Scramble String

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.

------------------------------------- thinking ------------------------------------


------------------------------------- recursive codes -----------------------------
class Solution {
public:
    bool isScramble(string s1, string s2) {
        return scrambleHelper(s1, 0, s1.size() - 1, s2, 0, s2.size() - 1);
    }
    bool scrambleHelper(string s1, int start1, int end1, string s2, int start2, int end2) {
        if (end1 - start1 != end2 - start2) {
            return false;
        }
        if (end1 == start1 && s1[end1] == s2[end2]) {
            return true;
        }
        //important part here, don't miss it
        //or it will cause ETL
        int cnt[256] = {0};
        for (int ind=start1; ind <= end1; ind++) {
            cnt[s1[ind]]++;
        }
        for (int ind=start2; ind <= end2; ind++) {
            cnt[s2[ind]]--;
        }
        for (int ind=0; ind < 256; ind++) {
            if (cnt[ind] != 0) {
                return false;
            }
        }
        for (int i = start1; i < end1; i++) {
            if (scrambleHelper(s1, start1, i, s2, start2, i - start1 + start2) &&
                scrambleHelper(s1, i+1, end1, s2, i - start1 + start2 + 1, end2)) {
                return true;
            }
            if (scrambleHelper(s1, start1, i, s2, start1 - i + end2, end2) &&
                scrambleHelper(s1, i+1, end1, s2, start2, end1 - i - 1 + start2)) {
                return true;
            }
        }
        return false;
    }
};