const regex = /^((?|[1-9](?=.*1\t+(?<x>0+))|[1-9]\d(?=.*2\t+(?<x>0+))|[1-9]\d{2}(?=.*3\t+(?<x>0+))|[1-9]\d{3}(?=.*4\t+(?<x>0+))|[1-9]\d{4}(?=.*5\t+(?<x>0+))|[1-9]\d{5}(?=.*6\t+(?<x>0+))|[1-9]\d{6}(?=.*7\t+(?<x>0+))|[1-9]\d{7}(?=.*8\t+(?<x>0+))))\b/gsm;
// Alternative syntax using RegExp constructor
// const regex = new RegExp('^((?|[1-9](?=.*1\\t+(?<x>0+))|[1-9]\\d(?=.*2\\t+(?<x>0+))|[1-9]\\d{2}(?=.*3\\t+(?<x>0+))|[1-9]\\d{3}(?=.*4\\t+(?<x>0+))|[1-9]\\d{4}(?=.*5\\t+(?<x>0+))|[1-9]\\d{5}(?=.*6\\t+(?<x>0+))|[1-9]\\d{6}(?=.*7\\t+(?<x>0+))|[1-9]\\d{7}(?=.*8\\t+(?<x>0+))))\\b', 'gsm')
const str = `1
12
123
12345678
123456789
1 00000000
2 0000000
3 000000
4 00000
5 0000
6 000
7 00
8 0`;
const subst = `${x}$1`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for JavaScript, please visit: https://developer.mozilla.org/en/docs/Web/JavaScript/Guide/Regular_Expressions