Two strings are called isomorphic if there exists a one-to-one mapping from characters in one string to characters in the other.

For example, "aab" and "xxy" are isomorphic, because the one-to-one mapping {'a': 'x', 'b': 'y'}, when applied to "aab", yields "xxy". However, "acb" and "xxy" are not isomorphic, because the mapping {'a': 'x', 'c': 'x', 'b': 'y'} is not one-to-one, because 'x' appears twice as a value.

One way to think about it is that the mapping should allow us to transform back and forth between the two strings. After transforming "aab" to "xxy", we can invert the mapping to undo the transformation. However, after transforming "acb" to "xxy", we cannot undo the transformation.

Create a public class IsomorphicStrings that provides a single public static method areIsomorphic. It accepts two Strings and returns true if the Strings are isomorphic, and false otherwise. If either String is null, throw an IllegalArgumentException. Also note that you can begin with a simple length check to rule out certain cases.

A natural way to approach this problem is to use a Map<Character, Character>. However, this is not necessary, and is somewhat inefficient.

Instead, approach the problem this way. At a given position i, what do we need to check for all later positions in both Strings?

1. That if first[i] maps to second[i], then that mapping is consistent throughout the rest of the string: meaning for all j > i, first[i] == first[j] -> second[i] == second[j].
2. That there are no other mappings to second[i].

Building of these ideas, you can solve the problem without maps or sets. You may want to use a single boolean[] to avoid rechecking the same spot in the String twice.