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.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

AC code:

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1 == s2) return true;
        int len = s1.length();
        vector<int> v(26, 0);
        for (int i = 0; i < len; ++i) {
            v[s1[i]-'a']++;
            v[s2[i]-'a']--;
        }
        for (int i = 0; i < 26; ++i) {
            if (v[i] != 0) return false;
        }
        for (int i = 1; i < len; ++i) {
            if (isScramble(s1.substr(0, i), s2.substr(0, i)) && 
                isScramble(s1.substr(i), s2.substr(i))) return true;
            if (isScramble(s1.substr(0, i), s2.substr(len-i)) &&
                isScramble(s1.substr(i), s2.substr(0, len-i))) return true;
        }
        return false;
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Scramble String.