<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:
        def is_palindrome(sub: str) -> bool:
            return sub == sub[::-1]
        
        n = len(s)
        
        # Precompute palindrome table
        dp = [[False] * n for _ in range(n)]
        
        for i in range(n):
            dp[i][i] = True
        
        for i in range(n - 1):
            dp[i][i + 1] = (s[i] == s[i + 1])
        
        for length in range(3, n + 1):
            for i in range(n - length + 1):
                j = i + length - 1
                dp[i][j] = (s[i] == s[j] and dp[i + 1][j - 1])
        
        # Check for three palindromic partitions
        for i in range(1, n - 1):
            for j in range(i, n - 1):
                if dp[0][i - 1] and dp[i][j] and dp[j + 1][n - 1]:
                    return True
        
        return False
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