Given two strings s and t of the same length, return true if they're isomorphic — meaning the characters of s can be replaced, one-for-one and consistently, to get exactly t. No two characters may map to the same character, but a character may map to itself. Comparing every pair of positions directly confirms consistency but costs O(n²). The faster approach walks both strings together while tracking the mapping both ways — s-to-t and t-to-s — in two hash maps. Checking both directions at every step is what catches both kinds of broken mapping in a single pass.