {
        const go = ws =>
            ws.some(isNull) ? (
                []
            ) : [ws.map(head)].concat(
                go(ws.map(tail))
            );

        return takeWhile(allSame)(
                go(xs.map(s => [...s]))
            )
            .map(head)
            .join("");

    };


    // ---------------------- TEST -----------------------

    // main :: IO ()
    const main = () => [
        ["interspecies", "interstellar", "interstate"],
        ["throne", "throne"],
        ["throne", "dungeon"],
        ["cheese"],
        [""],
        ["prefix", "suffix"],
        ["foo", "foobar"]
    ].map(showPrefix).join("\n");


    // showPrefix :: [String] -> String
    const showPrefix = xs =>
        `${show(xs)}  -> ${show(lcp(xs))}`;


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

    // allSame :: [a] -> Bool
    const allSame = xs =>
        // True if xs has less than 2 items, or every item
        // in the tail of the list is identical to the head.
        2 > xs.length || (() => {
            const [h, ...t] = xs;

            return t.every(x => h === x);
        })();


    // head :: [a] -> a
    const head = xs =>
        xs.length ? (
            xs[0]
        ) : undefined;


    // isNull :: [a] -> Bool
    // isNull :: String -> Bool
    const isNull = xs =>
        // True if xs is empty.
        1 > xs.length;


    // show :: a -> String
    const show = JSON.stringify;


    // tail :: [a] -> [a]
    const tail = xs =>
        0 < xs.length ? (
            xs.slice(1)
        ) : [];


    // takeWhile :: (a -> Bool) -> [a] -> [a]
    const takeWhile = p =>
        xs => {
            const i = xs.findIndex(x => !p(x));

            return -1 !== i ? (
                xs.slice(0, i)
            ) : xs;
        };


    // MAIN ---
    return main();
})();

Output:

["interspecies","interstellar","interstate"]  -> "inters"
["throne","throne"]  -> "throne"
["throne","dungeon"]  -> ""
["cheese"]  -> "cheese"
[""]  -> ""
["prefix","suffix"]  -> ""
["foo","foobar"]  -> "foo"

jq

Works with: jq version 1.4

See #Scala for a description of the approach used in this section.

# If your jq includes until/2
# then feel free to omit the following definition:
def until(cond; next):
  def _until: if cond then . else (next|_until) end;  _until;

def longest_common_prefix:
 if length == 0 then ""        # by convention
 elif length == 1 then .[0]    # for speed
 else sort 
 | if .[0] == "" then ""       # for speed
   else .[0] as $first
   |    .[length-1] as $last
   | ([$first, $last] | map(length) | min) as $n
   | 0 | until( . == $n or $first[.:.+1] != $last[.:.+1]; .+1)
   | $first[0:.]
   end
 end;

Test Cases

def check(ans): longest_common_prefix == ans;

(["interspecies","interstellar","interstate"] | check("inters")) and
(["throne","throne"]                          | check("throne")) and
(["throne","dungeon"]                         | check("")) and
(["throne", "", "throne"]                     | check("")) and
(["cheese"]                                   | check("cheese")) and
([""]                                         | check("")) and
([]                                           | check("")) and
(["prefix","suffix"]                          | check("")) and
(["foo","foobar"]                             | check("foo"))

Output:

$ jq -n -f longest_common_prefix.jq
true

Julia

Works with: Julia version 0.6

function lcp(str::AbstractString...)
    r = IOBuffer()
    str = [str...]
    if !isempty(str)
        i = 1
        while all(i ≤ length(s) for s in str) && all(s == str[1][i] for s in getindex.(str, i))
            print(r, str[1][i])
            i += 1
        end
    end
    return String(r)
end

@show lcp("interspecies", "interstellar", "interstate")
@show lcp("throne","throne")
@show lcp("throne","dungeon")
@show lcp("throne", "", "throne")
@show lcp("cheese")
@show lcp("")
@show lcp()
@show lcp("prefix","suffix")
@show lcp("foo","foobar")

Output:

lcp("interspecies", "interstellar", "interstate") = "inters"
lcp("throne", "throne") = "throne"
lcp("throne", "dungeon") = ""
lcp("throne", "", "throne") = ""
lcp("cheese") = "cheese"
lcp("") = ""
lcp() = ""
lcp("prefix", "suffix") = ""
lcp("foo", "foobar") = "foo"

Kotlin

// version 1.0.6

fun lcp(vararg sa: String): String {
    if (sa.isEmpty()) return ""
    if (sa.size == 1) return sa[0]
    val minLength = sa.map { it.length }.min()!!
    var oldPrefix = "" 
    var newPrefix: String
    for (i in 1 .. minLength) { 
        newPrefix = sa[0].substring(0, i)
        for (j in 1 until sa.size) 
            if (!sa[j].startsWith(newPrefix)) return oldPrefix
        oldPrefix = newPrefix
    }
    return oldPrefix         
}

fun main(args: Array<String>) {
    println("The longest common prefixes of the following collections of strings are:\n")
    println("""["interspecies","interstellar","interstate"] = "${lcp("interspecies", "interstellar", "interstate")}"""")
    println("""["throne","throne"]                          = "${lcp("throne", "throne")}"""")
    println("""["throne","dungeon"]                         = "${lcp("throne", "dungeon")}"""")
    println("""["throne","","throne"]                       = "${lcp("throne", "", "throne")}"""")
    println("""["cheese"]                                   = "${lcp("cheese")}"""")
    println("""[""]                                         = "${lcp("")}"""")
    println("""[]                                           = "${lcp()}"""")
    println("""["prefix","suffix"]                          = "${lcp("prefix", "suffix")}"""")
    println("""["foo","foobar"]                             = "${lcp("foo", "foobar")}"""")
}

Output:

The longest common prefixes of the following collections of strings are:

["interspecies","interstellar","interstate"] = "inters"
["throne","throne"]                          = "throne"
["throne","dungeon"]                         = ""
["throne","","throne"]                       = ""
["cheese"]                                   = "cheese"
[""]                                         = ""
[]                                           = ""
["prefix","suffix"]                          = ""
["foo","foobar"]                             = "foo"

Lobster

Translation of: Go

// lcp finds the longest common prefix of the input strings

def lcp(l):
    // Special cases first
    let len = l.length
    if len == 0:
        return ""
    else: if len == 1:
        return l[0]
    // LCP of min and max (lexigraphically) is the LCP of the whole set
    var min, max = l[0], l[0]
    for(l) s, i:
        if i > 0:
            if s < min:
                min = s
            else: if s > max:
                max = s
    let slen = min(min.length, max.length)
    if slen == 0:
        return ""
    for(slen) i:
        if min[i] != max[i]:
            return min.substring(0, i)
    // In the case where lengths are not equal but all bytes
    // are equal, min is the answer ("foo" < "foobar")
    return min

for([["interspecies", "interstellar", "interstate"],
    ["throne", "throne"],
    ["throne", "dungeon"],
    ["throne", "", "throne"],
    ["cheese"],
    [""],
    [],
    ["prefix", "suffix"],
    ["foo", "foobar"]]):
        print("lcp" + _ + " = \"" + lcp(_) + "\"")

Output:

lcp["interspecies", "interstellar", "interstate"] = "inters"
lcp["throne", "throne"] = "throne"
lcp["throne", "dungeon"] = ""
lcp["throne", "", "throne"] = ""
lcp["cheese"] = "cheese"
lcp[""] = ""
lcp[] = ""
lcp["prefix", "suffix"] = ""
lcp["foo", "foobar"] = "foo"

Lua

function lcp (strList)
    local shortest, prefix, first = math.huge, ""
    for _, str in pairs(strList) do
        if str:len() < shortest then shortest = str:len() end
    end
    for strPos = 1, shortest do
        if strList[1] then
            first = strList[1]:sub(strPos, strPos)
        else
            return prefix
        end
        for listPos = 2, #strList do
            if strList[listPos]:sub(strPos, strPos) ~= first then
                return prefix
            end
        end
        prefix = prefix .. first
    end
    return prefix
end

local testCases, pre = {
    {"interspecies", "interstellar", "interstate"},
    {"throne", "throne"},
    {"throne", "dungeon"},
    {"throne", "", "throne"},
    {"cheese"},
    {""},
    {nil},
    {"prefix", "suffix"},
    {"foo", "foobar"}
}
for _, stringList in pairs(testCases) do
    pre = lcp(stringList)
    if pre == "" then print(string.char(238)) else print(pre) end
end

Output:

inters
throne
ε
ε
cheese
ε
ε
ε
foo

Maple

lcp := proc(arr)
	local A:
	if (arr = []) then return "": end if:
	A := sort(arr):
	return (A[1][1..(StringTools:-CommonPrefix(A[1],A[-1]))]):
end proc:

Test Cases

lcp(["interspecies","interstellar","interstate"]);
lcp(["throne","throne"]);
lcp(["throne","dungeon"]);
lcp(["throne","","dungeon"]);
lcp(["cheese"]);
lcp([""]);
lcp([]);
lcp(["prefix","suffix"]);
lcp(["foo","foobar"]);

Output:

inters
throne
""
""
cheese
""
""
""
foo

Mathematica /Wolfram Language

ClearAll[LCP]
LCP[x_List] := Module[{l, s},
  If[Length[x] > 0,
   l = Min[StringLength /@ x];
   s = Characters[StringTake[x, l]];
   s //= Transpose;
   l = LengthWhile[s, Apply[SameQ]];
   StringTake[First[x], l]
   ,
   ""
   ]
  ]
LCP[{"interspecies", "interstellar", "interstate"}]
LCP[{"throne", "throne"}]
LCP[{"throne", "dungeon"}]
LCP[{"throne", "", "throne"}]
LCP[{"cheese"}]
LCP[{""}]
LCP[{}]
LCP[{"prefix", "suffix"}]
LCP[{"foo", "foobar"}]

Output:

"inters"
"throne"
""
""
"cheese"
""
""
""
"foo"

MATLAB / Octave

function lcp = longest_common_prefix(varargin)
ca    = char(varargin);
ix    = [any(ca~=ca(1,:),1),1];
lcp   = ca(1,1:find(ix,1)-1);
end

longest_common_prefix('aa', 'aa', 'aac')

Output:

ans = aa

MiniScript

We find the shortest and longest strings (without sorting, which makes the code slightly longer but much more efficient), and then just compare those.

lcp = function(strList)
    if not strList then return null
    // find the shortest and longest strings (without sorting!)
    shortest = strList[0]
    longest = strList[0]
    for s in strList
        if s.len < shortest.len then shortest = s
        if s.len > longest.len then longest = s
    end for
    if shortest.len < 1 then return ""
    // now find how much of the shortest matches the longest
    for i in range(0, shortest.len-1)
        if shortest[i] != longest[i] then return shortest[:i]
    end for
    return shortest
end function

print lcp(["interspecies","interstellar","interstate"])
print lcp(["throne","throne"])
print lcp(["throne","dungeon"])
print lcp(["throne", "", "throne"])
print lcp(["cheese"])
print lcp([])
print lcp(["foo","foobar"])

Output:

inters
throne


cheese
null
foo

Miranda

main :: [sys_message]
main = [Stdout (lay (map test tests))]

test :: [[char]]->[char]
test strings = show strings ++ " = " ++ show (lcp strings)

tests :: [[[char]]]
tests = [["interspecies","interstellar","interstate"],
         ["throne","throne"],
         ["throne","dungeon"],
         ["throne","","throne"],
         [""],
         [],
         ["prefix","suffix"],
         ["foo","foobar"]]

lcp :: [[char]]->[char]
lcp strings = map hd (takewhile same (transpose truncated))
              where same (a:as) = and [c=a | c<-as]
                    truncated   = map (take length) strings
                    length      = min (map (#) strings)

Output:

main :: [sys_message]
main = [Stdout (lay (map test tests))]

test :: [[char]]->[char]
test strings = show strings ++ " = " ++ show (lcp strings)

tests :: [[[char]]]
tests = [["interspecies","interstellar","interstate"],
         ["throne","throne"],
         ["throne","dungeon"],
         ["throne","","throne"],
         [""],
         [],
         ["prefix","suffix"],
         ["foo","foobar"]]

lcp :: [[char]]->[char]
lcp strings = map hd (takewhile same (transpose truncated))
              where same (a:as) = and [c=a | c<-as]
                    truncated   = map (take length) strings
                    length      = min (map (#) strings)

Modula-2

Translation of: C#

MODULE LCP;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;

TYPE String = ARRAY[0..15] OF CHAR;

PROCEDURE Length(str : String) : CARDINAL;
VAR len : CARDINAL;
BEGIN
    len := 0;
    WHILE str[len] # 0C DO
        INC(len)
    END;
    RETURN len
END Length;

PROCEDURE LongestCommonPrefix(params : ARRAY OF String) : String;
VAR
    ret : String;
    idx,widx : CARDINAL;
    thisLetter : CHAR;
BEGIN
    ret := "";
    idx := 0;
    LOOP
        thisLetter := 0C;
        FOR widx:=0 TO HIGH(params) DO
            IF idx = Length(params[widx]) THEN
                (* if we reached the end of a word then we are done *)
                RETURN ret
            END;
            IF thisLetter = 0C THEN
                (* if this is the first word then note the letter we are looking for *)
                thisLetter := params[widx][idx]
            END;
            IF thisLetter # params[widx][idx] THEN
                RETURN ret
            END
        END;

        (* if we haven't said we are done then this position passed *)
        ret[idx] := thisLetter;
        INC(idx);
        ret[idx] := 0C
    END;
    RETURN ret
END LongestCommonPrefix;

(* Main *)
TYPE
    AS3 = ARRAY[0..2] OF String;
    AS2 = ARRAY[0..1] OF String;
    AS1 = ARRAY[0..0] OF String;
BEGIN
    WriteString(LongestCommonPrefix(AS3{"interspecies", "interstellar", "interstate"}));
    WriteLn;
    WriteString(LongestCommonPrefix(AS2{"throne", "throne"}));
    WriteLn;
    WriteString(LongestCommonPrefix(AS2{"throne", "dungeon"}));
    WriteLn;
    WriteString(LongestCommonPrefix(AS3{"throne", "", "throne"}));
    WriteLn;
    WriteString(LongestCommonPrefix(AS1{"cheese"}));
    WriteLn;
    WriteString(LongestCommonPrefix(AS1{""}));
    WriteLn;
    WriteString(LongestCommonPrefix(AS2{"prefix", "suffix"}));
    WriteLn;
    WriteString(LongestCommonPrefix(AS2{"foo", "foobar"}));
    WriteLn;

    ReadChar
END LCP.

Output:

inters
throne


cheese

foo

Nim

import sequtils, strformat, strutils

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

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


test("interspecies", "interstellar", "interstate")
test("throne", "throne")
test("throne", "dungeon")
test("cheese")
test("")
test()
test("prefix", "suffix")
test("foo", "foobar")

Output:

lcp("interspecies", "interstellar", "interstate") = "inters"
lcp("throne", "throne") = "throne"
lcp("throne", "dungeon") = ""
lcp("cheese") = "cheese"
lcp("") = ""
lcp() = ""
lcp("prefix", "suffix") = ""
lcp("foo", "foobar") = "foo"

Ol

(define (lcp . args)
   (if (null? args)
      ""
      (let loop ((args (map string->list args)) (out #null))
         (if (or (has? args #null)
               (not (apply = (map car args))))
            (list->string (reverse out))
            (loop (map cdr args) (cons (caar args) out))))))

(print "> " (lcp "interspecies" "interstellar" "interstate"))
(print "> " (lcp "throne" "throne"))
(print "> " (lcp "throne" "dungeon"))
(print "> " (lcp "throne" "" "throne"))
(print "> " (lcp "cheese"))
(print "> " (lcp ""))
(print "> " (lcp))
(print "> " (lcp "prefix" "suffix"))
(print "> " (lcp "foo" "foobar"))

{Out}

> inters
> throne
> 
> 
> cheese
> 
> 
> 
> foo

ooRexx

Translation of: REXX

Call assert lcp(.list~of("interspecies","interstellar","interstate")),"inters"
Call assert lcp(.list~of("throne","throne")),"throne"
Call assert lcp(.list~of("throne","dungeon")),""
Call assert lcp(.list~of("cheese")),"cheese"
Call assert lcp(.list~of("",""))
Call assert lcp(.list~of("prefix","suffix")),""
Call assert lcp(.list~of("a","b","c",'aaa')),""
Exit

assert:
  If arg(1)==arg(2) Then tag='ok'
                    Else tag='??'
  Say tag 'lcp="'arg(1)'"'
  Say ''
  Return

lcp:
Use Arg l
a=l~makearray()
s=l~makearray()~makestring((LINE),',')
say 'lcp('s')'
an=a~dimension(1)
If an=1 Then
  Return a[1]
s=lcp2(a[1],a[2])
Do i=3 To an While s<>''
  s=lcp2(s,a[i])
  End
Return s

lcp2:
Do i=1 To min(length(arg(1)),length(arg(2)))
  If substr(arg(1),i,1)<>substr(arg(2),i,1) Then
    Leave
  End
Return left(arg(1),i-1)

Output:

lcp(interspecies,interstellar,interstate)
ok lcp="inters"

lcp(throne,throne)
ok lcp="throne"

lcp(throne,dungeon)
ok lcp=""

lcp(cheese)
ok lcp="cheese"

lcp(,)
ok lcp=""

lcp(prefix,suffix)
ok lcp=""

lcp(a,b,c,aaa)
ok lcp=""

Perl

If the strings are known not to contain null-bytes, we can let the regex backtracking engine find the longest common prefix like this:

sub lcp {
    (join("\0", @_) =~ /^ ([^\0]*) [^\0]* (?:\0 \1 [^\0]*)* $/sx)[0];
}

Testing:

use Test::More;
plan tests => 8;

is lcp("interspecies","interstellar","interstate"), "inters";
is lcp("throne","throne"),                          "throne";
is lcp("throne","dungeon"),                         "";
is lcp("cheese"),                                   "cheese";
is lcp(""),                                         "";
is lcp(),                                           "";
is lcp("prefix","suffix"),                          "";
is lcp("foo","foobar"),                             "foo";

Output:

As in the Raku example.

Phix

with javascript_semantics
function lcp(sequence strings)
    string res = ""
    if length(strings) then
        res = strings[1]
        for i=2 to length(strings) do
            string si = strings[i]
            for j=1 to length(res) do
                if j>length(si) or res[j]!=si[j] then
                    res = res[1..j-1]
                    exit
                end if
            end for
            if length(res)=0 then exit end if
        end for
    end if
    return res
end function
 
constant tests = {{"interspecies", "interstellar", "interstate"},
                  {"throne", "throne"},
                  {"throne", "dungeon"},
                  {"throne", "", "throne"},
                  {"cheese"},
                  {""},
                  {},
                  {"prefix", "suffix"},
                  {"foo", "foobar"}
                 }
for i=1 to length(tests) do
    ?lcp(tests[i])
end for

Output:

"inters"
"throne"
""
""
"cheese"
""
""
""
"foo"

PL/I

Translation of: REXX

*process source xref attributes or(!);
 (subrg):
 lcpt: Proc Options(main);
 Call assert(lcp('interspecies interstellar interstate'),'inters');
 Call assert(lcp('throne throne'),'throne');
 Call assert(lcp('throne dungeon'),'');
 Call assert(lcp('cheese'),'cheese');
 Call assert(lcp(' '),' ');
 Call assert(lcp('prefix suffix'),'');
 Call assert(lcp('a b c aaa'),'');

 assert: Proc(result,expected);
   Dcl (result,expected) Char(*) Var;
   Dcl tag Char(2) Init('ok');
   If result^=expected Then tag='??';
   Put Edit(tag,' lcp="',result,'"','')(Skip,4(a));
   End;

 lcp: Proc(string) Returns(Char(50) Var);
   Dcl string Char(*);
   Dcl xstring Char(50) Var;
   Dcl bn Bin Fixed(31) Init(0);
   Dcl bp(20) Bin Fixed(31);
   Dcl s Char(50) Var;
   Dcl i Bin Fixed(31);
   xstring=string!!' ';
   Put Edit('"'!!string!!'"')(Skip,a);
   Do i=2 To length(xstring);
     If substr(xstring,i,1)=' ' Then Do;
       bn+=1;
       bp(bn)=i;
       End;
     End;
   If bn=1 Then Return(substr(string,1,bp(1)-1));
   s=lcp2(substr(string,1,bp(1)-1),substr(string,bp(1)+1,bp(2)-bp(1)));
   Do i=3 To bn While(s^='');
     s=lcp2(s,substr(string,bp(i-1)+1,bp(i)-bp(i-1)));
     End;
   Return(s);
   End;

 lcp2: Proc(u,v) Returns(Char(50) Var);
   Dcl (u,v) Char(*);
   Dcl s Char(50) Var;
   Dcl i Bin Fixed(31);
   Do i=1 To min(length(u),length(v));
     If substr(u,i,1)^=substr(v,i,1) Then
       Leave;
     End;
   Return(left(u,i-1));
   End;

 End;

Output:

"interspecies interstellar interstate"
ok lcp="inters"

"throne throne"
ok lcp="throne"

"throne dungeon"
ok lcp=""

"cheese"
ok lcp="cheese"

" "
ok lcp=" "

"prefix suffix"
ok lcp=""

"a b c aaa"
ok lcp=""

PowerShell

function lcp ($arr) {
    if($arr){
        $arr = $arr | sort {$_.length} | select -unique 
        if(1 -lt $arr.count) {
            $lim, $i, $test = $arr[0].length, 0, $true
            while (($i -lt $lim) -and $test) {
                $test = ($arr | group {$_[$i]}).Name.Count -eq 1
                if ($test) {$i += 1}
            }
            $arr[0].substring(0,$i)
        } else {$arr}
    } else{''}

}
function show($arr) {
    function quote($str) {"`"$str`""}
    "lcp @($(($arr | foreach{quote $_}) -join ', ')) = $(lcp $arr)"
}
show @("interspecies","interstellar","interstate")
show @("throne","throne")
show @("throne","dungeon")
show @("throne", "","throne")
show @("cheese")
show @("") 
show @()
show @("prefix","suffix")
show @("foo","foobar")

Output:

lcp @("interspecies", "interstellar", "interstate") = inters
lcp @("throne", "throne") = throne
lcp @("throne", "dungeon") = 
lcp @("throne", "", "throne") = 
lcp @("cheese") = cheese
lcp @("") = 
lcp @() = 
lcp @("prefix", "suffix") = 
lcp @("foo", "foobar") = foo

Prolog

Works with: SWI Prolog

common_prefix(String1, String2, Prefix):-
    string_chars(String1, Chars1),
    string_chars(String2, Chars2),
    common_prefix1(Chars1, Chars2, Chars),
    string_chars(Prefix, Chars).

common_prefix1([], _, []):-!.
common_prefix1(_, [], []):-!.
common_prefix1([C1|_], [C2|_], []):-
    C1 \= C2,
    !.
common_prefix1([C|Chars1], [C|Chars2], [C|Chars]):-
    common_prefix1(Chars1, Chars2, Chars).
    
lcp([], ""):-!.
lcp([String], String):-!.
lcp(List, Prefix):-
    min_member(Min, List),
    max_member(Max, List),
    common_prefix(Min, Max, Prefix).

test(Strings):-
    lcp(Strings, Prefix),
    writef('lcp(%t) = %t\n', [Strings, Prefix]).

main:-
    test(["interspecies", "interstellar", "interstate"]),
    test(["throne", "throne"]),
    test(["throne", "dungeon"]),
    test(["throne", "", "throne"]),
    test(["cheese"]),
    test([""]),
    test([]),
    test(["prefix", "suffix"]),
    test(["foo", "foobar"]).

Output:

lcp(["interspecies","interstellar","interstate"]) = "inters"
lcp(["throne","throne"]) = "throne"
lcp(["throne","dungeon"]) = ""
lcp(["throne","","throne"]) = ""
lcp(["cheese"]) = "cheese"
lcp([""]) = ""
lcp([]) = ""
lcp(["prefix","suffix"]) = ""
lcp(["foo","foobar"]) = "foo"

Python

Note: this makes use of the error in os.path.commonprefix where it computes the longest common prefix regardless of directory separators rather than finding the common directory path.

import os.path

def lcp(*s):
    return os.path.commonprefix(s)

assert lcp("interspecies","interstellar","interstate") == "inters"
assert lcp("throne","throne") == "throne"
assert lcp("throne","dungeon") == ""
assert lcp("cheese") == "cheese"
assert lcp("") == ""
assert lcp("prefix","suffix") == ""
assert lcp("foo","foobar") == "foo"

Python: Functional

To see if all the n'th characters are the same I compare the min and max characters in the lambda function.

from itertools import takewhile

def lcp(*s):
    return ''.join(ch[0] for ch in takewhile(lambda x: min(x) == max(x),
					     zip(*s)))

assert lcp("interspecies","interstellar","interstate") == "inters"
assert lcp("throne","throne") == "throne"
assert lcp("throne","dungeon") == ""
assert lcp("cheese") == "cheese"
assert lcp("") == ""
assert lcp("prefix","suffix") == ""
assert lcp("foo","foobar") == "foo"

The above runs without output.

Alternative Functional

An alternative solution that takes advantage of the observation that the longest common prefix of a set of strings must be the same as the longest common prefix of the lexicographically minimal string and the lexicographically maximal string, since moving away lexicographically can only shorten the common prefix, never lengthening it. Finding the min and max could do a lot of unnecessary work though, if the strings are long and the common prefix is short.

from itertools import takewhile

def lcp(*s):
    return ''.join(a for a,b in takewhile(lambda x: x[0] == x[1],
					  zip(min(s), max(s))))

Or, defined in terms of a generic transpose function:

from itertools import (takewhile)


# lcp :: [String] -> String
def lcp(xs):
    return ''.join(
        x[0] for x in takewhile(allSame, transpose(xs))
    )


# TEST --------------------------------------------------

# main :: IO ()
def main():
    def showPrefix(xs):
        return ''.join(
            ['[' + ', '.join(xs), '] -> ', lcp(xs)]
        )

    print (*list(map(showPrefix, [
        ["interspecies", "interstellar", "interstate"],
        ["throne", "throne"],
        ["throne", "dungeon"],
        ["cheese"],
        [""],
        ["prefix", "suffix"],
        ["foo", "foobar"]])), sep='\n'
    )


# GENERIC FUNCTIONS -------------------------------------


# allSame :: [a] -> Bool
def allSame(xs):
    if 0 < len(xs):
        x = xs[0]
        return all(map(lambda y: x == y, xs))
    else:
        return True


# transpose :: [[a]] -> [[a]]
def transpose(xs):
    return map(list, zip(*xs))


# TEST ---
if __name__ == '__main__':
    main()

Output:

[interspecies, interstellar, interstate] -> inters
[throne, throne] -> throne
[throne, dungeon] -> 
[cheese] -> cheese
[] -> 
[prefix, suffix] -> 
[foo, foobar] -> foo

Quackery

  [ dup [] = iff
      [ drop true ] done
    true swap
    behead swap
    witheach
      [ over != if
          [ dip not conclude ] ] 
    drop ]                       is allsame       ( [ --> b )

  [ dup [] = iff
      [ drop 0 ] done
    behead size swap
    witheach [ size min ] ]      is minsize       ( [ --> n )
 
  [ over [] = iff 
      drop done
    [] unrot
    swap witheach
      [ over split 
        drop nested
        rot swap join swap ]
    drop ]                       is truncall    ( [ n --> ] )

  [ dup [] = if done
    dup minsize truncall
    [ dup allsame not while
      -1 truncall
      again ] 
    0 peek ]                    is commonprefix (   [ --> $ )

  [ dup $ "" = if
      [ drop 
        $ "** empty string" ]
    echo$
    cr ]                        is echoresult   (   $ --> $ )

  $ "interspecies interstellar interstate"
  nest$ commonprefix echoresult
 
  $ "throne throne"
  nest$ commonprefix echoresult
 
  $ "throne throne"
  nest$ $ "" swap 1 stuff 
  commonprefix echoresult
 
  $ "throne dungeon"
  nest$ commonprefix echoresult
 
  $ "cheese"
  nest$ commonprefix echoresult
 
  $ ""
  nest$ commonprefix echoresult
 
  ' [ ] commonprefix echoresult
 
  $ "prefix suffix"
  nest$ commonprefix echoresult
 
  $ "foo foobar"
  nest$ commonprefix echoresult

Output:

inters
throne
** empty string
** empty string
cheese
** empty string
** empty string
** empty string
foo

Racket

Note that there are three cases to the match, because zip needs at least one list, and char=? needs at least 2 characters to compare.

#lang racket
(require srfi/1)

(define ε "")  
(define lcp
  (match-lambda*
    [(list) ε]
    [(list a) a]
    [ss (list->string
         (reverse
          (let/ec k
            (fold (lambda (a d) (if (apply char=? a) (cons (car a) d) (k d))) null
                  (apply zip (map string->list ss))))))]))

(module+ test
  (require tests/eli-tester)
  (test
   (lcp "interspecies" "interstellar" "interstate") => "inters"
   (lcp "throne" "throne") => "throne"
   (lcp "throne" "dungeon") => ""
   (lcp "cheese") => "cheese"
   (lcp ε) => ε
   (lcp) => ε
   (lcp "prefix" "suffix") => ε))

All tests pass.

Raku

(formerly Perl 6)

Works with: rakudo version 2015-11-28

This should work on infinite strings (if and when we get them), since .ords is lazy. In any case, it does just about the minimal work by evaluating all strings lazily in parallel. A few explanations of the juicy bits: @s is the list of strings, and the hyper operator » applies the .ords to each of those strings, producing a list of lists. The | operator inserts each of those sublists as an argument into an argument list so that we can use a reduction operator across the list of lists, which makes sense if the operator in question knows how to deal with list arguments. In this case we use the Z ('zip') metaoperator with eqv as a base operator, which runs eqv across all the lists in parallel for each position, and will fail if not all the lists have the same ordinal value at that position, or if any of the strings run out of characters. Then we count up the leading matching positions and carve up one of the strings to that length.

multi lcp()    { '' }
multi lcp($s)  { ~$s }
multi lcp(*@s) { substr @s[0], 0, [+] [\and] [Zeqv] |@s».ords }

use Test;
plan 8;

is lcp("interspecies","interstellar","interstate"), "inters";
is lcp("throne","throne"), "throne";
is lcp("throne","dungeon"), '';
is lcp("cheese"), "cheese";
is lcp(''), '';
is lcp(), '';
is lcp("prefix","suffix"), '';
is lcp("foo","foobar"), 'foo';

Output:

1..8
ok 1 - 
ok 2 - 
ok 3 - 
ok 4 - 
ok 5 - 
ok 6 - 
ok 7 - 
ok 8 -

REXX

version 1

/* REXX */
Call assert lcp("interspecies","interstellar","interstate"),"inters"
Call assert lcp("throne","throne"),"throne"
Call assert lcp("throne","dungeon"),""
Call assert lcp("cheese"),"cheese"
Call assert lcp("","")
Call assert lcp("prefix","suffix"),""
Call assert lcp("a","b","c",'aaa'),""
Call assert lcp("foo",'foobar'),"foo"
Call assert lcp("ab","","abc"),""
Exit

assert:
  If arg(1)==arg(2) Then tag='ok'
                    Else tag='??'
  Say tag 'lcp="'arg(1)'"'
  Say ''
  Return

lcp: Procedure
ol='test lcp('
Do i=1 To arg()
  ol=ol||""""arg(i)""""
  If i<arg() Then ol=ol','
             Else ol=ol')'
  End
Say ol
If arg()=1 Then
  Return arg(1)
s=lcp2(arg(1),arg(2))
Do i=3 To arg() While s<>''
  s=lcp2(s,arg(i))
  End
Return s

lcp2: Procedure
Do i=1 To min(length(arg(1)),length(arg(2)))
  If substr(arg(1),i,1)<>substr(arg(2),i,1) Then
    Leave
  End
Return left(arg(1),i-1)

Output:

test lcp("interspecies","interstellar","interstate")
ok lcp="inters"

test lcp("throne","throne")
ok lcp="throne"

test lcp("throne","dungeon")
ok lcp=""

test lcp("cheese")
ok lcp="cheese"

test lcp("","")
ok lcp=""

test lcp("prefix","suffix")
ok lcp=""

test lcp("a","b","c","aaa")
ok lcp=""

test lcp("foo","foobar") ok lcp="foo"

test lcp("ab","","abc") ok lcp=""

version 2

This REXX version makes use of the compare BIF.

/*REXX program  computes the   longest common prefix  (LCP)   of any number of  strings.*/
say LCP('interspecies',  "interstellar",  'interstate')
say LCP('throne',  "throne")                     /*2 strings, they are exactly the same.*/
say LCP('throne',  "dungeon")                    /*2 completely different strings.      */
say LCP('throne',  '',   "throne")               /*3 strings, the middle string is null.*/
say LCP('cheese')                                /*just a single cheesy argument.       */
say LCP('')                                      /*just a single  null  argument.       */
say LCP()                                        /*no arguments are specified at all.   */
say LCP('prefix',  "suffix")                     /*two mostly different strings.        */
say LCP('foo',     "foobar")                     /*two mostly similar strings.          */
say LCP('a',  "b",  'c',  "aaa")                 /*four strings, mostly different.      */
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
LCP: @= arg(1);    m= length(@);    #= arg();    say copies('▒', 50)
                                 do i=1  for #;  say '────────────── string'  i":"  arg(i)
                                 end   /*i*/
                   do j=2  to #;    x= arg(j);     t= compare(@, x)   /*compare to next.*/
                   if t==1 | x==''  then do;   @= ;   leave;    end   /*mismatch of strs*/
                   if t==0 & @==x   then t= length(@) + 1             /*both are equal. */
                   if t>=m  then iterate                              /*not longest str.*/
                   m= t - 1;        @= left(@,  max(0, m) )           /*define maximum. */
                   end   /*j*/
     return  '  longest common prefix='    @                          /*return answer.  */

output when using the default inputs:

▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1: interspecies
────────────── string 2: interstellar
────────────── string 3: interstate
  longest common prefix= inters
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1: throne
────────────── string 2: throne
  longest common prefix= throne
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1: throne
────────────── string 2: dungeon
  longest common prefix=
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1: throne
────────────── string 2:
────────────── string 3: throne
  longest common prefix=
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1: cheese
  longest common prefix= cheese
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1:
  longest common prefix=
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
  longest common prefix=
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1: prefix
────────────── string 2: suffix
  longest common prefix=
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1: foo
────────────── string 2: foobar
  longest common prefix= foo
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
────────────── string 1: a
────────────── string 2: b
────────────── string 3: c
────────────── string 4: aaa
  longest common prefix=

version 3

This REXX version explicitly shows null values and the number of strings specified.

/*REXX program  computes the   longest common prefix  (LCP)   of any number of  strings.*/
say LCP('interspecies',  "interstellar",  'interstate')
say LCP('throne',  "throne")                     /*2 strings, they are exactly the same.*/
say LCP('throne',  "dungeon")                    /*2 completely different strings.      */
say LCP('throne',  '',   "throne")               /*3 strings, the middle string is null.*/
say LCP('cheese')                                /*just a single cheesy argument.       */
say LCP('')                                      /*just a single  null  argument.       */
say LCP()                                        /*no arguments are specified at all.   */
say LCP('prefix',  "suffix")                     /*two mostly different strings.        */
say LCP('foo',     "foobar")                     /*two mostly similar strings.          */
say LCP('a',  "b",  'c',  "aaa")                 /*four strings, mostly different.      */
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
LCP: @= arg(1);  m= length(@);      #=arg();       say copies('▒', 60)
                 say '──────────────     number of strings specified:'  #
                        do i=1  for #;  say '────────────── string' i":"  showNull(arg(i))
                        end   /*i*/

                    do j=2  to #;    x= arg(j);    t= compare(@, x)   /*compare to next.*/
                    if t==1 | x==''  then do;   @=;   leave;   end    /*mismatch of strs*/
                    if t==0 & @==x   then t= length(@) + 1            /*both are equal. */
                    if t>=m          then iterate                     /*not longest str.*/
                    m= t - 1;        @= left(@, max(0, m) )           /*define maximum. */
                    end   /*j*/
     return  '  longest common prefix='    shownull(@)                /*return answer.  */
/*──────────────────────────────────────────────────────────────────────────────────────*/
showNull: procedure;   parse arg z;        if z==''  then z= "«null»";         return z

output when using the default inputs:

▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 3
────────────── string 1: interspecies
────────────── string 2: interstellar
────────────── string 3: interstate
  longest common prefix= inters
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 2
────────────── string 1: throne
────────────── string 2: throne
  longest common prefix= throne
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 2
────────────── string 1: throne
────────────── string 2: dungeon
  longest common prefix= «null»
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 3
────────────── string 1: throne
────────────── string 2: «null»
────────────── string 3: throne
  longest common prefix= «null»
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 1
────────────── string 1: cheese
  longest common prefix= cheese
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 1
────────────── string 1: «null»
  longest common prefix= «null»
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 0
  longest common prefix= «null»
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 2
────────────── string 1: prefix
────────────── string 2: suffix
  longest common prefix= «null»
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 2
────────────── string 1: foo
────────────── string 2: foobar
  longest common prefix= foo
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
──────────────     number of strings specified: 4
────────────── string 1: a
────────────── string 2: b
────────────── string 3: c
────────────── string 4: aaa
  longest common prefix= «null»

Ring

# Project : Longest common prefix

aList1 = ["interspecies","interstellar","interstate"]
aList2 = list(len(aList1))
flag = 1
comp=""
for n=1 to len(aList1[1])
    aList2 = list(len(aList1))
    flag=1 
    for m=1 to len(aList1)
        aList2[m] = left(aList1[m], n )
        compare =  left(aList1[1], n )
    next
    for p=1 to len(aList1)
        if aList2[p] != compare
           flag = 0
           exit
        ok
    next
    if flag=1
       if len(compare) > comp
          comp=compare
        ok
     ok
next
if comp=""
   see "none"
else   
   see comp + nl 
ok

Output:

inters

RPL

≪ DUP SIZE → n
  ≪ CASE 
       n NOT  THEN DROP "" END
       n 1 == THEN 1 GET END
       DUP ≪ SIZE ≫ DOLIST ≪ MIN ≫ STREAM           @ get the size of the smallest string
       IF DUP NOT THEN DROP2 "" ELSE
          1 OVER FOR j
             OVER 1 ≪ 1 j SUB ≫ DOLIST
             IF ≪ == ≫ DOSUBS 1 + ΠLIST NOT THEN 
                j 1 - SWAP ‘j’ STO END
          NEXT 
       SWAP 1 GET 1 ROT SUB
    END END 
≫ ≫ 'LCP' STO

{ { "interspecies" "interstellar" "interstate" } { "throne" "throne" } { "throne" "dungeon" }{ "throne" "" "throne" } { "cheese" } { "" } {  } { "prefix" "suffix" } { "foo" "foobar" } }  ≪ LCP ≫ DOLIST

Output:

1: { "inters" "throne" "" "" "cheese" "" "" "" "foo" }

Ruby

def lcp(*strs)
  return "" if strs.empty?
  min, max = strs.minmax
  idx = min.size.times{|i| break i if min[i] != max[i]}
  min[0...idx]
end

data = [
  ["interspecies","interstellar","interstate"],
  ["throne","throne"],
  ["throne","dungeon"],
  ["throne","","throne"],
  ["cheese"],
  [""],
  [],
  ["prefix","suffix"],
  ["foo","foobar"]
]

data.each do |set|
  puts "lcp(#{set.inspect[1..-2]}) = #{lcp(*set).inspect}"
end

Output:

lcp("interspecies", "interstellar", "interstate") = "inters"
lcp("throne", "throne") = "throne"
lcp("throne", "dungeon") = ""
lcp("throne", "", "throne") = ""
lcp("cheese") = "cheese"
lcp("") = ""
lcp() = ""
lcp("prefix", "suffix") = ""
lcp("foo", "foobar") = "foo"

Rust

Rust String by default is utf-8 encoded. Since utf-8 is variable width, indexing in constant time is not possible. This example therefore uses byte strings (slices of u8) for the strings. The implementation shown here is similar to the Java implementation.

fn main() {
    let strs: [&[&[u8]]; 7] = [
        &[b"interspecies", b"interstellar", b"interstate"],
        &[b"throne", b"throne"],
        &[b"throne", b"dungeon"],
        &[b"cheese"],
        &[b""],
        &[b"prefix", b"suffix"],
        &[b"foo", b"foobar"],
    ];
    strs.iter().for_each(|list| match lcp(list) {
        Some(prefix) => println!("{}", String::from_utf8_lossy(&prefix)),
        None => println!(),
    });
}

fn lcp(list: &[&[u8]]) -> Option<Vec<u8>> {
    if list.is_empty() {
        return None;
    }
    let mut ret = Vec::new();
    let mut i = 0;
    loop {
        let mut c = None;
        for word in list {
            if i == word.len() {
                return Some(ret);
            }
            match c {
                None => {
                    c = Some(word[i]);
                }
                Some(letter) if letter != word[i] => return Some(ret),
                _ => continue,
            }
        }
        if let Some(letter) = c {
            ret.push(letter);
        }
        i += 1;
    }
}

Output:

inters
throne

cheese


foo

Scala

Take the first and last of the set of sorted strings; zip the two strings into a sequence of tuples ('view' makes this happen laziliy, on demand), until the two characters in the tuple differ, at which point, unzip the sequence into two character sequences; finally, arbitarily take one of these sequences (they are identical) and convert back to a string

"interspecies" \                                                                 / i, n, t, e, r, s \
                > zip takeWhile: (i,i), (n,n), (t,t), (e,e), (r,r), (s,s) unzip <                     > "inters"
"intesteller"  /                                                                 \ i, n, t, e, r, s

class TestLCP extends FunSuite {
  test("shared start") {
    assert(lcp("interspecies","interstellar","interstate") === "inters")
    assert(lcp("throne","throne") === "throne")
    assert(lcp("throne","dungeon").isEmpty)
    assert(lcp("cheese") === "cheese")
    assert(lcp("").isEmpty)
    assert(lcp(Nil :_*).isEmpty)
    assert(lcp("prefix","suffix").isEmpty)
  }

  def lcp(list: String*) = list.foldLeft("")((_,_) =>
    (list.min.view,list.max.view).zipped.takeWhile(v => v._1 == v._2).unzip._1.mkString)
}

sed

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

Output:

$ printf '%s\n' interspecies interstellar interstate | sed -f lcp.sed
inters
$ printf '%s\n' throne throne | sed -f lcp.sed
throne
$ printf '%s\n' throne dungeon | sed -f lcp.sed

$ printf '%s\n' throne '' throne | sed -f lcp.sed

$ printf '%s\n' cheese | sed -f lcp.sed
cheese
$ printf '%s\n' '' | sed -f lcp.sed

$ printf '%s\n' prefix suffix | sed -f lcp.sed

$ printf '%s\n' foo foobar | sed -f lcp.sed
foo

Sidef

# Finds the first point where the tree bifurcates
func find_common_prefix(hash, acc) {
    if (hash.len == 1) {
        var pair = hash.to_a[0]
        return __FUNC__(pair.value, acc+pair.key)
    }
    return acc
}

# Creates a tree like: {a => {b => {c => {}}}}
func lcp(*strings) {
    var hash = Hash()

    for str in (strings.sort_by{.len}) {
        var ref = hash
        str.is_empty && return ''
        for char in str {
            if (ref.contains(char)) {
                ref = ref{char}
                ref.len == 0 && break
            }
            else {
                ref = (ref{char} = Hash())
            }
        }
    }

    return find_common_prefix(hash, '')
}

Demonstration:

var data = [
  ["interspecies","interstellar","interstate"],
  ["throne","throne"],
  ["throne","dungeon"],
  ["throne","","throne"],
  ["cheese"],
  [""],
  [],
  ["prefix","suffix"],
  ["foo","foobar"]
];

data.each { |set|
    say "lcp(#{set.dump.substr(1,-1)}) = #{lcp(set...).dump}";
};

Output:

lcp("interspecies", "interstellar", "interstate") = "inters"
lcp("throne", "throne") = "throne"
lcp("throne", "dungeon") = ""
lcp("throne", "", "throne") = ""
lcp("cheese") = "cheese"
lcp("") = ""
lcp() = ""
lcp("prefix", "suffix") = ""
lcp("foo", "foobar") = "foo"

Smalltalk

Works with: Smalltalk/X

There is already a longestCommonPrefix method in Collection; however, if there wasn't, the following will do:

prefixLength := [:a :b | 
                    |end| 
                    end := (a size) min:(b size). 
                    ((1 to:end) detect:[:i | (a at:i) ~= (b at:i)] ifNone:end+1)-1].

lcp := [:words |
            words isEmpty 
                ifTrue:['']
                ifFalse:[
                    |first l|
                    first := words first.
                    l := (words from:2) 
                            inject:first size 
                            into:[:minSofar :w | minSofar min:(prefixLength value:first value:w)].
                    first copyTo:l]].

#(
    ('interspecies' 'interstellar' 'interstate') 
    ('throne' 'throne')      
    ('throne' 'dungeon')      
    ('throne' '' 'throne')   
    ('cheese') 
    ('')        
    ()        
    ('prefix' 'suffix')    
    ('foo' 'foobar')
) do:[:eachList |
    Transcript show:eachList storeString; show:' ==> '; showCR:(lcp value:eachList)
]

Output:

#('interspecies' 'interstellar' 'interstate') ==> inters
#('throne' 'throne') ==> throne
#('throne' 'dungeon') ==> 
#('throne' '' 'throne') ==> 
#('cheese') ==> cheese
#('') ==> 
#() ==> 
#('prefix' 'suffix') ==> 
#('foo' 'foobar') ==> foo

Standard ML

val lcp =
  let
    val longerFirst = fn pair as (a, b) =>
      if size a < size b then (b, a) else pair
    and commonPrefix = fn (l, s) =>
      case CharVector.findi (fn (i, c) => c <> String.sub (l, i)) s of
        SOME (i, _) => String.substring (s, 0, i)
      | _ => s
  in
    fn [] => "" | x :: xs => foldl (commonPrefix o longerFirst) x xs
  end

Test:

val test = [
  ["interspecies", "interstellar", "interstate"],
  ["throne", "throne"],
  ["throne", "dungeon"],
  ["throne", "", "throne"],
  ["cheese"],
  [""],
  [],
  ["prefix", "suffix"],
  ["foo", "foobar"]
]

val () = (print o concat o map (fn lst => "'" ^ lcp lst ^ "'\n")) test

Output:

'inters'
'throne'
''
''
'cheese'
''
''
''
'foo'

Swift

func commonPrefix(string1: String, string2: String) -> String {
    return String(zip(string1, string2).prefix(while: {$0 == $1}).map{$0.0})
}

func longestCommonPrefix(_ strings: [String]) -> String {
    switch (strings.count) {
    case 0:
        return ""
    case 1:
        return strings[0]
    default:
        return commonPrefix(string1: strings.min()!, string2: strings.max()!)
    }
}

func printLongestCommonPrefix(_ strings: [String]) {
    print("lcp(\(strings)) = \"\(longestCommonPrefix(strings))\"")
}

printLongestCommonPrefix(["interspecies", "interstellar", "interstate"])
printLongestCommonPrefix(["throne", "throne"])
printLongestCommonPrefix(["throne", "dungeon"])
printLongestCommonPrefix(["throne", "", "throne"])
printLongestCommonPrefix(["cheese"])
printLongestCommonPrefix([""])
printLongestCommonPrefix([])
printLongestCommonPrefix(["prefix", "suffix"])
printLongestCommonPrefix(["foo", "foobar"])

Output:

lcp(["interspecies", "interstellar", "interstate"]) = "inters"
lcp(["throne", "throne"]) = "throne"
lcp(["throne", "dungeon"]) = ""
lcp(["throne", "", "throne"]) = ""
lcp(["cheese"]) = "cheese"
lcp([""]) = ""
lcp([]) = ""
lcp(["prefix", "suffix"]) = ""
lcp(["foo", "foobar"]) = "foo"

Tcl

Since TIP#195 this has been present as a core command:

% namespace import ::tcl::prefix
% prefix longest {interstellar interspecies interstate integer} ""
inte

UNIX Shell

Works with: bash

#!/bin/bash

lcp () {
	local i=0 word c longest

	case $# in
		0)
			return 1
		;;
		1)
			printf %s "$1"
			return
		;;
	esac

	while :; do
		c=
		for word; do
			[[ $i == ${#word} ]] && break 2
			[[ -z $c ]] && c="${word:i:1}"
			[[ ${word:i:1} != "$c" ]] && break 2
		done
		longest+="$c"
		((i++))
	done

	printf %s "$longest"
}

mapfile -t tests <<'TEST'
interspecies	interstellar	interstate
throne	throne
throne	dungeon
throne		throne
cheese

prefix	suffix
foo	foobar
TEST

for test in "${tests[@]}"; do
	mapfile -t -d $'\t' words <<<"$test"
	words=("${words[@]%$'\n'}")
	printf '%s -> "%s"\n' "$(declare -p words)" "$(lcp "${words[@]}")"
done

Output:

declare -a words=([0]="throne" [1]="throne") -> "throne"
declare -a words=([0]="throne" [1]="dungeon") -> ""
declare -a words=([0]="throne" [1]="" [2]="throne") -> ""
declare -a words=([0]="cheese") -> "cheese"
declare -a words=() -> ""
declare -a words=([0]="prefix" [1]="suffix") -> ""
declare -a words=([0]="foo" [1]="foobar") -> "foo"

VBScript

Function lcp(s)
	'declare an array
	str = Split(s,",")
	'indentify the length of the shortest word in the array
	For i = 0 To UBound(str)
		If i = 0 Then
			l = Len(str(i))
		ElseIf Len(str(i)) < l Then
			l = Len(str(i))
		End If
	Next
	'check prefixes and increment index
	idx = 0
	For j = 1 To l
		For k = 0 To UBound(str)
			If UBound(str) = 0 Then
				idx = Len(str(0))
			Else
				If k = 0 Then
					tstr = Mid(str(k),j,1)
				ElseIf k <> UBound(str) Then
					If Mid(str(k),j,1) <> tstr Then
						Exit For
					End If
				Else
					If Mid(str(k),j,1) <> tstr Then
						Exit For
					Else
						idx = idx + 1
					End If
				End If	
			End If
		Next
		If idx = 0 Then
			Exit For
		End If
	Next
	'return lcp
	If idx = 0 Then
		lcp = "No Matching Prefix"
	Else
		lcp = Mid(str(0),1,idx)
	End If
End Function

'Calling the function for test cases.
test = Array("interspecies,interstellar,interstate","throne,throne","throne,dungeon","cheese",_
		"","prefix,suffix")
		
For n = 0 To UBound(test)
	WScript.StdOut.Write "Test case " & n & " " & test(n) & " = " & lcp(test(n))
	WScript.StdOut.WriteLine
Next

Output:

Test case 0 interspecies,interstellar,interstate = inters
Test case 1 throne,throne = throne
Test case 2 throne,dungeon = No Matching Prefix
Test case 3 cheese = cheese
Test case 4  = No Matching Prefix
Test case 5 prefix,suffix = No Matching Prefix

Visual Basic .NET

Translation of: C#

Module Module1

    Function LongestCommonPrefix(ParamArray sa() As String) As String
        If IsNothing(sa) Then
            Return "" REM special case
        End If
        Dim ret = ""
        Dim idx = 0

        While True
            Dim thisLetter = Nothing
            For Each word In sa
                If idx = word.Length Then
                    REM if we reach the end of a word then we are done
                    Return ret
                End If
                If IsNothing(thisLetter) Then
                    REM if this is the first word, thennote the letter we are looking for
                    thisLetter = word(idx)
                End If
                If thisLetter <> word(idx) Then
                    Return ret
                End If
            Next

            REM if we haven't said we are done the this position passed
            ret += thisLetter
            idx += 1
        End While

        Return ""
    End Function

    Sub Main()
        Console.WriteLine(LongestCommonPrefix("interspecies", "interstellar", "interstate"))
        Console.WriteLine(LongestCommonPrefix("throne", "throne"))
        Console.WriteLine(LongestCommonPrefix("throne", "dungeon"))
        Console.WriteLine(LongestCommonPrefix("throne", "", "throne"))
        Console.WriteLine(LongestCommonPrefix("cheese"))
        Console.WriteLine(LongestCommonPrefix(""))
        Console.WriteLine(LongestCommonPrefix(Nothing))
        Console.WriteLine(LongestCommonPrefix("prefix", "suffix"))
        Console.WriteLine(LongestCommonPrefix("foo", "foobar"))
    End Sub

End Module

Output:

inters
throne


cheese



foo

V (Vlang)

Translation of: go

// lcp finds the longest common prefix of the input strings.
fn lcp(l []string) string {
    // Special cases first
    match l.len {
        0 {
            return ""
        }
        1 {
            return l[0]
        }
        else {}
    }
    // LCP of min and max (lexigraphically)
    // is the LCP of the whole set.
    mut min, mut max := l[0], l[0]
    for s in l[1..] {
        if s < min {
            min = s
        } else if s > max {
            max = s
        }
    }
    for i := 0; i < min.len && i < max.len; i++ {
        if min[i] != max[i] {
            return min[..i]
        }
    }
    // In the case where lengths are not equal but all bytes
    // are equal, min is the answer ("foo" < "foobar").
    return min
}
 
// Normally something like this would be a TestLCP function in *_test.go
// and use the testing package to report failures.
fn main() {
    for l in [
        ["interspecies", "interstellar", "interstate"],
        ["throne", "throne"],
        ["throne", "dungeon"],
        ["throne", "", "throne"],
        ["cheese"],
        [""],
        []string{},
        ["prefix", "suffix"],
        ["foo", "foobar"],
     ] {
        println("lcp($l) = ${lcp(l)}")
    }
}

Output:

Same as Go entry

Wren

Translation of: Kotlin

Library: Wren-fmt

import "./fmt" for Fmt

var lcp = Fn.new { |sa|
    var size = sa.count
    if (size == 0) return ""
    if (size == 1) return sa[0]
    var minLen = sa.skip(1).reduce(sa[0].count) { |min, s|  s.count < min ? s.count : min }
    var oldPrefix = ""
    for (i in 1..minLen) {
        var newPrefix = sa[0][0...i]
        for (j in 1...size) if (!sa[j].startsWith(newPrefix)) return oldPrefix
        oldPrefix = newPrefix
    }
    return oldPrefix
}

var lists = [
    ["interspecies","interstellar","interstate"],
    ["throne","throne"],
    ["throne","dungeon"],
    ["throne","","throne"],
    ["cheese"],
    [""],
    [],
    ["prefix","suffix"],
    ["foo","foobar"]
]

System.print("The longest common prefixes of the following collections of strings are:\n")
for (sa in lists) {
    Fmt.print("  $-46s = $q", Fmt.v("q", 0, sa), lcp.call(sa))
}

Output:

The longest common prefixes of the following collections of strings are:

  ["interspecies", "interstellar", "interstate"] = "inters"
  ["throne", "throne"]                           = "throne"
  ["throne", "dungeon"]                          = ""
  ["throne", "", "throne"]                       = ""
  ["cheese"]                                     = "cheese"
  [""]                                           = ""
  []                                             = ""
  ["prefix", "suffix"]                           = ""
  ["foo", "foobar"]                              = "foo"

XPL0

proc LCP(N, R);         \Show longest common prefix
int  N;  char R;
int  I, J;
string 0;
[ChOut(0, ^");
if N <= 0 then [ChOut(0, ^");  CrLf(0);  return];
for J:= 0 to -1>>1 do
    [for I:= 0 to N-1 do
        if R(I,J) # R(0,J) or R(0,J) = 0 then
            [ChOut(0, ^");  CrLf(0);  return];
    ChOut(0, R(0,J));
    ];
];
[
LCP(3, ["interspecies", "interstellar", "interstate"]);
LCP(2, ["throne", "throne"]);
LCP(2, ["throne", "dungeon"]);
LCP(3, ["throne", "", "throne"]);
LCP(1, ["cheese"]);
LCP(1, [""]);
LCP(0);
LCP(2, ["prefix", "suffix"]);
LCP(2, ["foo", "foobar"]);
]

Output:

"inters"
"throne"
""
""
"cheese"
""
""
""
"foo"

XProfan

Proc lcp
   Parameters long liste
   Declare int longest, j, L, string s,t
   var int cnt = GetCount(liste)
   WhileLoop 0, cnt-1
      L = Len(GetString$(liste,&loop))
      case &loop == 0 : longest = L
      case L < longest : longest = L
   EndWhile
   s = GetString$(liste,0)
   WhileLoop 1, cnt-1
      t = GetString$(liste,&loop)
      For j,1,longest
         If SubStr$(s,j) <> SubStr$(t,j)
            longest = j - 1
            BREAK
         EndIf
      EndFor
      If longest < 1
        Clear longest
        BREAK
      EndIf
      s = t
   EndWhile
   Return Left$(s,longest)
EndProc

cls
declare string s[7]
s[0] = "interspecies,interstellar,interstate"
s[1] = "throne,throne"
s[2] = "throne,dungeon"
s[3] = "throne,,throne"
s[4] = "cheese"
s[5] = ""
s[6] = "prefix,suffix"
s[7] = "foo,foobar"

WhileLoop 0,7
   ClearList 0
   Move("StrToList",s[&loop],",")
   Print "list: ("+s[&loop]+") = "+chr$(34) + lcp(0) + chr$(34)
EndWhile

ClearList 0
WaitKey
end

Output:

list: (interspecies,interstellar,interstate) = "inters"
list: (throne,throne) = "throne"
list: (throne,dungeon) = ""
list: (throne,,throne) = ""
list: (cheese) = "cheese"
list: () = ""
list: (prefix,suffix) = ""
list: (foo,foobar) = "foo"

zkl

The string method prefix returns the number of common prefix characters.

fcn lcp(s,strings){ s[0,s.prefix(vm.pasteArgs(1))] }

Or, without using prefix:

Translation of: Scala

fcn lcp(strings){
   vm.arglist.reduce(fcn(prefix,s){ Utils.Helpers.zipW(prefix,s) // lazy zip
      .pump(String,fcn([(a,b)]){ a==b and a or Void.Stop })
   })
}

tester:=TheVault.Test.UnitTester.UnitTester();
tester.testRun(lcp.fp("interspecies","interstellar","interstate"),Void,"inters",__LINE__);
tester.testRun(lcp.fp("throne","throne"),Void,"throne",__LINE__);
tester.testRun(lcp.fp("throne","dungeon"),Void,"",__LINE__);
tester.testRun(lcp.fp("cheese"),Void,"cheese",__LINE__);
tester.testRun(lcp.fp(""),Void,"",__LINE__);
tester.testRun(lcp.fp("prefix","suffix"),Void,"",__LINE__);
tester.stats();

The fp (partial application) method is used to delay running lcp until the tester actually tests.

Output:

===================== Unit Test 1 =====================
Test 1 passed!
===================== Unit Test 2 =====================
Test 2 passed!
===================== Unit Test 3 =====================
Test 3 passed!
===================== Unit Test 4 =====================
Test 4 passed!
===================== Unit Test 5 =====================
Test 5 passed!
===================== Unit Test 6 =====================
Test 6 passed!
6 tests completed.
Passed test(s): 6 (of 6)