87 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.

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1.size()!=s2.size())
            return false;
        if (s1==s2)
            return true;
        string t1=s1,t2=s2;
        sort(t1.begin(),t1.end());
        sort(t2.begin(),t2.end());
        if (t1!=t2)
            return false;
        
        for (int i=1;i<s1.size();i++)
        {
            if (isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i,s1.size()-i),s2.substr(i,s2.size()-i)))
                return true;
            if (isScramble(s1.substr(0,i),s2.substr(s2.size()-i,i))&&isScramble(s1.substr(i,s1.size()-i),s2.substr(0,s2.size()-i)))
                return true;
        }
        return false;
    }
};

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.equals(s2))
            return true;
        if (s1.length()!=s2.length())
            return false;
        char[] t1 = s1.toCharArray(), t2=s2.toCharArray();
        Arrays.sort(t1);
        Arrays.sort(t2);
        if (!(new String(t1)).equals(new String(t2)))
            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;
    }
}