Leetcode 87. Scramble String

Posted 2023-02-18 zhangwj0101

Question

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.

Code

  public boolean isScramble(String s1, String s2) 
        if (s1 == null || s2 == null || s1.length() != s2.length()) 
            return false;
        
        if (s1.equals(s2)) 
            return true;
        
        char[] chars1 = s1.toCharArray();
        char[] chars2 = s2.toCharArray();
        Arrays.sort(chars1);
        Arrays.sort(chars2);
        if (!Arrays.equals(chars1, chars2)) 
            return false;
        
        for (int i = 1; i < s1.length(); ++i) 
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i))) 
                return true;
            
            if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i)) && isScramble(s1.substring(i), s2.substring(0, s2.length() - i))) 
                return true;
            
        
        return false;