```
"""LOGIC: Assign elements in s as key, and elements in t as its corresponding values
1. Iterate through the length of the list. (as it is given that the lengths are equal)
2. If s[i] is not present in the dictionary,
a. To overcome s = "aa" t ="ab", check if t[i] is already present in dict.values(). If it is, then there is a duplicate, meaning, 2 keys have the same value, and that should not be the case. return False
b. Else, add it as a key and assign t[i] as its value
3. If it is present, check whether the value for that key is equal to t[i], if not return false
4. Finally return True
"""
class Solution(object):
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
dict = {}
for i in range(len(s)):
if s[i] not in dict:
if t[i] in dict.values():
return False
dict[s[i]] = t[i]
else:
if dict[s[i]] != t[i]:
return False
return True
```