Iterating the same method on a string

Question

I have a string and I want to iterate the .replace(re,'') until I achieve and empty sting "". How to do this rather than repeating over and over again the same method

var isValid = (s)=> {
   const re = /(\(\))|(\[\])|(\{\})/gi
   return s.replace(re,'').replace(re,'').replace(re,'').replace(re,'') !='' ? false :true
  
};

console.log(isValid("[({(())}[()])]"))

Would [replaceAll](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replaceAll) help here? — evolutionxbox, Jan 04 '21 at 14:12
To rephrase (if I'm not mistaken) you want to **recursively replace/remove** brackets that have a ***"valid"*** closing bracket? — Roko C. Buljan, Jan 04 '21 at 14:12
There are many posts online that discuss solving this type of problem. Just Google for "balanced parentheses algorithm". Most of them use a stack to do this. Here is an SO question that discusses this: https://stackoverflow.com/questions/16874176/parenthesis-brackets-matching-using-stack-algorithm. — Francis Gagnon, Jan 04 '21 at 14:41
Note that the stack approach solves this in O(n) instead of your O(n^2). — Francis Gagnon, Jan 04 '21 at 15:05

Unmitigated · Accepted Answer · 2021-01-04T14:21:17.750

1

You can use a while loop.

var isValid = (s)=> {
   const re = /(\(\))|(\[\])|(\{\})/gi
   while(re.test(s)) s = s.replace(re, '');
   return s === '';
};

edited Jan 04 '21 at 14:21

answered Jan 04 '21 at 14:12

Unmitigated

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Even better (IMO) `while (re.test(s))` since `test` returns a Boolean. – Heretic Monkey Jan 04 '21 at 14:14

Iterating the same method on a string

1 Answers1