<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<int> > dp(n, vector<int> (n,0));
        for(int i=0;i<n;++i) dp[i][i] = 1;

        for(int u=1;u<n;++u) {
            for(int i=0;i+u<n;++i) {
                int j = i+u;

                if(s[i] == s[j] && (u == 1 || dp[i+1][j-1])) {
                    dp[i][j] = 1;
                } 
            }
        }

        for(int i=0;i<n;++i) {
            for(int j=i+1;j+1<n;++j) {
                if(dp[0][i] && dp[i+1][j] && dp[j+1][n-1]) {
                    return true;
                }
            }
        }
        return false;
    }
};
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