<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 isPalindrome(int start,int end,string &s)
    {
        while(start<=end)
        {
            if(s[start++]!=s[end--])
            return false;

        }
        return true;
    }
    int f(int index,int cuts,string &s,vector<vector<int>>&dp)
    {
        int n=s.size();
        if(cuts==0 && index<n)
        return isPalindrome(index,n-1,s) ? 1:0;
        if(index==n)
        return 0;

        if(dp[index][cuts]!=-1)return dp[index][cuts];

        int flag=0;

        for(int i=index;i<n;i++)
        {
            if(isPalindrome(index,i,s))
            {
                flag=(flag | f(i+1,cuts-1,s,dp));
            }
        }
        return dp[index][cuts] = flag;
    }
    bool checkPartitioning(string s) {
        int n=s.size();
        vector<vector<int>>dp(n+1,vector<int>(3,-1));

        // for(int i=0;i<n;i++)
        // {
        //     if(isPalindrome(0,i,s))
        //     {
        //         for(int j=i+1;j<n-1;j++)
        //         {
        //             if(isPalindrome(i+1,j,s) && isPalindrome(j+1,n-1,s))
        //              return true;
        //         }
        //     }

        // }
        return f(0,2,s,dp);
        
    }
};
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