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.

Think - DP Version

• Ignore two situations: both length not equal and the characters not the same
• Two sequence but 3-D memorized array
• memo[i][j][k] means state: for s1.substring(i, i + k) and s2.substring(j, j + k), if they are scramble string
• Two conditions we can regard as scramble, for range of word1(i -> i+k) or word2(j -> j+k):
  • i -> i + split = j -> j + split (len = split) and split + i -> i + k = split + i -> j + k (len = k - split)
  • i -> i + split = j + (k - split) -> j+k [len = split] and i + split -> i+k = j -> j + (k - split)(len = k - split)
• Consider about the initialization:
  • for k == 1, we only check if word1[i] == word2[j]

Solution

    /**
     * @param s1 A string
     * @param s2 Another string
     * @return whether s2 is a scrambled string of s1
     */
    public boolean isScramble(String s1, String s2) {
        // check length
        if(s1==null||s2==null||s1.length()!=s2.length())
            return false;
        // check anagram
        char[] c1 = s1.toCharArray();
        char[] c2 = s2.toCharArray();
        Arrays.sort(c1);
        Arrays.sort(c2);
        if (!Arrays.equals(c1, c2))
            return false;

        if(s1.length() != s2.length())
            return false;
        int len = s1.length();

        // state: for s1.substring(i, i + k) and s2.substring(j, j + k), if they are scramble string
        boolean[][][] memo = new boolean[len][len][len+1];

        // initial, only check if s1[i] == s2[j] 
        for(int i=0;i<s1.length();i++)
            for(int j=0;j<s2.length();j++)
                memo[i][j][1] = (s1.charAt(i) == s2.charAt(j));

        for(int k = 2; k <= len; k++) {

            for(int i = 0; i <= len - k; i++) {
                for(int j = 0; j <= len - k; j++) {
                    // split point should start from 1 to k - 1
                    for(int split = 1; split < k; split++) {
                        memo[i][j][k] |= (memo[i][j][split]&&memo[i+split][j+split][k-split])||(memo[i][j+(k - split)][split]&&memo[i+split][j][k-split]);
                    }
                }
            }
        }
        return memo[0][0][len];
    }