<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:
    def checkPartitioning(self, s: str) -> bool:
        n = len(s)
        dp = [[False] * n for _ in range(n)]
        for i in range(n):
            dp[i][i] = True
        
        for j in range(n):
            for i in range(j - 1, -1, -1):
                if s[i] != s[j]:
                    continue
                    
                if i == j - 1:
                    dp[i][j] = True
                else:
                    dp[i][j] = dp[i + 1][j - 1]
        
        cuts = [[False] * 3 for _ in range(n)]
        for i in range(n):
            for k in range(min(3, i + 1)):
                if k == 0:
                    cuts[i][k] = dp[0][i]
                    continue
    
                for j in range(i - 1, -1, -1):
                    if cuts[j][k - 1] and dp[j + 1][i]:
                        cuts[i][k] = True
        return cuts[-1][-1]
2364