RegEx that gives letters not enclosed by parentheses

I want to write a regular expression that gives me letters that are not enclosed by parentheses. It should not give me any letters in any parentheses.

For example:

If the input is x+y^(a+b*c)+z
it should give me x, y, and z, but not a, b, and c

I’ve tried this but it didn’t work:

/[^\(][a-z][^\)]/g

Test strings:

"x+y^(a+b)+z"         // should return ["x", "y", "z"]
"x+y^(a+b)+z*(c-d)/w" // should return ["x", "y", "z", "w"]
"x+y^(a+b*(c))+z"     // should return ["x", "y", "z"]

Also, the indexes of all the letters in the original string are needed.

Please answer with an explanation

>Solution :

Here are two solutions:

strings = [
  'x+y^(a+b)+z',
  'x+y^(a+b+c)+q+z',
  'x+y^(a+b)+p+q^(c+d)+z'
].forEach(str => {
  let match1 = str.match(/(?<!\([^\)]*)[a-z]+/g);
  let match2 = str.replace(/\^\([^\)]*/g, '').match(/[a-z]+/g);
  console.log(str + ' =>'
    + '\n  match1: ' + match1
    + '\n  match2: ' + match2);
});

Output:

x+y^(a+b)+z =>
  match1: x,y,z
  match2: x,y,z
x+y^(a+b+c)+q+z =>
  match1: x,y,q,z
  match2: x,y,q,z
x+y^(a+b)+p+q^(c+d)+z =>
  match1: x,y,p,q,z
  match2: x,y,p,q,z

Explanation for match1:

• .match(/(?<!\([^\)]*)[a-z]+/g) — negative lookbehind for ^(... pattern
• note that negative lookbehind does not work in all browsers, notably Safari

Explanation for match2:

• .replace(/\^\([^\)]*/g, '') — remove all ^(...) patterns
• .match(/[a-z]+/g) — simple match for letters

Do you have nested parenthesis? This is possible too with pre-tagging parenthesis with nesting level. Let me know.