<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 checkPartitioning(string s) {
        const int n = s.size();
        const int K = 3;
        vector<vector<int>> palin(n,vector<int>(n));
        for(int i=n-1;i>=0;i--){
            for(int j=i;j<n;j++){
                if(s[i]==s[j] and (j-i+1 <= 3 or palin[i+1][j-1])) palin[i][j]=1;
            }
        }
        vector<vector<bool>> dp(n+1, vector<bool>(K+1));
        dp[0][0] = true;
        for(int k = 1; k<=3; k++){
            for(int i=1;i<=n;i++){
                for(int j=i; j>=1; j--) {
                    if(palin[j-1][i-1]) {
                        dp[i][k] = dp[i][k] || dp[j-1][k-1]; 
                    }
                }
            }
        }
        return dp[n][3];
    }
};
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