Longest Common Prefix of 2 Strings

Question

Write a program that takes 2 strings as input, and returns the longest common prefix. This is code-golf, so the answer with the shortest amount of bytes wins.

Test Case 1:

"global" , "glossary"
"glo"


Test Case 2:

"department" , "depart"
"depart"

Test Case 3:

"glove", "dove"
""

Answer 1

Python 3, 54 bytes

Thanks Python for having a built-in function for this task! :D

import os;print(os.path.commonprefix(input().split()))

Takes input as two words separated by a space such as glossary global.

Answer 2

Haskell, 29 bytes

(c:x)%(d:y)|c==d=c:x%y;_%_=""

Usage:

>> "global"%"glossary"
"glo"

Recursively defines the binary function % by pattern matching. On two strings with equal first letters, takes that first letters, and prepends it to the function of the remainder of the strings. On anything else, gives the empty string.

Answer 3

Pyth, 8 7 bytes

e@F._MQ

Thanks @isaacg for 1 byte off

Takes input quoted and comma separated, like "abc", "acc". This exits on an error (but leaves stdout empty) when the result is the empty string. If that is unacceptable, add 2 bytes for #e@F._MQq

Test Suite

Explanation

e@F._MQ        : implicit Q = eval(input)
   ._MQ        : Map the prefix operator onto both inputs
 @F            : Fold the setwise intersection operator over those lists
e              : Take the last such element, the prefixes are always made from shortest
               : to longest, so this always gives the longest matching prefix

Answer 4

Haskell, 38 bytes

((map fst.fst.span(uncurry(==))).).zip

Usage example: ( ((map fst.fst.span(uncurry(==))).).zip ) "global" "glossary" -> "glo".

Zip both input string into a list of pairs of characters. Make two lists out of it: the first one with all pairs from the beginning as long as both characters are equal, the second one with all the rests. Drop the second list and extract all characters from the first list.

Answer 5

C++, 101 100 99 bytes

#include<iostream>
int i;main(){std::string s,t;std::cin>>s>>t;for(;s[i]==t[i];std::cout<<s[i++]);}

Reads two strings from stdin, prints the character at the current position from one of the strings while the character at the current position is equal to the character at the same position in the other string.

Thanks to Zereges for saving one byte.

Answer 6

CJam, 12 11 9 bytes

l_q.-{}#<

This reads the strings on two separate lines with Unix-style line ending, i.e., <string>\n<string>\n.

Thanks to @MartinBüttner for -1 byte, and to @jimmy23013 for -2 bytes!

Try it online in the CJam interpreter.

How it works

l_         e# Read a line (w/o trailing LF) from STDIN and push a copy.
  q        e# Read another line from STDIN (with trailing LF).
           e# The trailing linefeed makes sure that the lines are not equal.
   .-      e# Perform vectorized character subtraction. This yields 0 for equal
           e# characters, a non-zero value for two different characters, and the
           e# characters themselves (truthy) for the tail of the longer string.
     {}#   e# Find the index of the first truthy element.
        <  e# Keep that many characters from the first string.

Answer 7

APL, 13

{⊃↓K/⍨=⌿K←↑⍵}

This is a function that takes an array of two strings, and returns the prefix:

      {⊃↓K/⍨=⌿K←↑⍵}'glossary' 'global'
glo
      {⊃↓K/⍨=⌿K←↑⍵}'department' 'depart'
depart

Answer 8

AppleScript, 215 Bytes

And I tried so hard... ;(

set x to(display dialog""default answer"")'s text returned
set a to(display dialog""default answer"")'s text returned
set n to 1
set o to""
repeat while x's item n=a's item n
set o to o&x's item n
set n to n+1
end
o

I wanted to see how well AppleScript could pull this off, and man is it not built for string comparisons.

Answer 9

sed, 18

I had something much longer and more complicated in mind, so credit for this idea goes to @kirbyfan64sos.

s/(.*).* \1.*/\1/

Includes +1 for the -r option to sed.

Answer 10

C#, 201 147 bytes

using System.Linq;class a{static void Main(string[]a){a[0].Take(a[1].Length).TakeWhile((t,i)=>a[1][i]==t).ToList().ForEach(System.Console.Write);}}

I know it isn't terribly competitive. I just wanted to see what it would look like.

EDIT: Thanks Ash Burlakzenko, Berend, and Dennis_E

Answer 11

rs, 14 bytes

(.*).* \1.*/\1

Live demo and test cases.

This is pretty simple. It just matches the...longest common prefix and removes the rest of the string. If there is no longest common prefix, it just clears everything.

Answer 12

Common Lisp, 39

(lambda(a b)(subseq a 0(mismatch a b)))

Takes two string arguments, determines the index i where they differ, and returns a substring from 0 to i.

Answer 13

CJam, 12 8 26

r:AAr:B.=0#_W={;;ABe<}{<}?

Try it Online.

(Got idea to use .= instead of .- after looking at Dennis's answer.)

With all the edge cases, it became to hard for a CJam beginner like me to keep it short. Hopefully, this at least works for all cases.

Answer 14

Perl 5, 20 19 18 bytes

19 bytes, plus 1 for the -E flag instead of -e:

say<>=~/^(.*).* \1/

This is copied shamelessly from Digital Trauma's sed answer. It assumes the input is a couple of words without spaces in them (or before the first) and with one space between them.

---

Update:

ThisSuitIsBlackNot suggested using -pe as follows, to save a byte (thanks!):

($_)=/^(.*).* \1/

And then Luk Storms suggested using -nE as follows to save another byte (thanks!):

say/^(.*).* \1/

(I'm counting -E as one byte instead of the standard -e, but -n or -p as two. My impression is that that's SOP around here.)

Answer 15

Javascript ES6, 52 bytes

f=(a,b)=>[...a].filter((e,i)=>e==b[i]?1:b='').join``

Usage:

>> f("global","glossary")
"glo"

Answer 16

Python 3, 47

def f(w):[print(end=c[c!=d])for c,d in zip(*w)]

A function that takes a list w of two words, and prints the common prefix before terminating with an error.

Python 3's print function lets you prints strings flush against each other with print(end=c) (thanks to Sp3000 for saving 3 bytes with this shorter syntax). This repeatedly take two letters from the words, and prints the first of the letters. The indexing c[c!=d] gives an out-of-bounds error where c!=d, terminating the execution when two unequal letters are encountered.

An explicit for loop is one char longer than the list comprehension:

def f(w):
 for c,d in zip(*w):print(end=c[c!=d])

Answer 17

Python 3, 72

31 bytes saved thanks to FryAmTheEggman. 8 saved thanks to DSM.

r=''
for x,y in zip(input(),input()):
 if x==y:r+=x
 else:break
print(r)

Answer 18

05AB1E, 6 bytes

ø€ËγнÏ

Try it online or verify all test cases.

Or alternatively:

ηR.ΔÅ?

Might be invalid, since it outputs -1 instead of an empty string if the strings have no common prefix.

Try it online or verify all test cases.

Explanation:

ø      # Zip the two (implicit) inputs together, creating pairs
       #  i.e. "abc1de98f" and "abc2de87f"
       #   → ["aa","bb","cc","12","dd","ee","98","87","ff"]
 €Ë    # Check for each if all inner items (the two characters in this case) are the same
       #  → [1,1,1,0,1,1,1,0,0,1]
   γ   # Split it into parts, grouping the same subsequent values together
       #  → [[1,1,1],[0],[1,1],[0,0],[1]]
    н  # Pop and only leave the first item
       #  → [1,1,1]
     Ï # Only leave the characters at the truthy indices in the (implicit second) input
       #  i.e. "abc2de87f" and [1,1,1] → "abc"
       # (after which the result is output implicitly)

η      # Get the prefixes of the first (implicit) input
       #  i.e. "abc1de98f"
       #   → ["a","ab","abc","abc1","abc1d","abc1de","abc1de9","abc1de98","abc1de98f"]
 R     # Reverse this list
       #  → ["abc1de98f","abc1de98","abc1de9","abc1de","abc1d","abc1","abc","ab","a"]
  .Δ   # Then find the first item in this list which is truthy for:
       # (which will result in -1 if none are found)
    Å? #  Check if the second (implicit) input starts with the current prefix
       #   i.e. "abc2de87f" and "abc1d" → 0 (falsey)
       #   i.e. "abc2de87f" and "abc" → 1 (truthy)
       # (after which the result is output implicitly)

Answer 19

K, 24 bytes

{(+/&\=/(&/#:'x)#'x)#*x}

Find the minimum of the length of each string. ((&/#:'x)). Trim each string to that length (#'x). Then compare, smear and sum the resulting sequence:

  =/("globaa";"glossa")
1 1 1 0 0 1
  &\=/("globaa";"glossa")
1 1 1 0 0 0
  +/&\=/("globaa";"glossa")
3

Finally, take that many characters from the first of the strings provided (#*x).

In action:

 f: {(+/&\=/(&/#:'x)#'x)#*x};
 f'(("global";"glossary")
    ("department";"depart")
    ("glove";"dove")
    ("aaa";"aaaaa")
    ("identical";"identical")
    ("aca";"aba"))
("glo"
 "depart"
 ()
 "aaa"
 "identical"
 ,"a")

Answer 20

MATL, 11 9 bytes

y!=XdYpf)

Try it online!

(-2 bytes thanks to Giuseppe)

 y  % implicitly input the two strings, then duplicate the
    %  first one into the stack again
    %  stack: ['department' 'deported' 'department']
 !  % transpose the last string into a column vector
 =  % broadcast equality check - gives back a matrix comparing
    %  every letter in first input with the letters in the second
 Xd % diagonal of the matrix - comparison result of each letter with
    %  only corresponding letter in the other string
    %  stack: ['department' [1; 1; 1; 0; 1; 1; 0; 0;]]
 Yp % cumulative product (so only initial sequence of 1s remains
    %  1s, others become 0)
    %  stack: ['department' [1; 1; 1; 0; 0; 0; 0; 0;]]
 f  %  find the indices of the 1s
 )  % index at those elements so we get those letters out
    % (implicit) convert to string and display

Answer 21

Uiua ^SBCS, 5 bytes

▽\×∊,

Try it!

Explanation

▽\×∊,­⁡​‎‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏⁠‎⁡⁠⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌­
   ∊,  # ‎⁡mask of where letters match
 \×    # ‎⁢cumulative product
▽      # ‎⁣keep

Answer 22

Retina, 14 bytes

Uses the same idea as kirbyfan64sos. Unfortunately, despite Martin's claim that eventually Match mode will feature a way to print capturing groups, it hasn't been implemented yet. Otherwise, (.*).* \1 could be used along with 2 bytes or so for some not-yet-existing configuration string option.

(.*).* \1.*
$1

Each line would go in its own file, with 1 byte added per additional file. Alternatively, run in a single file with the -s flag.

Answer 23

Powershell, 65 bytes

Compare the strings, shrinking the first until it either matches (print and exit) or the string is null and the loop terminates.

param($a,$b)while($a){if($b-like"$a*"){$a;exit}$a=$a-replace".$"}

Answer 24

Julia, 62 bytes

f(a,b)=(c="";for(i,j)=zip(a,b) i!=j?break:(c*=string(i))end;c)

Ungolfed:

function f(a::AbstractString, b::AbstractString)
    # Initialize an output string
    c = ""

    # Iterate over the pairs of characters in a and b,
    # truncated to the shorter of the two lengths
    for (i, j) in zip(a, b)
        if i == j
            # If they match, append to the output string
            c *= string(i)
        else
            # Otherwise stop everything!
            break
        end
    end

    return c
end

Fixed an issue (at the hefty cost of 14 bytes) thanks to xnor!

Answer 25

C99, 73 bytes

main(int c,char *a[]){for(char *x=a[1],*y=a[2];*x==*y++;putchar(*x++));}

Similar to this answer, but shorter and meets spec (takes input from stdin).

Answer 26

MATLAB, 50 40 bytes

Defines a function that accepts 2 strings as input, outputs to command window

function t(a,b);a(1:find([diff(char(a,b)) 1],1)-1)

This solution will work for any string, outputs

ans =

   Empty string: 1-by-0

if no match is given.

Can be golfed by using a script instead of a function (using local variables a, b) (-16 bytes).

so getting 34 Bytes

a(1:find([diff(char(a,b)) 1],1)-1)

The function style (which seems to be the accepted style), yields

@(a,b)a(1:find([diff(char(a,b)) 1],1)-1)

(Thanks @Stewie Griffin)

Answer 27

Perl 6, 28 bytes

I came up with two that take their values from STDIN which are based on the Perl 5 answer.

lines~~/(.*).*' '$0/;say ~$0

lines~~/:s(.*).* $0/;say ~$0

The first requires exactly one space between the inputs, while the other requires at least one whitespace character between the inputs.

---

That is quite a bit shorter than the first thing I tried which takes the values from the command line.

say [~] map ->($a,$b){$a eq$b&&$a||last},[Z] @*ARGS».comb # 58 bytes

or even the lambda version of it:

{[~] map ->($a,$b){$a eq$b&&$a||last},[Z] @_».comb} # 52 bytes

Though this is much easier to adjust so that it accepts any number of input strings, at the cost of only one stroke.

{[~] map ->@b {([eq] @b)&&@b[0]||last},[Z] @_».comb} # 53 bytes
#          ┗━┛ ┗━━━━━━━┛  ┗━━━┛

my &common-prefix = {[~] map ->@b {([eq] @b)&&@b[0]||last},[Z] @_».comb}

say common-prefix <department depart>; # "depart"
say common-prefix; # ""
say common-prefix <department depart depot deprecated dependant>; # "dep"

# This code does not work directly with a single argument, so you have
# to give it an itemized List or Array, containing a single element.

say common-prefix $('department',); # "department"

# another option would be to replace `@_` with `(@_,)`

Answer 28

Dyalog APL, 12 bytes

{⊥⍨⌽=⌿↑⍵}↑∊

That's two bytes less than the previous APL solution!

The overall function is ↑, which takes n elements (characters) from the flattened (∊) argument, where n is the result of applying the function {⊥⍨⌽=⌿↑⍵} to the argument:

↑⍵ convert list of strings to table (padding with spaces to form rectangle)
=⌿ compare down (columns) giving boolean list
⌽ reverse
⊥⍨ count trailing trues*

---

*Literally it is a mixed-base to base-10 conversion, using the boolean list as both number and base:

⊥⍨0 1 0 1 1 is the same as 0 1 0 1 1⊥⍨0 1 0 1 1 which is 0×(0×1×0×1×1) 1×(1×0×1×1) 0×(0×1×1) 1×(1×1) + 1×(1) which again is two (the number of trailing 1s).

Answer 29

Brainfuck, 61 bytes

+
[
  ,[<+> >+<-]
  ++++[>--------<-]
  >
]
<<[<]
<+
[
  ,[>+>-<<-]
  >>[<]
  <[.>]
  <
]

Expects two words separated by a space.

Try it online.

Answer 30

Japt, 27 bytes

Japt is a shortened version of JavaScript. Interpreter

Um$(X,Y)=>$A&&X==VgY ?X:A=P

(The strings go into the Input box like so: "global" "glossary")

This code is exactly equivalent to the following JS:

A=10;(U,V)=>U.split``.map((X,Y)=>A&&X==V[Y]?X:A="").join``

I have not yet implemented anonymous functions, which is what the $...$ is for: anything between the dollar signs is left untouched in the switch to JS. After I add functions, this 21-byte code will suffice:

UmXY{A&&X==VgY ?X:A=P

And after I implement a few more features, it will ideally be 18 bytes:

UmXY{AxX=VgY ?X:AP

Suggestions welcome!

---

So it turns out that this program is only 15 bytes in modern Japt:

¡A©X¥VgY ?X:A=P

Try it online!