I'm trying to understand regex as much as I can, so I came up with this regex-based solution to codingbat.com repeatEnd:

Given a string and an int N, return a string made of N repetitions of the last N characters of the string. You may assume that N is between 0 and the length of the string, inclusive.

public String repeatEnd(String str, int N) {
  return str.replaceAll(
    ".(?!.{N})(?=.*(?<=(.{N})))|."
      .replace("N", Integer.toString(N)),
    "$1"
  );
}

Explanation on its parts:

• .(?!.{N}): asserts that the matched character is one of the last N characters, by making sure that there aren't N characters following it.

• (?=.*(?<=(.{N}))): in which case, use lookforward to first go all the way to the end of the string, then a nested lookbehind to capture the last N characters into \1. Note that this assertion will always be true.

• |.: if the first assertion failed (i.e. there are at least N characters ahead) then match the character anyway; \1 would be empty.

• In either case, a character is always matched; replace it with \1.

My questions are:

• Is this technique of nested assertions valid? (i.e. looking behind during a lookahead?)
• Is there a simpler regex-based solution?