Scramble String

ScrambleString.java

/**
 * 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.
 */

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.length() != s2.length()) 
            return false;
        if(s1.length() == 0 && s2.length() == 0) 
            return true;
        boolean[][][] scramble = new boolean[s1.length()][s2.length()][s1.length()];
        for(int i = 0; i < s1.length(); i++) {
            for(int j = 0; j < s2.length(); j++) {
                scramble[i][j][0] = (s1.charAt(i) == s2.charAt(j));
            }
        }
        for(int l = 1; l < s1.length(); l++) {
            for(int i = 0; i < s1.length() - l; i++) {
                for(int j = 0; j < s2.length() - l; j++) {
                    for(int k = 0; k < l; k++) {
                        if((scramble[i][j][k] && scramble[i + k + 1][j + k + 1][l - k - 1]) || (scramble[i][j + l - k][k] && scramble[i + k + 1][j][l - k - 1]))
                            scramble[i][j][l] = true;
                    }
                }
            }
        }
        return scramble[0][0][s1.length() - 1];
    }
}

ScrambleStringRecursion.java

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.length() != s2.length())
			return false;
		if (s1.equals(s2))
			return true;
        String ts1, ts2;
        char[] c_s1 = s1.toCharArray();
        Arrays.sort(c_s1);
        ts1 = new String(c_s1);
        char[] c_s2 = s2.toCharArray();
        Arrays.sort(c_s2);
        ts2 = new String(c_s2);
        if(!ts1.equals(ts2))
            return false;
        for(int isep = 1; isep < s1.length(); isep++) {
            String seg11 = s1.substring(0, isep);
            String seg12 = s1.substring(isep);
            // see if forward order is OK
            String seg21 = s2.substring(0, isep);
            String seg22 = s2.substring(isep);
            if(isScramble(seg11, seg21) && isScramble(seg12, seg22))
                return true;
            // see if backward order is OK
            seg21 = s2.substring(s2.length() - isep);
            seg22 = s2.substring(0, s2.length() - isep);
            if(isScramble(seg11, seg21) && isScramble(seg12, seg22))
                return true;
        }
        return false;
    }
}