## Description

Given two strings s*** and **t*, determine if they are isomorphic.

Two strings are isomorphic if the characters in s*** can be replaced to get **t*.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

Example 1:

Input: s = "egg", t = "add"
Output: true


Example 2:

Input: s = "foo", t = "bar"
Output: false


Example 3:

Input: s = "paper", t = "title"
Output: true


Note: You may assume both s*** and **t* have the same length.

## Solutions

### 1. Hash Table

# Time: O(n)
# Space: O(2n)
class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if (not s and t) or (s and not s) or len(s) != len(t):
return False
n = len(s)
hash_s = collections.defaultdict()
hash_t = collections.defaultdict()
for i in range(n):
if s[i] in hash_s:
if hash_s[s[i]] != t[i]:
return False
else:
hash_s[s[i]] = t[i]
if t[i] in hash_t:
if hash_t[t[i]] != s[i]:
return False
else:
hash_t[t[i]] = s[i]
return True
# 30/30 cases passed (40 ms)
# Your runtime beats 57.07 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.7 MB)