<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 {
private:
    bool solve(int i, int n, int k, string &s, vector<vector<int>>&dp2, vector<vector<bool>>&dp1){
        //base case
        if(i==s.size()){
            if(n==k) return true;
            return false;
        }

        if(dp2[i][n]!=-1) return dp2[i][n];

        bool isPossible = false;
        for(int j=i; j<s.size(); j++){
            if(dp1[i][j] && (n+1<=k)){
                isPossible |= solve(j+1,n+1,k,s,dp2,dp1);
            }
        }

        return dp2[i][n] = isPossible;
    }
public:
    bool checkPartitioning(string s) {
        int n = s.size();
        vector<vector<bool>> dp1(n, vector<bool>(n,false));
        for(int i=0; i<n; i++) dp1[i][i]=true;
        for(int len=2; len<=n; len++){
            for(int i=0; i<(n-len+1); i++){
                int j=i+len-1;
                if(s[i]==s[j]){
                    if(len==2) dp1[i][j]=true;
                    else dp1[i][j]=dp1[i+1][j-1];
                }
            }
        }

        int k=3;
        vector<vector<int>> dp2(n,vector<int>(n,-1));
        return solve(0,0,k,s,dp2,dp1);
    }
};
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