Rotate brackets until they're balanced

Question

Take the string of brackets ]][][[. When you rotate it to the right once, you get []][][. If you rotate it again, you get [[]][]. All brackets in this string are balanced.

The Task:

Your program (or function) will be given a string of brackets, represented in any reasonable format (including using other things in place of the brackets, like -1 and 1). The numbers of opening and closing brackets will always be equal, so [ or ][] won't be given as inputs.

Output should be a rotation of those brackets which is balanced. You can check this by repeatedly removing pairs of brackets ([]). With a balanced string of brackets, none will be left over.

Rotating a string to the right involves taking the last character, and moving it to the beginning. For example, 01234567 rotated right 3 times would be 56701234. The direction of rotation doesn't matter, but no brackets should be added, discarded, mirrored, etc. If multiple solutions are possible, such as [][[]] or [[]][], you can return any of them.

Test Cases:

[]          ->  []
]][[        ->  [[]]
[][]][      ->  [[][]]
][][[]      ->  [[]][] OR [][[]]
[[[][][]]]  ->  [[[][][]]]
]]][][][[[  ->  [[[]]][][] OR [][[[]]][] OR [][][[[]]]

Other:

This is code-golf, so shortest answer in bytes per language wins!

Answer 1

Charcoal, 25 bytes

≔Ｅθ⁻№…θκ[№…θκ]η⭆θ§θ⁺κ⌕η⌊η

Try it online! Link is to verbose version of code. Explanation:

≔Ｅθ⁻№…θκ[№…θκ]η

Calculate the imbalance between [s and ]s for every prefix.

⭆θ§θ⁺κ⌕η⌊η

Switch the prefix with the greatest imbalance with its suffix.

Answer 2

Excel, 131 bytes

=LET(s,LEN(A1),x,SEQUENCE(s),y,INDEX(SORTBY(MOD(x,s),MMULT((x>=TRANSPOSE(x))*1,(MID(A1,x,1)="[")*2-1)),1),MID(A1,y+1,s)&LEFT(A1,y))

Converts brackets to 1, -1; finds the minimum cumulative sum y; then moves the first y characters to the end.

Answer 3

Python 3, 123 bytes

[n:=range(l:=len(s)),next(r for i in n if [r:=(2*s)[i:i+l],min(sum("]-[".index(c)-1 for c in r[:p]) for p in n)][1]>=0)][1]

Takes in string of square brackets. Should return first valid rotation:

>>> s="]][][["
>>> [n:=range(l:=len(s)),next(r for i in n if [r:=(2*s)[i:i+l],min(sum("]-[".index(c)-1 for c in r[:p]) for p in n)][1]>=0)][1]
'[][[]]'

Answer 4

Japt v2.0a0, 11 8 bytes

Uses 1 for [ & 0 for ].

ÈeA}f@éY

Try it - header converts input from brackets and footer converts back to brackets.

ÈeA}f@éY     :Implicit input of string U
È            :Left function taking a string as argument
 e           :Recursively replace
  A          : "10"
   }         :End function
    f        :Return the first result of the right function that returns falsey (empty string) when passed through the left function
     @       :Right function with argument Y, 0-based Iteration index
      éY     :  Rotate U right Y times