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"
""
ᵃp=al
ᵃp=al
ᵃp Non-deterministically choose a prefix of both inputs
= Check if the two prefixes are equal
al Find the last possible solution, which is the longest
s=scan(,"");`?`=\(a,b)`if`(regexpr(b,a)-1,a?sub(".$","",b),b);s[1]?s[2]
s=scan(,"");a=s[1];b=s[2];`+`=\(x)substring(x,1,1:nchar(a));max((+a)[+a==+b],"")
Creates all possible sequential substrings that start from the 1st character for the both strings, subset for those that match and output the longest one.
abbreviatesub(".$","",max(abbreviate(scan(,""),1,F,,,,F)))
Note: unfortunately, this is not a valid answer, as it would fail if the provided strings contain a " ", "\t" or similar. We assume that, if not otherwise specified, an input string might contain at least a space symbol.
{*|(*v)@{&y~'x}.#[&/#:'v;]'v:{#[;x]'1+!#x}'x}
Takes input as a 2-element list.
Swift, 34 bytes
import UIKit
"global".commonPrefixWith("glossary")
But with Swift 2 it is actually more like:
"global".commonPrefixWithString("glossary",options:.CaseInsensitiveSearch)
class P{static void Main(string[]a){try{for(int i=0;a[0][i]==a[1][i];)System.Console.Write(a[0][i++]);}catch{}}}
Newlines and indentation for clarity:
class P{
static void Main(string[]a){
try{
for(int i=0;a[0][i]==a[1][i];)
System.Console.Write(a[0][i++]);
}
catch{}
}
}
(od" "=,)x(0gdo=?.O1)
Expects input as two words, space-separated, like so: department depart. Try it here.
(od" "=,) Loops through input until a space is encountered
x Dumps extraneous space
(0gdo= 1) Loops through second word and compares letters
?.O Halts if two letters are not equal, outputs them otherwise
^w[B!32]{>}>w[B!0]{t[B]vb[1]<[X]w[B!0]{>}b[T]w[B!1]{>}b[0]^>}v<[X]<t[0]w[T=0]{>t[B]^t[T-B]v}w[B!0]{b[0]>}
Takes two words separated by a single space. (I can save a byte by using a tab instead but that feels like cheating.)
In pb, the area that can be written to is thought of as a 2D space, with (0, 0) in the upper left. Additionally, input is initially kept at Y=-1. This program copies the second word of the input to Y=0 (starting at (0, 0)). Then, each letter is compared to the letter immediately above it until one is found that doesn't match. The rest of the word is erased and the desired output is already on the canvas so it's printed when execution halts.
Ungolfed:
^w[B!32]{>}> # Go to the first letter of the second word
w[B!0]{ # For each letter in the second word:
t[B] # Save the letter to T
vb[1] # Put a flag below that letter so it can be found later
<[X]w[B!0]{>} # Go to the first empty space on Y=0
b[T] # Write the contents of T
w[B!1]{>}b[0] # Go back to the flag and erase it
^> # Restart loop from next letter
}
v<[X]< # Go to (-1, 0)
t[0] # Set T to 0
w[T=0]{ # While T is 0:
>t[B] # Save the next letter of the second word to T
^t[T-B]v # Subtract the equivalent letter of the first word from T
# If they were the same, T is 0 and the loop continues.
}
w[B!0]{b[0]>} # Erase the rest of the second word
->a,b{i=0;i+=1while a[i]&&a[i]==b[i];a[0,i]}
Sample run:
2.1.5 :001 > ->a,b{i=0;i+=1while a[i]&&a[i]==b[i];a[0,i]}["global , "glossary"]
=> "glo"
2.1.5 :002 > ->a,b{i=0;i+=1while a[i]&&a[i]==b[i];a[0,i]}["department", "depart"]
=> "depart"
2.1.5 :003 > ->a,b{i=0;i+=1while a[i]&&a[i]==b[i];a[0,i]}["glove", "dove"]
=> ""
<?=substr($t=$argv[1],0,strspn($t^$argv[2],"\0"));
Replace \0 with the actual byte.
$m=$args-join"`n"-match"(^.*).*`n\1";$Matches[1]
one line input strings only.
param($a,$b)for($i=0;$a[$i]-eq$b[$i]){$c+=$a[$i++]};"$c"
Test script:
$f = {
param($a,$b)for($i=0;$a[$i]-eq$b[$i]){$c+=$a[$i++]};"$c"
}
"glo" -eq (&$f "global" "glossary")
"depart" -eq (&$f "department" "depart")
"" -eq (&$f "glove" "dove")
Output:
True
True
True
(a,b)=>Concat(a.Zip(b,(x,y)=>x==y?x:'$').TakeWhile(x=>x!='$'))
Zip! This byte count includes only the lambda expression, and some necessary using static directives are not counted.
It is assumed that no word will contain the magical char value $ (otherwise program may fail). Can use \0 instead (but that is longer to type).
{while((x=substr($1,++i,1))~substr($2,i,1))printf x}
{while((x=substr($1,++i,1)) # grab each char from first string
~substr($2,i,1)) # compare to each char of second
printf x} # print match
func[a b][copy/part a find a first difference a b]
copy/part a ; copy the head of the first string
find a ; that ends at the position
first ; of the first symbol
difference a b ; in the special difference of the two strings
String a="aa",b="ab";char[]c=a.toCharArray(),d=b.toCharArray();int e=0,f=Math.min(c.length,d.length);for(;e<f&&c[e]==d[e];e++);return new String(c,0,e);
globalandGLOSSARYreturngloor''? \$\endgroup\$