[Leetcode] 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.

Analysis:

for "rgtae" and "great", compare the following Recursively

r|gtae, g|reat, r|gtae, grea|t, 

Rg|tae, gr|eat, rg|tae, gre|at, 

rgt|ae, gre|at, rgt|ae, gr|eat, 

Rgta|e, grea|t, rgta|e, g|reat 

    bool isScramble(string s1, string s2) {
            if (s1.size() != s2.size())
                return false;
            else if (s1 == s2)
                return true;
            else
            {
                unordered_map<char, int> m;
                for (int i=0; i<s1.size(); i++)
                    m[s1[i]]++;
                for (int i=0; i<s2.size(); i++)
                {
                    if (m.find(s2[i]) == m.end() || m[s2[i]] <= 0)
                        return false;
                    m[s2[i]]--;
        
                }
                
                for(int i=1; i<s1.size();i++)
                {
                    if ((isScramble(s1.substr(0,i),s2.substr(0,i)) && isScramble(s1.substr(i),s2.substr(i))) || 
                        (isScramble(s1.substr(0,i),s2.substr(s1.size() - i)) && isScramble(s1.substr(i),s2.substr(0,s1.size() - i))))
                        return true;
                }
                return false;                
            }
    }