Day039 - 205. Isomorphic Strings

205. Isomorphic Strings

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

Two strings s and t 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 Explanation: The strings s and t can be made identical by:

• Mapping 'e' to 'a'.
• Mapping 'g' to 'd'.

Example 2:

Input: s = “foo”, t = “bar” Output: false Explanation: The strings s and t can not be made identical as 'o' needs to be mapped to both 'a' and 'r'.

Example 3:

Input: s = “paper”, t = “title” Output: true

Constraints:

• 1 <= s.length <= 5 * 104
• t.length == s.length
• s and t consist of any valid ascii character.

내 풀이

from collections import defaultdict

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        if len(set(s)) != len(set(t)): return False

        s_checker = defaultdict(list)
        t_checker = defaultdict(list)

        current_idx = 0

        s_same = False
        t_same = False

        for s_one, t_one in zip(s, t):          
            if (s_one in s_checker) != (t_one in t_checker):
                return False

            if current_idx >= 1:
                s_same = s_one == s[current_idx-1]
                t_same = t_one == t[current_idx-1]

            if s_same != t_same:
                return False

            if s_checker[s_one] != t_checker[t_one]:
                return False

            s_checker[s_one].append(current_idx)
            t_checker[t_one].append(current_idx)
            current_idx += 1

        return True

# Time Complexity : \(O(N)\)