Solutions

1. Hash Map (Two Pass)

class Solution:
    def helper(self, s: str, t: str) -> bool:
        mp = {}
        for i in range(len(s)):
            if (s[i] in mp) and (mp[s[i]] != t[i]):
                return False
            mp[s[i]] = t[i]
        return True

    def isIsomorphic(self, s: str, t: str) -> bool:
        return self.helper(s, t) and self.helper(t, s)

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(m)$

Where $n$ is the length of the input string and $m$ is the number of unique characters in the strings.

2. Hash Map (One Pass)

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        mapST, mapTS = {}, {}

        for i in range(len(s)):
            c1, c2 = s[i], t[i]
            if ((c1 in mapST and mapST[c1] != c2) or
                (c2 in mapTS and mapTS[c2] != c1)):
                return False
            mapST[c1] = c2
            mapTS[c2] = c1

        return True

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(m)$

Where $n$ is the length of the input string and $m$ is the number of unique characters in the strings.