<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 check(string& s){
        for(int i=0;i<s.size()/2;i++){
            if (s[i]!=s[s.size()-1-i]) return false;
        }
        return true;
    }
    bool help(int ind, string& s, int k, vector<vector<int>>& dp, vector<vector<bool>> &dp1){
        if (k>3) return dp[ind][k]=false;
        if (ind>=s.size() && k==3) return dp[ind][k]=true;
        if(dp[ind][k]!=-1) return dp[ind][k];
        string temp="";
        for(int i=ind;i<s.size();i++){
            temp+=s[i];
            if (dp1[ind][i]){
                if (help(i+1,s,k+1,dp,dp1)){
                    return dp[i][k]=true;
                }
            }
        }
        return dp[ind][k]=false;
    }
    bool checkPartitioning(string s) {
        int n=s.size();
        vector<vector<int>> dp(n+1,vector<int>(5,-1));
        vector<vector<bool>> dp1(n+2,vector<bool>(n+2,false));
    for (int j = 1; j <= n; j++){
        for (int i = 0; i <= n - j; i++) {
            if (j <= 2) {
                if (s[i] == s[i + j - 1])
                    dp1[i][i + j - 1] = true;
            }
            else if (s[i] == s[i + j - 1])
                dp1[i][i + j - 1] = dp1[i + 1][i + j - 2];
        }
    }
        return help(0,s,0,dp,dp1);
    }
};
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