Problem Description

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 = "f11", t = "b23"
Output: false
Explanation:
The strings s and t can not be made identical as '1' needs to be mapped to both '2' and '3'.

Example 3:

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

Constraints:

• 1 <= s.length <= 5 * 10^4
• t.length == s.length
• s and t consist of any valid ascii character.

Approach

Use two TreeMaps to count the frequencies of characters in both strings. Then, iterate through the strings again to ensure that at each index i, the character in string s has the same total frequency as the character in string t.

Solution

class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s.length()!=t.length()){
            return false;
        }
        int n = s.length();
        TreeMap<Character, Integer> hashs = new TreeMap<>();
        TreeMap<Character, Integer> hasht = new TreeMap<>();
        for(int i=0;i<n;i++){
            hashs.put(s.charAt(i), hashs.getOrDefault(s.charAt(i), 0)+1);
            hasht.put(t.charAt(i), hasht.getOrDefault(t.charAt(i), 0)+1);
        }
        for(int i=0;i<n;i++){
            if(hashs.get(s.charAt(i))!=hasht.get(t.charAt(i))){
                return false;
            }
        }
        return true;
    }
}