Regexp to match all the characters except spaces and 4 last characters

You can use

.replace(/\d(?=.*\d(?:\s*\d){3}\s*$)/g, '*')

See the regex demo

Regex details

• \d - a digit
• (?=.*\d(?:\s*\d){3}\s*$) - that is followed with any amount of any chars other than line break chars and then a digit, followed with three occurrences of 0+ whitespaces + a digit and then 0+ whitespaces at the end of string.

You could update the pattern to assert 4 digits at the right until the end of the string.

\d(?=(?:[ \d]*\d){4}$)

Explanation

• \d match a digit
• (?= Positive lookahead
  • (?: Non capture group
    • [ \d]*\d Match 0+ times a space or digits followed by a single digit
  • ){4}$ and repeat that 4 times ans assert the end of the string
• ) Close lookahead

Regex demo

[
  "1234 4567 8901 2345",
  "123445678901 23 4 5"
].forEach(s => console.log(s.replace(/\d(?=(?:[ \d]*\d){4}$)/gm, "*")));

I'm not sure you need a negative lookahead. If you have control over the input format then try this:

const number = '1234 5678 9012 3456';
const masked = number.replace(/\d{4}\s/g, '**** ');
console.log(masked);

If you want to accept also random spaces, try this:

const number = '1234 5678 9 012 3 45 6';
const masked = number.replace(/(\d\s*){4}\s/g, '**** ');
console.log(masked);