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.

Solution

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if ( s1 == s2 ) return true;
        int n1 = s1.size(), n2 = s2.size();
        if ( n1 != n2 ) return false;
        string sort1 = s1, sort2 = s2;
        sort(sort1.begin(), sort1.end());
        sort(sort2.begin(), sort2.end());
        if ( sort1 != sort2 ) return false;
        for ( int i = 1; i <= n1-1; i++ ) {
            string s11 = s1.substr(0, i), s12 = s1.substr(i);
            string s21 = s2.substr(0, i), s22 = s2.substr(i);
            if ( isScramble(s11, s21) and isScramble(s12, s22) ) return true;
            s21 = s2.substr(0, n1-i);
            s22 = s2.substr(n1-i);
            if ( isScramble(s11, s22) and isScramble(s12, s21) ) return true;
        }
        return false;
    }
};

Notes
A very smart solution!