<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:
  int dp[2001][2001];
  int dp1[2001][6];
   bool solve(int i,string& s,int j)
   {
        if(i>j) return true;
        if(dp[i][j]!=-1)return dp[i][j];
        solve(i+1,s,j);
        solve(i,s,j-1);
        if(s[i]==s[j])
        {
            dp[i][j]=solve(i+1,s,j-1);
        }
        else
         dp[i][j]=false;
         return dp[i][j];
   }
    bool solve1(int i,int k,int n,string& s)
    {   
        if(i>=n)
        {
            if(k==0)return true;
            return false;
        }
        if(k==0)return false;
        if(dp1[i][k]!=-1)return dp1[i][k];
        bool ok=false;
        for(int j=i;j<n;j++)
        {
            if(dp[i][j])
            {
                  ok=ok||solve1(j+1,k-1,n,s);
            }
        }
        return dp1[i][k]=ok;
    }
    bool checkPartitioning(string s) {
        int n=s.size();
        memset(dp,-1,sizeof(dp));
        solve(0,s,n-1);
        memset(dp1,-1,sizeof(dp1));
        return solve1(0,3,n,s);
    }
};
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