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.
Input: s = "egg", t = "add" Output: true
Input: s = "foo", t = "bar" Output: false
Input: s = "paper", t = "title" Output: true
Note: You may assume both s*** and **t* have the same length.
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)