<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>

int dp[2005][2005];
int dp2[2005][4];
bool helper(int index,int k,string &str){
 if(index==str.size() && k==0) return true;
 if(index>=str.size() || k<=0) return false;
 if(dp2[index][k]!=-1) return dp2[index][k];
 bool check=false;
 for(int i=index;i<str.size();i++){
    if(dp[index][i]){
     check=check|helper(i+1,k-1,str);   
    }
 }
 return  dp2[index][k]=check;

}


class Solution {
public:
    bool checkPartitioning(string str) {
        int n=str.size();
       memset(dp,0,sizeof(dp));
       memset(dp2,-1,sizeof(dp2));
       for(int i=n-1;i>=0;i--){
        for(int j=i;j<n;j++){
            if(i==j){
           dp[i][j]=1; 
           continue;
            }
            else if((j-i)==1  && str[i]==str[j]) {
                dp[i][j]=1;
                continue;
            }
            else{
                if(dp[i+1][j-1]>0 && str[i]==str[j]) dp[i][j]=1;
            }
            
        }
       }
       return helper(0,3,str);
    }
};
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