<p>We can scramble a string s to get a string t using the following algorithm:</p>

<ol>
	<li>If the length of the string is 1, stop.</li>
	<li>If the length of the string is &gt; 1, do the following:
	<ul>
		<li>Split the string into two non-empty substrings at a random index, i.e., if the string is <code>s</code>, divide it to <code>x</code> and <code>y</code> where <code>s = x + y</code>.</li>
		<li><strong>Randomly</strong>&nbsp;decide to swap the two substrings or to keep them in the same order. i.e., after this step, <code>s</code> may become <code>s = x + y</code> or <code>s = y + x</code>.</li>
		<li>Apply step 1 recursively on each of the two substrings <code>x</code> and <code>y</code>.</li>
	</ul>
	</li>
</ol>

<p>Given two strings <code>s1</code> and <code>s2</code> of <strong>the same length</strong>, return <code>true</code> if <code>s2</code> is a scrambled string of <code>s1</code>, otherwise, return <code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;great&quot;, s2 = &quot;rgeat&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> One possible scenario applied on s1 is:
&quot;great&quot; --&gt; &quot;gr/eat&quot; // divide at random index.
&quot;gr/eat&quot; --&gt; &quot;gr/eat&quot; // random decision is not to swap the two substrings and keep them in order.
&quot;gr/eat&quot; --&gt; &quot;g/r / e/at&quot; // apply the same algorithm recursively on both substrings. divide at random index each of them.
&quot;g/r / e/at&quot; --&gt; &quot;r/g / e/at&quot; // random decision was to swap the first substring and to keep the second substring in the same order.
&quot;r/g / e/at&quot; --&gt; &quot;r/g / e/ a/t&quot; // again apply the algorithm recursively, divide &quot;at&quot; to &quot;a/t&quot;.
&quot;r/g / e/ a/t&quot; --&gt; &quot;r/g / e/ a/t&quot; // random decision is to keep both substrings in the same order.
The algorithm stops now, and the result string is &quot;rgeat&quot; which is s2.
As one possible scenario led s1 to be scrambled to s2, we return true.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;abcde&quot;, s2 = &quot;caebd&quot;
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;a&quot;, s2 = &quot;a&quot;
<strong>Output:</strong> true
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>s1.length == s2.length</code></li>
	<li><code>1 &lt;= s1.length &lt;= 30</code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
</ul>

class Solution {
    unordered_map<string , bool>temp;
    bool solve(string s1, string s2, int i){
      if(s1.size() == 1){
        return s1[0] == s2[i];
      }
      string key = s1 + '#' +to_string(i) ;
      if(temp.count(key))return temp[key];
      bool ans = false;
      for(int k = 0;k<s1.size()-1;k++){
        string a = s1.substr(0,k+1);
        string b = s1.substr(k+1,(s1.size()-1)-k);
        ans |= (solve(a,s2,i) && solve(b,s2,i+k+1)||
                solve(b,s2,i) && solve(a,s2,i+b.size())
        );
        if(ans)break;
      }
      return temp[key] = ans;
    }
public:
    bool isScramble(string s1, string s2) {
      return solve(s1,s2,0);
    }
};
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