<p>Given a string <code>s</code>, return <code>true</code> <em>if it is possible to split the string</em> <code>s</code> <em>into three <strong>non-empty</strong> palindromic substrings. Otherwise, return </em><code>false</code>.​​​​​</p>

<p>A string is said to be palindrome if it the same string when reversed.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abcbdd&quot;
<strong>Output:</strong> true
<strong>Explanation: </strong>&quot;abcbdd&quot; = &quot;a&quot; + &quot;bcb&quot; + &quot;dd&quot;, and all three substrings are palindromes.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;bcbddxy&quot;
<strong>Output:</strong> false
<strong>Explanation: </strong>s cannot be split into 3 palindromes.
</pre>

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

<ul>
	<li><code>3 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code>​​​​​​ consists only of lowercase English letters.</li>
</ul>

class Solution
{
public:
    bool done[2001][2001]={},palins[2001][2001]={};
    string s;
    int n;
    bool isPalin(int i,int j)
    {
        if(i==j)            return true;
        if(i+1==j)          return s[i]==s[j];
        if(done[i][j])      return palins[i][j];
        done[i][j]=true;
        return palins[i][j]=(s[i]==s[j])&&isPalin(i+1,j-1);
    }
    bool checkPartitioning(string s)
    {    
        this->s=s,n=s.size();
        for(int i=0;i<n;i++)
            if(isPalin(0,i))
                for(int j=i+1;j<n-1;j++)
                    if(isPalin(i+1,j)&&isPalin(j+1,n-1))
                        return true;
        return false;
    }
};
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