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# 205. Isomorphic Strings

### Description

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.

### Constraints

### Approach

### Links

* [GeeksforGeeks](https://www.geeksforgeeks.org/check-if-two-given-strings-are-isomorphic-to-each-other/)
* [Leetcode](https://leetcode.com/problems/isomorphic-strings/)
* [ProgramCreek](https://www.programcreek.com/2014/05/leetcode-isomorphic-strings-java/)
* YouTube

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** s = "egg", t = "add"

**Output:** true
{% endtab %}

{% tab title="Example 2" %}
**Input:** s = "foo", t = "bar"

**Output:** false
{% endtab %}

{% tab title="Example 3" %}
**Input:** s = "paper", t = "title"

**Output:** true
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s.length() != t.length()) return false;
        
        Map<Character, Character> map = new HashMap();
        
        for(int i = 0; i < s.length(); i++) {
            char ch1 = s.charAt(i);
            char ch2 = t.charAt(i);
            
            if(map.containsKey(ch1)) {
                if(map.get(ch1) != ch2) return false;
            } else {
                map.put(ch1, ch2);
            }
        }
        
        Set<Character> values = new HashSet(map.values());
        if(values.size() == map.values().size()) {
            return true;
        }
        
        return false;
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s.length() != t.length()) return false;
        
        int[] map = new int[256];
        
        for(int i = s.length()-1; i >= 0; i--) {
            int p1 = (int) s.charAt(i);
            int p2 = (int) t.charAt(i);

            if(map[p1] != map[p2+128]) {
                return false;
            } else {
                map[p1] = i;
                map[p2+128] = i;
            }
        }
        
        return true;
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*