<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) {
        int n=s.size();
        vector<vector<bool>> check(n,vector<bool>(n));
        for(int i=0;i<n;++i){
            check[i][i]=true;
            for(int j=0;j<i;++j){
                if(s[i]==s[j]){
                    if(i-j>1){
                        check[j][i]= (check[j][i] | check[j+1][i-1]);
                    }else{
                        check[j][i]=true;
                    }
                }
            }
        }
        for(int i=0;i+1<n;++i){
            for(int j=i+1;j+1<n;++j){
                if(check[0][i] && check[i+1][j] && check[j+1][n-1]){
                    return true;
                }
            }
        }
        return false;
    }
};
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