205. Isomorphic strings

string

https://leetcode.com/problems/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 Example 2: Input: s = “foo”, t = “bar” Output: false 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.

  • code
class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s.length() != t.length()) return false;
        Map<Character, Character> m = new HashMap<>();
        Set<Character> v = new HashSet<>();//mapped values 

        for (int i = 0; i < s.length(); i++){
            if (!m.containsKey(s.charAt(i))){
                if (v.contains(t.charAt(i))) return false;
                m.put(s.charAt(i), t.charAt(i));
                v.add(t.charAt(i));
            }else{
                if (m.get(s.charAt(i)) != t.charAt(i)) return false;
            }
        }
        return true;
        
    }
}
  • code
class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        return len(set(zip(s, t))) == len(set(s)) == len(set(t))
  • code
class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        
        if len(s) != len(t):
            return False
        
        match = {}
        for i in range(len(s)):
            if s[i] not in match:
                if t[i] in match.values():
                    return False
                match[s[i]] = t[i]
            else:
                if match[s[i]] != t[i]:
                    return False
        return True
  • code
class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        d= {}
        if len(set(s)) != len(set(t)): return False

        for i in range(len(s)):
            if s[i] in d and d[s[i]] != t[i]:
                return False
            else:
                d[s[i]] = t[i]
        return True