Detecting loops in in in strings

Question

Given a string, mark which parts in the string repeat in box brackets, and write how many times it does so before the box.

Example:st st st Hello ThereThere
st(with a space after) repeats 3 times, There repeats 2 times, and l repeats 2 times in Hello, so the output will be:
3[st ]He2[l]o 2[There]
There is no space between [st ] and Hello because the space in the string is being taken up by the repeating part. [st ], 3 times with a Hello after it would be st st st Hello, which is what the string has.

Other examples:
st Hello st st Bye -> st He2[l]o 2[st ]Bye
Bookkeeper -> B2[o]2[k]2[e]per
/\/\/\\\\\\\\/\/\/ -> 3[/\]6[\]3[\/]

Shortestest code wins.

EDIT: To clarify the choice of selections, choose the largest possible repeating section, and try and leave as little non-repeating characters as possible. Do not nest boxed-in sections.

Answer 1

Perl, 36 bytes

#!perl -p
s!(.+)\1+!@{[$&=~/\Q$1/g]}."[$1]"!eg

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---

PHP 5.6, 68 bytes

Same code in PHP. Requires the command line option -F, and relies on deprecated /e modifier.

<?=preg_filter('/(.+)\1+/e','substr_count("\0","\1")."[\1]"',$argn);

Answer 2

scala, 102 chars

print("(?!\\s)(.+)\\1+".r.replaceAllIn(readLine,m=>m.matched.size/m.group(1).size+"["+m.group(1)+"]"))

Output

st st st Hello ThereThere ->    3[st ]He2[l]o 2[There]
st Hello st st Bye        ->    st He2[l]o 2[st ]Bye
Bookkeeper                ->    B2[o]2[k]2[e]per
/\/\/\\\\\\\\/\/\/        ->    3[/\]2[\\\]3[\/]

Answer 3

C, 224 205 203 199 195 193 chars

The string is given as program parameter. As usual, use double quotes if your string contains spaces, and for empty strings - calling the program without any parameter crashes, but calling with "" works.

Output for test cases is:

st st st Hello ThereThere => 3[st ]He2[l]o 2[There]
Bookkeeper                => B2[o]2[k]2[e]per
/\/\/\\\\\\\\/\/\/        => 3[/\]2[\\\]3[\/]

which looks correct to me regarding "choose the largest possible repeating section".

Code:

l,r,f,i;char*c;main(a,b)char**b;{l=strlen(c=b[1]);for(;putchar(*c);c++)for(a=l;--a;f=0)for(r=a;!f;r+=a){for(i=0;i<a;)f|=r>l||c[i]-c[r+i++];if(f*r>a)c[a]=0,printf("\b%i[%s]",r/a,c),c+=r-1,a=1;}}

Answer 4

Retina, 40 37 24 bytes

(.+)(\1)+
$.($#2*__)[$1]

-13 bytes thanks to @Neil.

Try it online or verify all test cases. (NOTE: Slightly different results for some test cases in comparison to the challenge description, but still the same as the accepted answer, so I assumed both are valid.)

Explanation:

(.+)(\1)+       # Look for the repeated sections (where the repeated string is captured in
                # group 1, and every time it's repeated in group 2)

Example: in the input "st st st Hello ThereThere" it will find the capture group 1 and 2 as: "st " & ["st ","st "]; "l" & ["l"]; and "There" & ["There"] respectively.

   $#2          # Get the amount of times group 2 was repeated
      *_        # Converted to unary (that amount of '_')
        _       # + 1 (with an additional '_' added)
$.(      )      # Convert that unary back to a number again by taking the length
          [$1]  # And append the match of the first capture group between block-quotes
                # after the number

Example: "st st st " will become "3[st ]"; "ll" will becomes "2[l]"; and "ThereThere" will become "2[There]". So "st st st Hello ThereThere" is now the intended result "3[st ]He2[l]o 2[There]" (which is output implicitly).

Answer 5

Perl 5 -p, 41 bytes

s|(.+)\1+|(length($&)/length$1)."[$1]"|ge

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Answer 6

PowerShell, 82 bytes

[regex]::Replace($args,'(.+)\1+',{$a,$g=$args|% gr*
"$($a.Length/$g.Length)[$g]"})

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