leetcode@ [87] Scramble String (Dynamic Programming)

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 check(string s1, string s2) {
        if(s1.length() != s2.length()) return false;
        
        string cp1 = s1, cp2 = s2;
        sort(cp1.begin(), cp1.end());
        sort(cp2.begin(), cp2.end());
        for(int i=0; i<cp1.length(); ++i) {
            if(cp1[i] != cp2[i]) return false;
        }
        return true;
    }
    bool dfs(string s1, string s2) {
        int m = s1.length(), n = s2.length();
        if(!check(s1, s2))  return false;
        if(m == 1) {
            if(s1 == s2)  return true;
            return  false;
        }
        
        string l, r, p, q;
        for(int le = 1; le < m; ++le) {
            l = s1.substr(0, le);
            r = s1.substr(le, m - le);
            p = s2.substr(0, le);
            q = s2.substr(le, m - le);
            if(dfs(l, p) && dfs(r, q))  return true;
            else {
                p = s2.substr(m - le, le);
                q = s2.substr(0, m - le);
                if(dfs(l, p) && dfs(r, q))  return true;
            }
        }
        
        return false;
    }

    bool isScramble(string s1, string s2) {
        int m = s1.length(), n = s2.length();
        
        if(m != n)  return false;
        
        return dfs(s1, s2);
    }
};