This task is to output the shortest path to a file, after glob expansion.

What is shell globbing? In most shells, you can use the * character in a path to represent any characters at the position. For example, If directory foo contains files bar baz and asdf, then foo/b* will expand to foo/bar foo/baz.

Now, lets say that the current directory contains a file called ihavealongname and nothing else. If I want to reference this file, I might type *, which will represent only that one file, rather than typing out the full name.

If the directory also contains a file called ialsohavealongname, I can't do *, as it will match both files. The I would have to do, at least, ih*.

The * pattern also works for matching directories above the file I'm looking for. If there are only two directories foo and bar, but foo contains only a file baz and bar contains file asdf, I can match foo/baz with */baz. Or, even more concisely, */b*. If bar was empty, */* would work.

Your task: Given a string array of paths that represent the "current directory" and a single target path, output the shortest possible string that would expand to only that target path after expanding *s.

The target path can be taken as it's own string, as an index into the array of paths, as the first item on the array of paths passed in, or some other convenient way that isn't hard-coding. Ask in comments if unsure.

The target path is guaranteed to be present in the "current directory".

You can assume that all paths contain only alphanumeric ASCII (and /s). You may take as input paths that are rooted (start with /) or relative (don't start with /).

If there are multiple equally short possibilities, return any or all of them.

This is , fewest bytes wins!

Test cases, thanks to Kevin Cruijssen.

  • 4
    \$\begingroup\$ So can we assume filenames don't contain spaces, newlines, backslashes, *, ?, [ etc ? It would maybe be easiest if you just state that file and directory names are alphanumeric \$\endgroup\$
    – Ton Hospel
    Feb 22, 2018 at 10:44
  • 3
    \$\begingroup\$ The actual disk I/O part will be boring in many languages. e.g. in perl I would just take the filename and replace all path components by * and run perl glob to get all filenames that can be relevant (e.g. foo/bar/baz becomes */*/*) . After that it becomes a string processing challenge. And that challenge is already hard enough. I think this challengwe would be cleaner as "given a list of alphanumeric (and /) relative paths, find the shortest glob that matches only this existing target path \$\endgroup\$
    – Ton Hospel
    Feb 22, 2018 at 14:22
  • 1
    \$\begingroup\$ @KevinCruijssen Sure, it's a fun challenge and should mostly keep the pure golfing languages away I think you will need a real program (unless you generate all possible strings until you hit the shortest that works which is boring and will exponentially explode) Your current examples barely start to cover the cases you need to handle. Here's anther case: use a*f to select azzf from azzf, azzg, bzzf. Extend at will to a*b*c etc.. \$\endgroup\$
    – Ton Hospel
    Feb 22, 2018 at 14:27
  • 2
    \$\begingroup\$ @TonHospel I'm convinced. You now take an array of paths as input. \$\endgroup\$
    – Pavel
    Feb 22, 2018 at 16:49
  • 4
    \$\begingroup\$ @WeijunZhou I've changed my mind about input. You can now take an array of paths. \$\endgroup\$
    – Pavel
    Feb 22, 2018 at 16:52

2 Answers 2


Perl 5, 136 107 102 bytes

Includes +2 for n0

Give list of files on STDIN. The first one is assumed to be the target file

perl -n0E '@a="";for$a(@a){s%%s/(?=$a

Just the code without making the newlines literal:


Crashes intentionally after printing the solution.

Still seems too long (the use of $a and 1/0 are very awkward), but it's a start and should be reasonably efficient.

Try it online!

How it works

The program builds candidate globs by growing them from the back to the front starting with the empty string. It does this in a breadth first way, so first globs of length 0 are tried (only ``), then length 1 (like t, i, *), next length 2 (like fb, i*, *g, **), next length 3 and so on until a glob is found that only matches the first path. This will then be the shortest glob that solves the problem (others of the same length may exist).

The globs of length n+1 are generated from the globs of length n by prepending each character from the list of paths and also * in front of each glob of length n. So e.g. length 3 glob *i* will contribute length 4 globs f*i*, o*i*, o*i*, /*i*, b*i* ... s*i*, t*i* and finally **i*. Notice that every character from the list of input paths is prepended even if it appears multiple times or makes no sense whatsoever because it leads to something that can never match.

Doing this naively would lead to a combinatorial explosion. That is why every candidate glob gets evaluated for how useful it is by determining at which points in the paths it could match if the glob was used at the end of a complete glob. I do this by inserting a ; at each place where a match is possible. For example for the glob t* I will get the string:


This represents the "distinguishing power" of the glob. Every glob that has exactly the same distinguishing power is equally good. If you replace them by each other at the end of a complete glob they will all match exactly the same paths. So you can just as well use the shortest one.

So when considering the length n globs I first look at its distinguishing power. If it has been seen before there was another glob of length n or shorter that was already considered and expanded, so this glob is pointless and gets pruned. This will for example get rid of candidates like **i* since the same distinguishing power will already have been seen as *i*. It also prunes impossible candidates like f*i* since the distinguishing string will have no ; and just be the original list of paths. Only the very first impossible glob will be accepted, all the others will be seen as having the same distinguishing power and will be pruned. And even that first one won't really get expanded since all expansions are still impossible and will be pruned when they are considered. Simularly in* will be pruned by i* etc.

The above leads to very aggressive pruning and the program is therefore able to handle complex cases in a very short time. A major inefficiency though is that it prefixes the candidate globs with all possible characters, not only the ones just before a ; in target path part of the distinguishing string. All added characters that are not in front of a ; are no problem since they lead to an impossible glob which will be pruned when it gets considered, but that still leaves the characters just before ; in the other paths. So in the end the program also builds globs that will be able match any combination of the given paths. It has no idea that it should be concentrating on the first path.

Now consider a solution to the problem. In the given example that could be */*er/t. This gives the following distinguishing string:


I recognize a solution by having a ; at the first position (so it matches the first path) and not having a ; at the start of any other path (so the others don't match)

With the algorithm explained I now get to the actual program:

The candidate globs will be in array @a which I loop over using variable $a which contains the glob currently under consideration. Instead of * in the glob I will however use \w* so $a is actually a regex instead of a glob. I'm going to abuse a weirdness of the perl for loop that you can append elements to the array being looped while the loop is running and these new elements will get picked up in the loop. Since when generating the length n+1 globs all length n globs are already on array @a this is breadth first.

Due to the -n0 option (implicit loop over the whole input) the list of paths is in $_ as one big string with each path terminated with a newline

@a="";                    Start everything with the length 0 glob
for$a(@a){    }           Loop over candidates in a breadth first way

Inside the { } we have:

s/(?=$a\n)/;/g            Loop over the paths and insert a ; at every
                          position that the suffix glob can match by
                          looking ahead and checking that the regex
                          under consideration can match up to the end of
                          the path we are in. The distinguishing sting is
                          now in `$_`.

Oops, I just destroyed $_ and I will need it for the next loop. So wrap the actual working code inside

s%%  ...code.. %e

This matches the empty string at the start of $_ and allows you to run code to determine what it gets replaced with. If I make sure that that code evaluates to the empty string $_ will at the end remain unchanged even if I change $_ during code.

Going back to just after I replaced $_ by the distinguishing string:

$$_//= expression

This is like:

$seen{$_} //= expression

// in perl is'defined or. It's like a short circuiting or where the second argument is only evaluated if the first one is undef. And it can be combined with an assignment just like += in some other languages. So if they key $_ in hash %seen is undef (which is is what you get when accessing a non-existing element) only then execute expression and assign it as value to key $_. So if I make sure expression does not return undef that basically means "evaluate expression if and only if this is the first time we see that particular distinguishing string". And because $_ is guaranteed to contain a \n it is in fact safe to abuse the perl global hash to store the distinguishing strings, so $$_ instead of $seen{$_}

For the expression I use:


Basically "For every character (except newline) in the distinguishing string and also * prepend it to the current glob and push that on the array of candidate globs". Execpt I use \w* for * to get a valid regex (I could use '' instead of "" to get rid of one backslash but then I couldn't run my code from the command line). Notice that this also picks up the ; and adds them to the candidate globs but when later testing them to the restored $_ which has no ; that will again be an impossible glob and get pruned.

/^;/>/\n;/ &&      If the distinguishing string corresponds to a solution

say$a=~s/\\w//gr   Then replace all \w* back to * and print the solution

1/!                Say returns 1 so this becomes a division by 0.
                   The program exits by crashing after solving it

Notice that /^;/>/\n;/ has a value equivalent to the empty string in case a solution was not yet found, so this will function as empty replacement string and $_ gets restored

  • \$\begingroup\$ I get an error on TIO but it works locally. Do you know why that is? \$\endgroup\$
    – Pavel
    Feb 23, 2018 at 2:46
  • 1
    \$\begingroup\$ @Pavel The -E activates the latest language level. You need at least perl 5.10.0 to be able to use say. So put use 5.10.0; in the header section and it will work. Options to set the language level count as free anyways even if you couldn't also do it using -E. In fact all options count as free nowadays (so I don't even have to count n0) but I consider that too lenient for perl \$\endgroup\$
    – Ton Hospel
    Feb 23, 2018 at 5:39
  • 2
    \$\begingroup\$ Exiting with an error is fine, so your 1/ solution is valid! I need to remember that too... \$\endgroup\$ Feb 23, 2018 at 7:49

Java 10, 854 824 796 738 728 703 688 655 652 647 624 bytes

import java.util.*;a->f->{var L=new Stack();List<String>s;int i=999,j,k;for(var t:f.split("/")){s=new java.util.concurrent.CopyOnWriteArrayList();s.add(t);for(k=1;k>0;){k=0;for(var x:s)for(j=0;j<x.length();)if(!s.contains(f=x.substring(0,j)+"~"+x.substring(++j))){s.add(f);k=1;}}for(var x:s)s.add(x.replaceAll("~+","\\*"));L.add(s);}p(L,s=new Stack(),0,f="");for(var y:s){k=0;for(var x:a)if(x.matches(y.replace("*",".*"))&x.split("/").length==y.split("/").length)k++;if(k==1&(j=y.length())<i){f=y;i=j;}}return f;};void p(List L,List r,int d,String c){if(d==L.size())r.add(c);else for(var o:(List)L.get(d))p(L,r,d+1,c+"/"+o);}

What a mess.. This is certainly not an easy challenge in Java. Can definitely be golfed by a couple hundred bytes, but I'm just glad it's finally working now. Told ya. :)
-5 bytes thanks to @ceilingcat.
-23 bytes switching from Java 8 to Java 10

Input as a String-array of file-paths (with directories as separated items, and all items containing a leading /), and a String with input file-path to grop.


Try it online. (The test cases with ialsohavealongname/ihavealongnameaswell are slightly reduced in length and s.add(x.replaceAll("~+","\\*")); has been replaced with {s.remove(x);s.add(x.replaceAll("~+","\\*"));} to work in 5-10 seconds on TIO, instead of timing out after 60+ sec.)

import java.util.*;   // Required import for List and Stack

// Method with String-array and String parameters and String return-type
  var L=new Stack();  //  Create a List of Lists
  List<String>s;      //  List of Strings (uninitialized)
  int i=999,j,k;      //  Three integers (`i` starting at 999,
                      //   because 260 is the maximum file-path length in Windows)
  for(var t:f.split("/")){
                      //  Loop over the input file-path split by "/":
    s=new java.util.concurrent.CopyOnWriteArrayList();
                      //  Create a List (which we can modify while iterating it)
    s.add(t);         //  Add the input to this List
    for(k=1;k>0;){    //  Loop as long as there are new items added to the List
      k=0;            //   Reset the newAdded-flag to false
      for(var x:s)    //   And inner loop over the List
                      //    Inner loop `j` in range [0,length-of-item):
                      //     Replace the character at index `j` with a '~'
                      //     And if it's a new item:
            s.add(f); //      Add it to the List
            k=1;}}    //      And set the newAdded-flag to true
    for(var x:s)      //  Loop over the List again
                      //   And replace all 1 or more '~' with a single asterisk
                      //   (NOTE: To reduce bytes it doesn't remove the existing items)
    L.add(s);}        //   Add this List to the List of Lists
  p(L,s=new Stack(),0,"");
                      //  Generate all permutations of the groppings
                      //  (List `s` now contains all groppings of the given file-path)
  for(var y:s){       //  Loop over the groppings in the String-List:
    k=0;              //   Reset integer `k` to 0
    for(var x:a)      //   Inner loop over the input file-paths:
                      //    If the current file-path matches the current gropping
                      //    and the amount of slashes are the same:
         k++;         //     Increase integer `k` by 1
    if(k==1           //   If only one of the file-paths matched,
                      //   and the length is shorter than `i`:
      f=y;            //    Replace the result with this gropping file-path
      i=j;}}          //    And also replace `i` with this shorter `j`
  return f;}          //  Finally return this shortest gropping file-path

// Separated method to generate gropping file-path permutations given a List of Lists
void p(List L,List r,int d,String c){
  if(d==L.size())     //  If we've reached the final depth
    r.add(c);         //   Add the current gropping-file path to the result-List
  else                //  Else:
    for(var o:(List)L.get(d))
                      //   Loop over the List of the current depth:
      p(L,r,d+1,      //    Recursive call with depth+1,
        c+"/"+o);}    //    and current + "/" + item of loop

Additional general explanation:

Example: Let's take /foo, /foo/bar, /foo/barber, /foo/bar/test, /foo/barber/test, /foo/barber/testing, /foo/barber/coding, /foo/test as given file-paths, and foo/bar/test as input file-path to grop.

1) I start by splitting the file-path input by /, and generate all file-groppings of these separated words:

foo: [foo, *oo, f*o, fo*, *o, *o*, f*, *]
bar: [bar, *ar, b*r, ba*, *r, *a*, b*, *]
test: [test, *est, t*st, te*t, tes*, *st, *e*t, *es*, t*t, t*s*, te*, *t, *s*, *e*, t*, *]

2) I then generate all permutations with these words in the same order (reapplying the / in between and at the front):

[/foo/bar/test, /foo/bar/*est, /foo/bar/t*st, /foo/bar/te*t, /foo/bar/tes*, /foo/bar/*st, /foo/bar/*e*t, /foo/bar/*es*, /foo/bar/t*t, /foo/bar/t*s*, /foo/bar/te*, /foo/bar/*t, /foo/bar/*s*, /foo/bar/*e*, /foo/bar/t*, /foo/bar/*, /foo/*ar/test, /foo/*ar/*est, /foo/*ar/t*st, /foo/*ar/te*t, /foo/*ar/tes*, /foo/*ar/*st, /foo/*ar/*e*t, /foo/*ar/*es*, /foo/*ar/t*t, /foo/*ar/t*s*, /foo/*ar/te*, /foo/*ar/*t, /foo/*ar/*s*, /foo/*ar/*e*, /foo/*ar/t*, /foo/*ar/*, /foo/b*r/test, /foo/b*r/*est, /foo/b*r/t*st, /foo/b*r/te*t, /foo/b*r/tes*, /foo/b*r/*st, /foo/b*r/*e*t, /foo/b*r/*es*, /foo/b*r/t*t, /foo/b*r/t*s*, /foo/b*r/te*, /foo/b*r/*t, /foo/b*r/*s*, /foo/b*r/*e*, /foo/b*r/t*, /foo/b*r/*, /foo/ba*/test, /foo/ba*/*est, /foo/ba*/t*st, /foo/ba*/te*t, /foo/ba*/tes*, /foo/ba*/*st, /foo/ba*/*e*t, /foo/ba*/*es*, /foo/ba*/t*t, /foo/ba*/t*s*, /foo/ba*/te*, /foo/ba*/*t, /foo/ba*/*s*, /foo/ba*/*e*, /foo/ba*/t*, /foo/ba*/*, /foo/*r/test, /foo/*r/*est, /foo/*r/t*st, /foo/*r/te*t, /foo/*r/tes*, /foo/*r/*st, /foo/*r/*e*t, /foo/*r/*es*, /foo/*r/t*t, /foo/*r/t*s*, /foo/*r/te*, /foo/*r/*t, /foo/*r/*s*, /foo/*r/*e*, /foo/*r/t*, /foo/*r/*, /foo/*a*/test, /foo/*a*/*est, /foo/*a*/t*st, /foo/*a*/te*t, /foo/*a*/tes*, /foo/*a*/*st, /foo/*a*/*e*t, /foo/*a*/*es*, /foo/*a*/t*t, /foo/*a*/t*s*, /foo/*a*/te*, /foo/*a*/*t, /foo/*a*/*s*, /foo/*a*/*e*, /foo/*a*/t*, /foo/*a*/*, /foo/b*/test, /foo/b*/*est, /foo/b*/t*st, /foo/b*/te*t, /foo/b*/tes*, /foo/b*/*st, /foo/b*/*e*t, /foo/b*/*es*, /foo/b*/t*t, /foo/b*/t*s*, /foo/b*/te*, /foo/b*/*t, /foo/b*/*s*, /foo/b*/*e*, /foo/b*/t*, /foo/b*/*, /foo/*/test, /foo/*/*est, /foo/*/t*st, /foo/*/te*t, /foo/*/tes*, /foo/*/*st, /foo/*/*e*t, /foo/*/*es*, /foo/*/t*t, /foo/*/t*s*, /foo/*/te*, /foo/*/*t, /foo/*/*s*, /foo/*/*e*, /foo/*/t*, /foo/*/*, /*oo/bar/test, /*oo/bar/*est, /*oo/bar/t*st, /*oo/bar/te*t, /*oo/bar/tes*, /*oo/bar/*st, /*oo/bar/*e*t, /*oo/bar/*es*, /*oo/bar/t*t, /*oo/bar/t*s*, /*oo/bar/te*, /*oo/bar/*t, /*oo/bar/*s*, /*oo/bar/*e*, /*oo/bar/t*, /*oo/bar/*, /*oo/*ar/test, /*oo/*ar/*est, /*oo/*ar/t*st, /*oo/*ar/te*t, /*oo/*ar/tes*, /*oo/*ar/*st, /*oo/*ar/*e*t, /*oo/*ar/*es*, /*oo/*ar/t*t, /*oo/*ar/t*s*, /*oo/*ar/te*, /*oo/*ar/*t, /*oo/*ar/*s*, /*oo/*ar/*e*, /*oo/*ar/t*, /*oo/*ar/*, /*oo/b*r/test, /*oo/b*r/*est, /*oo/b*r/t*st, /*oo/b*r/te*t, /*oo/b*r/tes*, /*oo/b*r/*st, /*oo/b*r/*e*t, /*oo/b*r/*es*, /*oo/b*r/t*t, /*oo/b*r/t*s*, /*oo/b*r/te*, /*oo/b*r/*t, /*oo/b*r/*s*, /*oo/b*r/*e*, /*oo/b*r/t*, /*oo/b*r/*, /*oo/ba*/test, /*oo/ba*/*est, /*oo/ba*/t*st, /*oo/ba*/te*t, /*oo/ba*/tes*, /*oo/ba*/*st, /*oo/ba*/*e*t, /*oo/ba*/*es*, /*oo/ba*/t*t, /*oo/ba*/t*s*, /*oo/ba*/te*, /*oo/ba*/*t, /*oo/ba*/*s*, /*oo/ba*/*e*, /*oo/ba*/t*, /*oo/ba*/*, /*oo/*r/test, /*oo/*r/*est, /*oo/*r/t*st, /*oo/*r/te*t, /*oo/*r/tes*, /*oo/*r/*st, /*oo/*r/*e*t, /*oo/*r/*es*, /*oo/*r/t*t, /*oo/*r/t*s*, /*oo/*r/te*, /*oo/*r/*t, /*oo/*r/*s*, /*oo/*r/*e*, /*oo/*r/t*, /*oo/*r/*, /*oo/*a*/test, /*oo/*a*/*est, /*oo/*a*/t*st, /*oo/*a*/te*t, /*oo/*a*/tes*, /*oo/*a*/*st, /*oo/*a*/*e*t, /*oo/*a*/*es*, /*oo/*a*/t*t, /*oo/*a*/t*s*, /*oo/*a*/te*, /*oo/*a*/*t, /*oo/*a*/*s*, /*oo/*a*/*e*, /*oo/*a*/t*, /*oo/*a*/*, /*oo/b*/test, /*oo/b*/*est, /*oo/b*/t*st, /*oo/b*/te*t, /*oo/b*/tes*, /*oo/b*/*st, /*oo/b*/*e*t, /*oo/b*/*es*, /*oo/b*/t*t, /*oo/b*/t*s*, /*oo/b*/te*, /*oo/b*/*t, /*oo/b*/*s*, /*oo/b*/*e*, /*oo/b*/t*, /*oo/b*/*, /*oo/*/test, /*oo/*/*est, /*oo/*/t*st, /*oo/*/te*t, /*oo/*/tes*, /*oo/*/*st, /*oo/*/*e*t, /*oo/*/*es*, /*oo/*/t*t, /*oo/*/t*s*, /*oo/*/te*, /*oo/*/*t, /*oo/*/*s*, /*oo/*/*e*, /*oo/*/t*, /*oo/*/*, /f*o/bar/test, /f*o/bar/*est, /f*o/bar/t*st, /f*o/bar/te*t, /f*o/bar/tes*, /f*o/bar/*st, /f*o/bar/*e*t, /f*o/bar/*es*, /f*o/bar/t*t, /f*o/bar/t*s*, /f*o/bar/te*, /f*o/bar/*t, /f*o/bar/*s*, /f*o/bar/*e*, /f*o/bar/t*, /f*o/bar/*, /f*o/*ar/test, /f*o/*ar/*est, /f*o/*ar/t*st, /f*o/*ar/te*t, /f*o/*ar/tes*, /f*o/*ar/*st, /f*o/*ar/*e*t, /f*o/*ar/*es*, /f*o/*ar/t*t, /f*o/*ar/t*s*, /f*o/*ar/te*, /f*o/*ar/*t, /f*o/*ar/*s*, /f*o/*ar/*e*, /f*o/*ar/t*, /f*o/*ar/*, /f*o/b*r/test, /f*o/b*r/*est, /f*o/b*r/t*st, /f*o/b*r/te*t, /f*o/b*r/tes*, /f*o/b*r/*st, /f*o/b*r/*e*t, /f*o/b*r/*es*, /f*o/b*r/t*t, /f*o/b*r/t*s*, /f*o/b*r/te*, /f*o/b*r/*t, /f*o/b*r/*s*, /f*o/b*r/*e*, /f*o/b*r/t*, /f*o/b*r/*, /f*o/ba*/test, /f*o/ba*/*est, /f*o/ba*/t*st, /f*o/ba*/te*t, /f*o/ba*/tes*, /f*o/ba*/*st, /f*o/ba*/*e*t, /f*o/ba*/*es*, /f*o/ba*/t*t, /f*o/ba*/t*s*, /f*o/ba*/te*, /f*o/ba*/*t, /f*o/ba*/*s*, /f*o/ba*/*e*, /f*o/ba*/t*, /f*o/ba*/*, /f*o/*r/test, /f*o/*r/*est, /f*o/*r/t*st, /f*o/*r/te*t, /f*o/*r/tes*, /f*o/*r/*st, /f*o/*r/*e*t, /f*o/*r/*es*, /f*o/*r/t*t, /f*o/*r/t*s*, /f*o/*r/te*, /f*o/*r/*t, /f*o/*r/*s*, /f*o/*r/*e*, /f*o/*r/t*, /f*o/*r/*, /f*o/*a*/test, /f*o/*a*/*est, /f*o/*a*/t*st, /f*o/*a*/te*t, /f*o/*a*/tes*, /f*o/*a*/*st, /f*o/*a*/*e*t, /f*o/*a*/*es*, /f*o/*a*/t*t, /f*o/*a*/t*s*, /f*o/*a*/te*, /f*o/*a*/*t, /f*o/*a*/*s*, /f*o/*a*/*e*, /f*o/*a*/t*, /f*o/*a*/*, /f*o/b*/test, /f*o/b*/*est, /f*o/b*/t*st, /f*o/b*/te*t, /f*o/b*/tes*, /f*o/b*/*st, /f*o/b*/*e*t, /f*o/b*/*es*, /f*o/b*/t*t, /f*o/b*/t*s*, /f*o/b*/te*, /f*o/b*/*t, /f*o/b*/*s*, /f*o/b*/*e*, /f*o/b*/t*, /f*o/b*/*, /f*o/*/test, /f*o/*/*est, /f*o/*/t*st, /f*o/*/te*t, /f*o/*/tes*, /f*o/*/*st, /f*o/*/*e*t, /f*o/*/*es*, /f*o/*/t*t, /f*o/*/t*s*, /f*o/*/te*, /f*o/*/*t, /f*o/*/*s*, /f*o/*/*e*, /f*o/*/t*, /f*o/*/*, /fo*/bar/test, /fo*/bar/*est, /fo*/bar/t*st, /fo*/bar/te*t, /fo*/bar/tes*, /fo*/bar/*st, /fo*/bar/*e*t, /fo*/bar/*es*, /fo*/bar/t*t, /fo*/bar/t*s*, /fo*/bar/te*, /fo*/bar/*t, /fo*/bar/*s*, /fo*/bar/*e*, /fo*/bar/t*, /fo*/bar/*, /fo*/*ar/test, /fo*/*ar/*est, /fo*/*ar/t*st, /fo*/*ar/te*t, /fo*/*ar/tes*, /fo*/*ar/*st, /fo*/*ar/*e*t, /fo*/*ar/*es*, /fo*/*ar/t*t, /fo*/*ar/t*s*, /fo*/*ar/te*, /fo*/*ar/*t, /fo*/*ar/*s*, /fo*/*ar/*e*, /fo*/*ar/t*, /fo*/*ar/*, /fo*/b*r/test, /fo*/b*r/*est, /fo*/b*r/t*st, /fo*/b*r/te*t, /fo*/b*r/tes*, /fo*/b*r/*st, /fo*/b*r/*e*t, /fo*/b*r/*es*, /fo*/b*r/t*t, /fo*/b*r/t*s*, /fo*/b*r/te*, /fo*/b*r/*t, /fo*/b*r/*s*, /fo*/b*r/*e*, /fo*/b*r/t*, /fo*/b*r/*, 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3) Then I loop over the items in this list above and validate if it only matches a single file-path in the input array of file-paths. (I do this by checking two things: are the amount of slashes the same, and does it match the regex where every * is replaced with .*.)
If it does: keep the (first) shortest, which we return in the end.

  • \$\begingroup\$ What's >>>? I know >> is bitwise right shift. \$\endgroup\$
    – Pavel
    Feb 23, 2018 at 15:22
  • 2
    \$\begingroup\$ @Pavel For positive integers, >>> acts the same as >>. But for negative integers it changes the parity bit to 0 (you can see some examples here at the section ">> vs >>>"). -1>>>1 is just a shorter variant of Integer.MAX_VALUE (and 1<<31 would be Integer.MIN_VALUE). \$\endgroup\$ Feb 23, 2018 at 15:27

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