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) {
        if (s1.equals(s2)) return true;

        int[] letters = new int[26];
        for (int i=0; i<s1.length(); i++) {
            letters[s1.charAt(i)-'a']++;
            letters[s2.charAt(i)-'a']--;
        }
        for (int i=0; i<26; i++) if (letters[i]!=0) return false;

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