<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) {
        vector<vector<bool>> isPal (2000, vector<bool> (2000, false));

        for(int i = s.length()-1; i >= 0; i--)
        {
            isPal[i][i] = true;
            for(int j = i+1; j < s.length(); j++)
            {
                isPal[i][j] = (s[i] == s[j])? (i+1 >= j || isPal[i+1][j-1]) : false;
            }
        }

        for(int i = 2; i < s.length(); i++)
        {
            if(!isPal[i][s.length()-1]) continue;
            for(int j = 1; j < i; j++)
            {
                if(isPal[0][j-1] && isPal[j][i-1]) return true;
            }
        }

        return false;
    }
};
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