Is there an unmatched parenthesis in this String?

Question

Introduction

Given a String containing an arithmetic expression, your task is to output a truthy or falsey value based on whether it contains unmatched parentheses.

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Input

Your program should take in a String containing an arithmetic expression. It may take input in any way except assuming it to be present in a pre-defined variable. Reading from file, input box, modal dialog box etc. is fine. Taking input as function argument is allowed as well!

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Output

Your program should output a truthy or falsey value based on whether the input String contains any unmatched parentheses. It may output in any way except except writing to a variable. Writing to file, screen, console, terminal etc. is allowed. Outputting with function return is allowed as well!

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Rules

• For the purpose of this challenge, a parentheses is defined as any one of [, {, (, ), }, ]. A parentheses is said to be unmatched if it does not have any corresponding opening/closing parentheses. For example, [1+3 does contain unmatched parentheses.

• Ordering of parentheses does not matter in this challenge. So, ]1+3[ does not contain any unmatched parentheses. {[)(}] doesn't, as well.

• The input String will only contain following characters : {, [, (, ), ], }, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, +, -, *, / and ^.

• Your program should output a truthy value if the input String contains an unmatched parentheses.

• Your program should output a falsey value if the input String does not contain an unmatched parentheses.

• You must specify the truthy and falsey values in your post.

• The input String will never be empty.

• If the input String does not contain any parentheses, your program should output falsey value.

• Standard loopholes apply.

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Test Cases

Input            ->            Output

"{"                            Truthy
"{{{{}}"                       Truthy
"{[]}"                         Falsey
"[]]["                         Falsey
")()()()("                     Falsey
"1+4"                          Falsey
"{)"                           Truthy
"([)]"                         Falsey
"(()"                          Truthy
"2*{10+4^3+(10+3)"             Truthy
"-6+[18-{4*(9-6)}/3]"          Falsey
"-6+[18-{4*)9-6](/3}"          Falsey

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Winning Criterion

This is code-golf, so the shortest code in bytes wins!

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Note

I'll be adding a similar challenge but in which order will matter, after some time! Stay Tuned!

Answer 1

05AB1E, 14 10 bytes

žuS¢2ô€ËP_

Try it online!

-4 thanks to Adnan.

1 for truthy, 0 for falsey.

Its main power point is the žu builtin.

Explanation:

žuS¢2ô€ËP_
žu         Push '()<>[]{}'
  S        Split into individual chars
   ¢       Count occurrences of each char in the input. For <>, the count will always be 0
    2      Push 2
     ô     Split the array into pieces of that length, corresponding to respective matching brackets
      €    Map next command over the chunks
       Ë   Check if all elements are equal. If equal, it means that the corresponding bracket type is matched. <> will always be matched, since they will never occur in the input, so they'll both always have 0 occurrences
        P  Take the product of the boolean values. <> won't affect this, since it would always be 1
         _ Logically negate, convert 1 to 0 and non-1 (always 0 in this case) to 1

Answer 2

Python 2, 47 bytes

lambda s:map(s.count,'([{')!=map(s.count,')]}')

Try it online! Test cases from musicman523.

Checks if list of counts for each type of open paren matches that for the corresponding close parens.

The program is one byte longer

q=input().count
print map(q,'([{')!=map(q,')]}')

Answer 3

Python 2, 61 bytes

Saved 4 bytes thanks to WheatWizard!

lambda s:any(s.count(i)^s.count(j)for i,j in['{}','[]','()'])

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Answer 4

Jelly, 17 bytes

ċЀ“[]{}()”s2E€ẠṆ

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1 is truthy, 0 is falsey.

Answer 5

Python 3, 62 61 bytes

Surprisingly pythonic for codegolf.

lambda x:all(x.count(y)-x.count(z)for y,z in("[]","()","{}"))

-1 byte thanks to @musicman523

Answer 6

MATL, 15 bytes

!'[](){}'=sHeda

This outputs 1 if unmatched, 0 if matched.

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Explanation

!          % Implicitly input string. Tranpose into vertical vector of chars
'[](){}'   % Push this string
=          % Compare for equalty, with broadcast
s          % Sum of each column. This gives the count of each char of '[](){}'
He         % Reshape as a two-row column (in column-major order)
d          % Difference of the two entries of each column
a          % Any: gives true (1) if some entry is nonzero, false (0) otherwise
           % Implicitly display

Answer 7

Javascript, 71 bytes:

x=>['()','[]','{}'].some(a=>x.split(a[0]).length!=x.split(a[1]).length)

Answer 8

Ruby, 58+1 = 59 bytes

+1 byte for the -n flag.

f=->c{$_.count c}
p %w"() [] {}".any?{|s|f[s[0]]!=f[s[1]]}

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Answer 9

Javascript 57 bytes

x=>(g=z=>x.split(z).length,g`(`-g`)`|g`[`-g`]`|g`{`-g`}`)

Starting from asgallant's solution.

Matching returns 0. Non-matching returns non-zero.

Answer 10

Python 2, 85 bytes

I can definitely golf this using subtraction instead but I'm running out of time. D:

lambda s,c=str.count:(c(s,'(')!=c(s,')'))or(c(s,'{')!=c(s,'}'))or(c(s,'[')!=c(s,']'))

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Answer 11

Retina, 26 bytes

O`.
+`\(\)|\[]|{}

[^*-9^]

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Explanation

O`.

Sort all characters so that corresponding brackets are adjacent (since neither \ nor | can appear in the input).

+`\(\)|\[]|{}

Repeatedly remove pairs of matching brackets.

[^*-9^]

Try to match any remaining brackets (the regex matches anything that isn't a ^ or in the ASCII range from * to 9 which, among the valid input characters, are only the 6 brackets). If we find any, that means that there was an unmatched one. The result will be a positive integer if the input was unbalanced, 0 otherwise.

Answer 12

PowerShell, 95 86 bytes

param($s)$h=@{};[char[]]$s|%{$h["$_"]++};$h.'['-$h.']'-or$h.'('-$h.')'-or$h.'{'-$h.'}'

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Previous version

param($s)switch([char[]]$s){'('{$k++}')'{$k--}'['{$l++}']'{$l--}'{'{$m++}'}'{$m--}}$k-or$l-or$m

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Answer 13

Javascript, 149 140 bytes

New solution incorporates bitwise operators (Thanks, arjun) and calculates if they are not equal, and uses or instead of and.

function f(x){return x.split("(").length!=x.split(")").length|x.split("{").length!=x.split("}").lengt|x.split("[").length!=x.split("]").length}

I don't yet have a snippet for this, so I'm not sure it works yet.

OLD SOLUTION:

function f(x){return !(x.split("(").length==x.split(")").length&&x.split("{").length==x.split("}").length&&x.split("[").length==x.split("]").length)}

Checks if there are the same amount of one bracket and another, and then returns the opposite.

function f(x){return !(x.split("(").length==x.split(")").length&&x.split("{").length==x.split("}").length&&x.split("[").length==x.split("]").length)}

/*"{"                            Truthy
"{[]}"                         Falsey
"[]]["                         Falsey
")()()()("                     Falsey
"1+4"                          Falsey
"{)"                           Truthy
"([)]"                         Falsey
"2*{10+4^3+(10+3)"             Truthy
"-6+[18-{4*(9-6)}/3]"          Falsey
"-6+[18-{4*)9-6](/3}"          Falsey*/

console.log(f("{"));
console.log(f("{[]}"));
console.log(f("[]]["));
console.log(f(")()()()("));
console.log(f("1+4"));
console.log(f("{)"));
console.log(f("([)]"));
console.log(f("2*{10+4^3+(10+3)"));
console.log(f("-6+[18-{4*(9-6)}/3]"));
console.log(f("-6+[18-{4*)9-6](/3}"));

Answer 14

C#, 87 bytes

using System.Linq;s=>s.Count(c=>c=='['|c=='{'|c=='(')==s.Count(c=>c==']'|c=='}'|c==')')

Pretty self explanatory, just test to see if the of left parenthesis equals the count of right parenthesis.

Answer 15

Visual Basic.Net, 218 192 Bytes

function t(m as string,v as string)
return Len(m)-Len(m.Replace(v,""))
end function
function b(q as String)
return not(t(q,"{")=t(q,"}")and t(q,"[")=t(q,"]")and t(q,"(")=t(q,")"))
end function

Answer 16

PHP>7.1, 124 Bytes

$t=!!$p=preg_replace("#[\d+/*^-]#","",$argn);for(;$p;$p=substr($p,1,-1))$t*=chr(($o=ord($p[0]))%5?$o+2:$o+1)==$p[-1];echo$t;

Instead of #[\d+/*^-] you can use [^]{()}[] or [^\p{Pe}\p{Ps}]

IntlChar::charMirror IntlChar::charMirror($p[0])==$p[-1] instead of chr(($o=ord($p[0]))%5?$o+2:$o+1)==$p[-1] checks not if the char is an openining or closing punctation so we can not use it without additional check &$p[0]<$p[-1] +8 Bytes

Online Version

PHP, 151 bytes

A full Regex Solution

for($c=$t=!!$p=($r=preg_replace)("#[^]{()}[]#","",$argn);$p&&$c;)$p=$r(["#^{(.*)}$#","#^\[(.*)]$#","#^\((.*)\)$#"],[$a="$1",$a,$a],$p,1,$c);echo$t&!$p;

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Answer 17

Clojure, 65 bytes

 #(let[f(frequencies %)](some not(map =(map f"[{(")(map f"]})"))))

So long keywords :/ Returns nil for falsy and true for truthy.