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Balanced brackets conditional issue

I've been looking into Ruby and felt I was learning quite a bit. I'm currently trying to solve the balanced brackets algorithm but struggling with a condition. Here's what I have:

def balanced?(list_of_brackets)
    if list_of_brackets.length % 2 != 0
        false
    else
        stack = []
        bracket_sets = {
            '{' => '}', 
            '[' => ']', 
            '(' => ')'
        }
        list_of_brackets.chars do |bracket|
            if bracket == "{" or "[" or "("
                puts "#{bracket} is an opening bracket"
            else
                puts "#{bracket} is a closing bracket"
            end
        end
    end
    stack.empty?
end

puts balanced?('{}()[]')

The result I get is:

{ is an opening bracket
} is an opening bracket
( is an opening bracket
) is an opening bracket
[ is an opening bracket
] is an opening bracket

The closing brackets are somehow getting through the first condition. I'm missing something here but I can't spot it. Maybe another set of eyes can help me here. I'd appreciate any help and/or advice!

over 4 years ago · Santiago Trujillo
4 Respuestas
Responde la pregunta

0

This

if bracket == "{" or "[" or "("

is basically checking if bracket is equal to "{" or if "[" is truthy or "(" is truthy, you missed the bracket == for the rest of them. That's why they take the if branch and you get all of them are opening brackets. The solution for that problem is to compare every of them:

if bracket == "{" || bracket == "[" || bracket == "("
  ...

PS: make sure to not confuse OR with ||, they have a different procedence.

over 4 years ago · Santiago Trujillo Denunciar

0

The if statement if bracket == "{" or "[" or "(" needs to have a bracket == X for each or condition in the statement. Change:

if bracket == "{" or "[" or "("

to

if bracket == "{" or bracket == "[" or bracket ==  "("

...and that should work. The full code would be:

def balanced?(list_of_brackets)
    if list_of_brackets.length % 2 != 0
        false
    else
        stack = []
        bracket_sets = {
            '{' => '}', 
            '[' => ']', 
            '(' => ')'
        }
        list_of_brackets.chars do |bracket|
            if bracket == "{" or bracket == "[" or bracket ==  "("
                puts "#{bracket} is an opening bracket"
            else
                puts "#{bracket} is a closing bracket"
            end
        end
    end
    stack.empty?
end

puts balanced?('{}()[]')
over 4 years ago · Santiago Trujillo Denunciar

0

I will use the term enclosures to refer to parentheses ('(' and ')'), brackets ('[' and ']') and braces ('{' and '}'). A string of enclosures is balanced if:

  • each left parenthesis has a matching right parenthesis, and vice-versa;
  • each left bracket has a matching right bracket, and vice-versa;
  • each left brace has a matching right brace, and vice-versa; and
  • all matched pairs are well-nested in the sense every each matched pair must be separated by a string comprised of zero or more matched pairs.

For example, '([{}])' is well-nested whereas '([{]})' is not, as the string separating the matched pair, '[' and ]', namely '{', is not a matched pair.

If, in the code in the question, the length of the string is even stack is initialized to an empty array that is never changed, so the return value stack.empty? will always be true. For example, balanced?(')[') #=> true. You need something like the following.

RIGHT_ENCLOSURE_PAIRS = { '}'=>'{', ']'=>'[', ')'=>'(' }
RIGHT_ENCLOSURES = RIGHT_ENCLOSURE_PAIRS.keys
def balanced?(str)
  stack = []
  str.each_char do |c|
    if RIGHT_ENCLOSURES.include?(c)
      return false if stack.empty? || stack.last != RIGHT_ENCLOSURE_PAIRS[c]
      stack.pop
    else
      stack << c
    end
  end
  stack.empty?
end
balanced? '{}()[]'             #=> true
balanced? '([{}])'             #=> true
balanced? '([{}]{})'           #=> true
balanced? '((([[[{{{}}}]]])))' #=> true
balanced? '{[]'                #=> false
balanced? '{{(([['             #=> false
balanced? '{()[{]}(([]))'      #=> false

I initially had the following guard clause as the first line of the method.

return false if str.length.odd?

After reflection, however, I removed it, as the time spent executing it unnecessarily for strings with even numbers of enclosures was probably much greater than the time saved for strings with odd numbers of enclosures.

I have not included puts statements such as puts "#{c} is an opening enclosure" as they would normally only be used for debugging, but they could of course be added if desired, in which case it might be more helpful to write

ENCLOSURE_TYPE = { '('=>'left parenthesis', ')'=>'right parenthesis',
                   '['=>'left bracket', ']'=>'right bracket',
                   '{'=>'left brace', '}'=>'right brace' }
puts "#{c} is a #{ENCLOSURE_TYPE[c]}"
over 4 years ago · Santiago Trujillo Denunciar

0

The syntax issue was addressed by others, but if you want to get rid of the redundant comparisons to bracket, here’s another more streamlined approach you might want to consider:

["{", "[" , "("].include? bracket

This places all your desired strings into a single array and then tests to see if bracket is included in that array. Or you could even take your existing bracket_sets hash and test to to see if your bracket is included in the keys of that group...

bracket_sets.key?(bracket)
over 4 years ago · Santiago Trujillo Denunciar
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