<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 {
public:
int dp[51][51][51][51];


int sl(int l, int r, int l1, int r1, string& a, string& b, int n){    
    if(l==r){
        if(a[l]==b[l1]) return 1;
        return 0;
    }
    
    if(dp[l][r][l1][r1]!=-1) return dp[l][r][l1][r1];
    
    int ans=0;
    int d = l1-l;
    
    for(int i=l;i<=r-1;i++){
        int c1 = sl(l, i,l+d,i+d, a, b,n) & sl(i+1,r,i+1+d,r+d,a,b,n);
        int c2 = sl( l, i, l+r-i+d,r+d, a, b,n) & sl(i+1,r,l+d,r-i+l-1+d,a,b,n);
        if(c1==1 || c2==1) ans=1;      
    }
    
    return dp[l][r][l1][r1]=ans;
}
    bool isScramble(string s1, string s2) {

        int n=s1.size();
        
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k<n;k++){
                for(int l=0;l<n;l++){
                    dp[i][j][k][l]=-1;
                }
            }
        }
    }
    // vector<vector<vector>> dp(n,vector<int>(n,-1));
    
    return sl(0,n-1,0,n-1,s1,s2,n);
        
    }
};
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