
"Brackets" Review
Welcome back for another weekly review series!
When you open first "Electronic Station" the first mission that you encounter is the classic "Brackets" mission. This is a simple, but interesting and useful mission. "Brackets" can be solved with a wide variety of methods, and CiO players have certainly come up with some clever and ingenious solutions to it.
Description
You are given an expression with numbers, brackets and operators. For this task only the brackets matter. Brackets come in three flavors: "{}" "()" or "[]". Brackets are used to determine scope or to restrict some expression. If a bracket is open, then it must be closed with a closing bracket of the same type. The scope of a bracket must not intersected by another bracket. In this task you should make a decision, whether to correct an expression or not based on the brackets. Do not worry about operators and operands.
Stack Solution
I've solved this mission myself, using a stack. Iterate a given text symbol by symbol.
 If a character is not a bracket then skip it,
 If a character is an opened bracket then put in the stack,
 If a character is a closed bracket, then try to take the top element of the stack and check whether they form a pair of identical brackets: "()", "[]" or "{}".
If we try to take an element from the stack and it's empty, or after checking all symbols the stack is not an empty, then our expression is wrong.
stack = # Yes, we can use dequepairs =for token in data:if token in :elif token in :if stack and token == :else:return Falsereturn not
You can examine this code execution with the Pythontutor service
Stackless Solution
This mission can be solved without using stack and instead with a string method as seen in @blabaster's "Stackless" solution
s = '.while s:s0, s = s, ..if s == s0:return Falsereturn True
As the first solution, we have to "clear" extra text and keep only the brackets. So a string like "{[(3+1)+2]+}+()" become to "{[()]}()". Next we remove all pairs of brackets which are close to each other like "()", "[]" or "{}". If for some step we don't remove anything and the string is not an empty, then the given text is wrong. You can examine this code with Pythontutor  "stackless".
Here's a short, slightly less readable and recursive version of the algorithm in @coells's solution.
import reEXPR =checkio = True if e == ' else False if e == f else
"Now They Have Two Problems"
How about regexp  @LuigiMoro wrote an almost regexp solution. He prepared three regexp expressions and checks if any of it matches, then returns False if they do. But before this we check that the quantity of opened brackets are equal to the closed for each type.
import repattern_1 =pattern_2 =pattern_3 =if ( != or!= or!= ) :return Falsereturn ( or or ) is None
Fython
I'm not sure why @astynax84 placed "Fython" in technically oneliner solution. Yes, this is a oneliner, but it's been indented for "readability".
checkio === (True, None)
And with this mind twisted solution I say goodbye for now  That's all I've got on brackets. If you need any more, go to the hardware store.
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