Scramble String

leetcode 87.

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.

Solution. This is a interesting problem. The idea is to compare isScramble(S1[0, …, i], S2[0, …, i]) and isScsramble(S1[i + 1, …, len – 1]), S2[i + 1, …, len – 1]), or isScramebl(S1[0, …, i], S2[len – i, len – i + 1, …, len]) and isScramble(S1[i + 1, i + 2, …, len – 1), S2[0, 1, …, len – i]).

For this case

S1 = X0 X1 X2 X3 X4 X5 X6
S2 = Y0 Y1 Y2 Y3 Y4 Y5 Y6

We check isScramble(X0X1X2, Y0Y1Y2) and isScramble(X3X4X5X6, Y3Y4Y5Y6)

Or check isScamble(X0X1X2, Y4Y5Y6) and isScramble(X3X4X5X6, Y0Y1Y2Y3)

public static boolean isScramble(String s1, String s2) {
    if (s1.equals(s2)) {
        return true;
    }
    int[] letters = new int[26];
    for (int i = 0; i < s1.length(); i++) { // this check to avoid enter useless loop.
        letters[s1.charAt(i) - 'a']++;
        letters[s2.charAt(i) - 'a']--;
    }
    for (int count : letters) {
        if (count != 0) {
            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(s1.length() - i)) && isScramble(s1.substring(i), s2.substring(0, s1.length() - i))) {
            return true;
        }
    }
    return false;
}

Check my code on github.