<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:
    int dp[2003][2003], dp1[2003][2003][4], n;
    void par(string& s) {
        for (int i = 0; i < n; i++) {
            dp[i][0] = 1;
            if (i + 1 < n & s[i] == s[i + 1])
                dp[i][1] = 1;
        }
        for (int j = 2; j < n; j++)
            for (int i = 0; i < n && i + j < n; i++)
                dp[i][j] = ((s[i] == s[i + j]) & (dp[i + 1][j - 2]));
    }
    bool checkPartitioning(string s) {
        n = s.length();
        memset(dp, 0, sizeof(dp));
        par(s);
        for (int i = 0; i < n - 1; i++)
            for (int j = i + 1; j < n - 1; j++)
                if (dp[0][i] & dp[i + 1][j - i - 1] & dp[j + 1][n - 1 - j - 1])
                    return true;
        return false;
    }
};
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