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

PHP 75 bytes

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

Sample input/output:

$ echo st Hello st st Bye | php string-loops.php
st He2[l]o2[ st] Bye

$ echo Bookkeeper | php string-loops.php
B2[o]2[k]2[e]per

$ echo /\/\/\\\\\\\\/\/\/ | php string-loops.php
3[/\]2[\\\]3[\/]

Perl 46 bytes

Same code in perl:

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

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

Javascript

f(s){return s.indexOf("loops")>=0}