[LeetCode] Scramble String

Problem:

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.

Solution:O(mn)

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1==null || s2==null || s1.length()!=s2.length()) 

            return false;
        char[] c1=s1.toCharArray();

        cahr[] c2=s2.toCharArray();
        Arrays.sort(c1);
        Arrays.sort(c2);
        if(!(new String(c1)).equals(new String(c2))) 

            return false;
        else if(s1.length()==1)

            return true;
    
        for(int i=0;i<s1.length()-1;i++){
            if(isScramble(s1.substring(0,i+1),s2.substring(s1.length()-i-1))
                  && isScramble(s1.substring(i+1),s2.substring(0,s1.length()-i-1))
                  || isScramble(s1.substring(0,i+1),s2.substring(0,i+1)) 
                  && isScramble(s1.substring(i+1),s2.substring(i+1)))

                return true;
        }
        return false;
    }
}