Leetcode–Scramble String

The Problem:

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.

Notes:

  1. A recursion problem, try each substring  see if a) s1.sub1 match s2.sub1, and s1.sub2 match s2.sub2; OR b) s1.sub1 match s2.sub2 and s1.sub2 match s2.sub1
  2. Add conditions pre-check length and characters before going to recursion.

Java Solution

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if((s1.length()==0 && s2.length()==0)|| s1.equals(s2))
            return true;
        if(s1.length()!=s2.length())
            return false;
        if(!sameChar(s1,s2))
            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)))||
            (isScramble(s1.substring(0, i),s2.substring(s2.length()-i)) && isScramble(s1.substring(i), s2.substring(0,s2.length()-i))))
                return true;
        }
        return false;
        
    }
    
    private boolean sameChar(String s1, String s2){
        char[] ch1 = s1.toCharArray();
        char[] ch2 = s2.toCharArray();
        Arrays.sort(ch1);
        Arrays.sort(ch2);
        return new String(ch1).equals(new String(ch2));
    }
}
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