Scramble String

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.

 1 public class Solution {
 2     /**
 3      * @param s1 A string
 4      * @param s2 Another string
 5      * @return whether s2 is a scrambled string of s1
 6      */
 7     public boolean isScramble(String s1, String s2) {
 8         if (s1.length() != s2.length()) return false;
 9
10         if (s1.equals(s2)) return true;
11
12         if (!isValid(s1, s2)) {
13             return false;
14         }
15
16         for (int i = 1; i < s1.length(); i++) {
17             String s11 = s1.substring(0, i);
18             String s12 = s1.substring(i);
19
20             String s21 = s2.substring(0, i);
21             String s22 = s2.substring(i);
22
23             String s23 = s2.substring(0, s2.length() - i);
24             String s24 = s2.substring(s2.length() - i);
25
26             if (isScramble(s11, s21) && isScramble(s12, s22) || isScramble(s11, s24) && isScramble(s12, s23)) {
27                 return true;
28             }
29         }
30         return false;
31     }
32
33     private boolean isValid(String s1, String s2) {
34         char[] arr1 = s1.toCharArray();
35         char[] arr2 = s2.toCharArray();
36         Arrays.sort(arr1);
37         Arrays.sort(arr2);
38         if (!(new String(arr1)).equals(new String(arr2))) {
39             return false;
40         }
41         return true;
42     }
43 }