# 87. Scramble String

[Link](https://leetcode.com/problems/scramble-string/description/)

Given a string *s1*, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of *s1* = `"great"`:

```
    great
   /    \
  gr    eat
 / \    /  \
g   r  e   at
           / \
          a   t
```

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node `"gr"` and swap its two children, it produces a scrambled string `"rgeat"`.

```
    rgeat
   /    \
  rg    eat
 / \    /  \
r   g  e   at
           / \
          a   t
```

We say that `"rgeat"` is a scrambled string of `"great"`.

Similarly, if we continue to swap the children of nodes `"eat"` and `"at"`, it produces a scrambled string `"rgtae"`.

```
    rgtae
   /    \
  rg    tae
 / \    /  \
r   g  ta  e
       / \
      t   a
```

We say that `"rgtae"` is a scrambled string of `"great"`.

Given two strings *s1* and *s2* of the same length, determine if *s2* is a scrambled string of *s1*.

**Example 1:**

```
Input: s1 = "great", s2 = "rgeat"
Output: true
```

**Example 2:**

```
Input: s1 = "abcde", s2 = "caebd"
Output: false
```

**Solution**

```
    public boolean isScramble(String s1, String s2) {
        
        //*check if equal
        if (s1.equals(s2)) return true;
        
        //check the number of each alphabet 
        int[] alphabet = new int[26];
        for(int i = 0; i < s1.length(); i++){
            alphabet[s1.charAt(i) - 'a']++;
            alphabet[s2.charAt(i) - 'a']--;
        }
        
        for(int i : alphabet){
            if(i != 0) return false;
        }
        
        //recursively check it's substring are scrambled
        for(int i = 1; i < s1.length(); i++){
            //s1's right, s1's right
            //s1's left, s2's left
            if(isScramble(s1.substring(0,i), s2.substring(0,i)) &&
              isScramble(s1.substring(i), s2.substring(i))){
                return true;
            }
    
            //check if these substring are scrambed
            //s1's right, s1's left
            //s1's left, s2's right
            if(isScramble(s1.substring(0,i), s2.substring(s2.length() - i)) &&
              isScramble(s1.substring(i), s2.substring(0, s2.length() - i))){
                return true;
            }
        }
        return false;
    }
```
