/*
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.
*/



public class ScrambleString {
    public boolean isScramble(String s1, String s2) {
        if( s1==null||s2==null){
            return false;
        }
        
        if (s1.length()==0){
            return s2.length()==0;
        }
        if (s1.length()!=s2.length()){
            return false;
        }
        if (s1.equals(s2)){
            return true;
        }
        
        int value1=0;
        int value2=0;
        
        // check is s1 and s2 has same chars
        for (int i=0; i<s1.length(); i++){
            value1+=s1.charAt(i)-'0';
            value2+=s2.charAt(i)-'0';
        }
        
        if (value1!=value2){
            return false;
        }
        
        for (int i =1; i<s1.length(); i++){
            String s1Left=s1.substring(0, i);
            String s1Right=s1.substring(i);
            
            String s2Left=s2.substring(0,i);
            String s2Right=s2.substring(i);
            
            if (isScramble(s1Left, s2Left)&&isScramble(s1Right,s2Right)){
                return true;
            }
            
           s2Left=s2.substring(s2.length()-i);
           s2Right=s2.substring(0, s2.length()-i);
            
            if (isScramble(s1Left, s2Left)&&isScramble(s1Right,s2Right)){
                return true;
            }
        }
        
        return false;
    }
}