Longest common prefix

Longest common prefix is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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:

${\displaystyle \forall x\ \in \ R:\ S\leq }$pref ${\displaystyle \ x}$ ~ "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, ${\displaystyle \varepsilon }$, is considered a prefix of all strings):

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


Task inspired by this stackoverflow question: Find the longest common starting substring in a set of strings

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


with Ada.Text_IO;

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
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
return;
end if;
end loop;
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");

Reset;   Consider ("throne");   Consider ("throne");

Reset;   Consider ("throne");   Consider ("dungeon");

Reset;   Consider ("prefix");   Consider ("suffix");

Reset;   Consider ("foo");   Consider ("foobar");

Reset;   Consider ("foo");

Reset;   Consider ("");

Reset;

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
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'


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)
.
.
.
,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"  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 (intercalate, transpose) lcp :: (Eq a) => [[a]] -> [a] lcp = fmap head . takeWhile ((all . (==) . head) <*> tail) . transpose main :: IO () main = do putStrLn$
intercalate
"\n"
(showPrefix <$> [ ["interspecies", "interstellar", "interstate"] , ["throne", "throne"] , ["throne", "dungeon"] , ["cheese"] , [""] , ["prefix", "suffix"] , ["foo", "foobar"] ]) putStrLn [] print$ 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;

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;

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
witheach
[ over != if
[ dip not conclude ] ]
drop ]                       is allsame       ( [ --> b )

[ dup [] = iff
[ drop 0 ] done
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)
`