<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(string &s,int i,int j){
    //     while(j>=i){
    //         if(s[i]!=s[j]){
    //             return false;
    //         }
    //         i++;
    //         j--;
    //     }
    //     return true;
    // }
   bool isPalindrome(string &s,int i,int j,vector<vector<int>>&dp){
        if(i>=j){
            return true;
        }
        if(dp[i][j]!=-1){
            return dp[i][j];
        }
        if(s[i]==s[j]){
            return dp[i][j]=isPalindrome(s,i+1,j-1,dp);
        }
        return dp[i][j]=false;
    }
    bool checkPartitioning(string s) {
        int n=s.size();
         vector<vector<int>>dp(s.size()+1,vector<int>(s.size()+1,-1));
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n-1;j++){
                if(isPalindrome(s,0,i,dp)&&isPalindrome(s,i+1,j,dp)&&isPalindrome(s,j+1,n-1,dp)){
                    return true;
                }
            }
        }
        return false;
    }
};
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