> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/dynamic-programming/scramble-string.md).

# 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
```

```java
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;
    }
}
```


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