<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 str) {
        int n = str.length();
        bool P[n+1][n+1];
    
        int i, j, k, L; 
    
        for (i = 1; i <= n; i++) {
            P[i][i] = true;
        }
    
        for (L = 2; L <= n; L++) {
            for (i = 1; i <= n - L + 1; i++) {
                j = i + L - 1;
    
                if (L == 2)
                    P[i][j] = (str[i-1] == str[j-1]);
                else
                    P[i][j] = (str[i-1] == str[j-1]) && P[i + 1][j - 1];
            }
        }
    
        vector<vector<int>> dp(n+1, vector<int>(4));
        for (i = 1; i <= n; i++){
            for (int j=1; j<=3; j++){
                if (j==1) dp[i][j] = P[1][i];
                else {
                    for (int st = 2; st <= i; st++){
                        dp[i][j] |= dp[st-1][j-1] && P[st][i];
                    }
                }
            }
        }
        return dp[n][3];
    }
};
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