Longest common prefix

From Rosetta Code
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:

pref ~ "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, , is considered a prefix of all strings):

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"

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

See Also:



Metrics: length

Sub-string search: Count occurrences of a substring

Multi-string operations: LCP, LCS, concatenation

Manipulation: reverse, lower- and uppercase

Aime[edit]

text
lcp(...)
{
integer n;
record r;
text h, l;
 
ucall(r_fix, 1, r, 0);
n = 0;
if (r_size(r)) {
h = r_high(r);
l = r_low(r);
while (l[n] && h[n] == l[n]) {
n += 1;
}
}
 
return cut(l, 0, n);
}
 
integer
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");
 
return 0;
}
Output:
"inters"
"throne"
""
""
"cheese"
""
""
""
"foo"

ALGOL 68[edit]

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

AutoHotkey[edit]

lcp(str*){
for k, v in str
w := v, list .= (list ? "`n" : "") v
return RegExReplace(list, "^(.*)\K(\V*\R\1\V*)+$")
}
Examples:
MsgBox % lcp("interspecies","interstellar","interstate")
Outputs:
inters

AWK[edit]

 
# 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++[edit]

#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

EchoLisp[edit]

 
;; 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")""
 
 

Elixir[edit]

Translation of: Ruby
defmodule RC do
def lcp([]), do: ""
def lcp(strs) do
min = Enum.min(strs)
max = Enum.max(strs)
index = Enum.find_index(0..String.length(min), fn i -> String.at(min,i) != String.at(max,i) end)
if index, do: String.slice(min, 0, index), else: min
end
end
 
data = [
["interspecies","interstellar","interstate"],
["throne","throne"],
["throne","dungeon"],
["throne","","throne"],
["cheese"],
[""],
[],
["prefix","suffix"],
["foo","foobar"]
]
 
Enum.each(data, fn strs ->
IO.puts "lcp(#{inspect strs}) = #{inspect RC.lcp(strs)}"
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"

FreeBASIC[edit]

' 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[edit]

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[edit]

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

import Data.List (intercalate)
 
lcp
:: (Eq a)
=> [[a]] -> [a]
lcp = (head <$>) . takeWhile allEqual . truncTranspose
where
-- Similar to transpose, but stops on end of shortest list.
truncTranspose :: [[a]] -> [[a]]
truncTranspose xs
| any null xs = []
| otherwise = (head <$> xs) : truncTranspose (tail <$> xs)
allEqual
:: (Eq a)
=> [a] -> Bool
allEqual (x:xs) = all (== x) xs
 
showPrefix :: [String] -> String
showPrefix xs = show xs ++ " -> " ++ show (lcp xs)
 
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'] -- prints "abcd"
Output:
["interspecies","interstellar","interstate"] -> "inters"
["throne","throne"] -> "throne"
["throne","dungeon"] -> ""
["cheese"] -> "cheese"
[""] -> ""
["prefix","suffix"] -> ""
["foo","foobar"] -> "foo"

"abcd"

J[edit]

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 (well designed) handling of edge cases here.

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[edit]

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[edit]

ES 5[edit]

(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 [];
}

jq[edit]

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

Kotlin[edit]

// 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"

Lua[edit]

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

ooRexx[edit]

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[edit]

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 Perl 6 example.

Perl 6[edit]

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 - 

Phix[edit]

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[edit]

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[edit]

 
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

Python[edit]

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[edit]

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))))

Racket[edit]

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.

REXX[edit]

version 1[edit]

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[edit]

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 /*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[edit]

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=' @ /*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[edit]

 
# Project : Longest common prefix
# Date  : 2017/09/26
# Author  : Gal Zsolt (~ CalmoSoft ~)
# Email  : <[email protected]>
 
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

Ruby[edit]

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"

Scala[edit]

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)
}

Sidef[edit]

# 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"

VBScript[edit]

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

Tcl[edit]

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

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

zkl[edit]

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)