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.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

public class Solution {
    public boolean isScramble(String s1, String s2) {
        // DP of strings combination -> ans (1 -> true, 0 -> false)
        HashMap<String, Integer> memoization = new HashMap<>();
        return isScrambleRecursion(s1, s2, memoization);
    }

    public boolean isScrambleRecursion(String s1, String s2, HashMap<String, Integer> memoization) {
        int ret = memoization.getOrDefault(s1 + "#" + s2, -1);
        if (ret != -1)
            return ret == 0 ? false : true;
        if (s1.equals(s2)) {
            memoization.put(s1 + "#" + s2, 1);
            return true;
        }
        // Checking if the 2 strings are anagrams of each other
        int[] letters = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            letters[s1.charAt(i) - 'a']++;
            letters[s2.charAt(i) - 'a']--;
        }
        for (int i = 0; i < 26; i++)
            if (letters[i] != 0) {
                memoization.put(s1 + "#" + s2, 0);
                return false;
            }
        // Calling the substrings combination for answers
        for (int i = 1; i < s1.length(); i++) {
            // Direct matching
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i))) {
                memoization.put(s1 + "#" + s2, 1);
                return true;
            }
            // Opposite matching
            if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i))
                    && isScramble(s1.substring(i), s2.substring(0, s2.length() - i))) {
                memoization.put(s1 + "#" + s2, 1);
                return true;
            }
        }
        memoization.put(s1 + "#" + s2, 0);
        return false;
    }
}