Is there an unmatched parenthesis in this String?

Question 1

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.

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!

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!

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.

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

Winning Criterion

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

Note

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

Question 2

I'm 99% sure this is a dupe but I can't find the older one.

Question 3

@StephenS This one is the one that came to my mind, but I don't think it's quite a dupe...

Question 4

Possible duplicate of Are the brackets fully matched?

Question 5

@StephenS I think the fact that this is entirely orderless makes it quite different. Neither ][ or ([)] would be valid in the other challenge but are valid here.

Question 6

This question's title should really make it clearer that order is irrelevant. That's not what most people think about when talking about unmatched parentheses.

Question 7

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,')]}')

Question 8

05AB1E, (削除) 14 (削除ここまで) 10 bytes

žuS2ドルô€ËP_

Try it online!

-4 thanks to Adnan.

1 for truthy, 0 for falsey.

Its main power point is the žu builtin.

Explanation:

žuS2ドルô€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

Question 9

Who doesn't have a command that pushes ()<>[]{}?

Question 10

@cairdcoinheringaahing CJam doesn't, but you can get it with [{(<>)}]` .

Question 11

@Challenger5 You just correctly ordered the brackets so that the (<>) won't affect the stack since it's in a code block, and then the codeblock goes into the list? Well, this challenge doesn't have <>, so you could just use [{()}]` or {[()]}` or even {([])}`. Oh wait a sec, you can actually use {()}a`.

Question 12

@EriktheOutgolfer Yeah, I forgot about that.

Question 13

@Arjun They're not handled, since they'll never be in the input.

Question 14

Python 2, 61 bytes

Saved 4 bytes thanks to WheatWizard!

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

Try it online!

Question 15

This can be made slightly shorter by ditching r. Try it online

Question 16

Thanks! I changed it to match the two characters into separate variables, removing the indexing and therefore saving 4 more bytes.

Question 17

Python 3, (削除) 62 (削除ここまで) 61 bytes

Surprisingly pythonic for codegolf.

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

-1 byte thanks to @musicman523

Question 18

You can use musicman's trick to make this shorter. TIO

Question 19

And now we have the same answers :) Although your post came before my edit, I did come up with it independently

Question 20

Yeah, I'll leave it as is.

Question 21

You could change == to - and keep all, then we have the same byte count but different answers

Question 22

@musicman523 Alright then!

Question 23

Jelly, 17 bytes

ċÐ€"[]{}()"s2E€ẠṆ

Try it online!

1 is truthy, 0 is falsey.

Question 24

MATL, 15 bytes

!'[](){}'=sHeda

This outputs 1 if unmatched, 0 if matched.

Try it online!

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

Question 25

Javascript, 71 bytes:

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

Question 26

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]]}

Try it online!

Question 27

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.

Question 28

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,']'))

Try it online!

Question 29

All instances of != can be -.

Question 30

This answer has already been out-golfed by a far better answer. Hence, I'd rather not try to golf this.

Question 31

PowerShell, (削除) 95 (削除ここまで) 86 bytes

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

Try it online!

Previous version

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

Try it online!

Question 32

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.

Question 33

Please check on {).

Question 34

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ドル",$a,$a],$p,1,$c);echo$t&!$p;

Try it online!

Question 35

Visual Basic.Net, (削除) 218 (削除ここまで) 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

Question 36

"around 218 bytes" I count 217. Also, if (expr) then return false else return true can be return not (expr).

Question 37

"around 218 bytes" I count 192. Do you have a trailing newline?

Question 38

I dont know how to put line through 218

Question 39

<s>218</s> works.

Question 40

well I don't have trailing newline

Question 41

Clojure, 65 bytes

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

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

Question 42

Retina, 26 bytes

O`.
+`\(\)|\[]|{}
[^*-9^]

Try it online!

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.

Question 43

{{{{}} is matched? Your code return 0 on that input.

Question 44

@user202729 Thanks, it should be fixed now (luckily at the same byte count).

Question 45

Javascript, (削除) 149 (削除ここまで) 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}"));

Question 46

I can save 3 bytes by making every == into a != And changing && to || but I'm on mobile

Question 47

I think, bitwise operators instead of logical ones will do fine. Like & instead of && and | instead of ||.

Question 48

You can remove the outermost paren after return !.

Question 49

Do you even need a space after return?

Question 50

I think you need that space otherwise JavaScript will count it as rerurnx and throw an error

xnor xnor 149k26 gold badges286 silver badges673 bronze badges · Accepted Answer · 2017-06-04 03:39:22Z

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,')]}')

Stack Exchange Network

Is there an unmatched parenthesis in this String?

Introduction

Input

Output

Rules

Test Cases

Winning Criterion

Note

17 Answers 17

Python 2, 47 bytes

05AB1E, (削除) 14 (削除ここまで) 10 bytes

Python 2, 61 bytes

Python 3, (削除) 62 (削除ここまで) 61 bytes

Jelly, 17 bytes

MATL, 15 bytes

Explanation

Javascript, 71 bytes:

Ruby, 58+1 = 59 bytes

Javascript 57 bytes

Python 2, 85 bytes

PowerShell, (削除) 95 (削除ここまで) 86 bytes

C#, 87 bytes

PHP>7.1, 124 Bytes

PHP, 151 bytes

Visual Basic.Net, (削除) 218 (削除ここまで) 192 Bytes

Clojure, 65 bytes

Retina, 26 bytes

Explanation

Javascript, (削除) 149 (削除ここまで) 140 bytes

OLD SOLUTION:

Your Answer

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Hot Network Questions

Is there an unmatched parenthesis in this String?

Introduction

Input

Output

Rules

Test Cases

Winning Criterion

Note

17 Answers 17

Python 2, 47 bytes

05AB1E, (削除) 14 (削除ここまで) 10 bytes

Python 2, 61 bytes

Python 3, (削除) 62 (削除ここまで) 61 bytes

Jelly, 17 bytes

MATL, 15 bytes

Explanation

Javascript, 71 bytes:

Ruby, 58+1 = 59 bytes

Javascript 57 bytes

Python 2, 85 bytes

PowerShell, (削除) 95 (削除ここまで) 86 bytes

C#, 87 bytes

PHP>7.1, 124 Bytes

PHP, 151 bytes

Visual Basic.Net, (削除) 218 (削除ここまで) 192 Bytes

Clojure, 65 bytes

Retina, 26 bytes

Explanation

Javascript, (削除) 149 (削除ここまで) 140 bytes

OLD SOLUTION:

Your Answer

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