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.

Solution

public class Solution {
       public boolean isScramble(String s1, String s2) {
        int len  = s1.length();

         char[] chars1 = s1.toCharArray();
        char[] chars2 = s2.toCharArray();

        Arrays.sort(chars1);
        Arrays.sort(chars2);

        for(int i = 0 ; i < chars1.length; i ++) {
            if (chars1[i] != chars2[i])
                return false;
        }

        if (s1 == null && s2 == null) {
            return true;
        } else if (s1 == null || s2 == null) {
            return false;
        } else if (s1.length() != s2.length()) {
            return false;
        } else {
            if (s1.length() == 1) {
                return s1.equals(s2);
            }

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

                if (isScramble(s1.substring(0, i), s2.substring(len - i, len)) && isScramble(s1.substring(i, len), s2.substring(0, len - i))) {
                    return true;
                }
            }


            return false;

        }
    }
}