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# 87     Scramble String

## 87. [Scramble String](https://leetcode.com/problems/scramble-string/description/)

## 1. Question

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
```

## 2. Implementation

**(1) Recursion**

思路: 如果s1和s2长度不同，显然return false。然后由于题目的输入string都只含小写字母，所以我们为了判断两个string是否含有一样的character时，我们利用一个size为26的count数字计数，然后再扫一遍count数组，如果count数组的元素不为0，说明存在不同或者数量不一样的character，此时返回false. 注意这步是必要的，否则会超时。

最后就通过递归的方式调用isScramble, 找出scramble的分割点即可

```java
class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.equals(s2)) {
            return true;
        }

        if (s1.length() != s2.length()) {
            return false;
        }

        int[] count = new int[26];

        // Check if s1 and s2 has the same characters
        for (int i = 0; i < s1.length(); i++) {
            ++count[s1.charAt(i) - 'a'];
            --count[s2.charAt(i) - 'a'];
        }

        for (int i = 0; i < 26; i++) {
            if (count[i] != 0) {
                return false;
            }
        }

        int n = s1.length();
        // Check if s1 and s2 are scrambled strings using recursion
        for (int i = 1; i < n; i++) {
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i))
                || (isScramble(s1.substring(0, i), s2.substring(n - i)) && isScramble(s1.substring(i), s2.substring(0, n - i)))) {
                return true;
            }
        }
        return false;
    }
}
```

**(2) DP**

```
```

## 3. Time & Space Complexity

**Recursion:** 时间复杂度O(4 ^ n), 空间复杂度O(n) 递归的深度是O(n)