[📝LeetCode #205] Isomorphic Strings

#leetcode #java #100daysofcode #hashtable

🎀 The Problem

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.

Example:

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'.

👩‍💻 My Answer

class Solution { public boolean isIsomorphic(String s, String t) { HashMap<Character, Character> map = new HashMap(); for (int i = 0; i < s.length(); i++) { if (!map.containsKey(s.charAt(i)) && !map.containsValue(t.charAt(i))) map.put(s.charAt(i), t.charAt(i)); else if (map.containsKey(s.charAt(i)) && map.get(s.charAt(i)) != t.charAt(i)) return false; else if (!map.containsKey(s.charAt(i)) && map.containsValue(t.charAt(i))) return false; } return true; } }

Pro & Con

✖️ Runtime & Memory

💋 Ideal Answer

Approach

I was not sure, and I am not sure why my code takes a long time, although I use a HashMap. My favorite YouTuber also suggested the HashMap. Nikhil Lohia's Isomorphic Strings (LeetCode 205)

In LeetCode, I found that the array method is faster than HashMap, so I would try with this array method.

New Code

class Solution { public boolean isIsomorphic(String s, String t) { int[] slist = new int[256]; int[] tlist = new int[256]; for (int i = 0; i < s.length(); i++) { if (slist[s.charAt(i)] != tlist[t.charAt(i)]) return false; slist[s.charAt(i)]=i+1; tlist[t.charAt(i)]=i+1; } return true; } }