<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)
        isPalindrome = [[True for _ in range(n)] for _ in range(n)]

        for length in range(2, n + 1):
            for start in range(0, n - length + 1):
                end = start + length - 1

                if length == 2:
                    isPalindrome[start][end] = (s[start] == s[end])
                else:
                    isPalindrome[start][end] = isPalindrome[start + 1][end - 1] and (s[start] == s[end])
                
        start = 0

        left = 0
        right = 0

        end = n

        for left in range(1, n - 1):
            for right in range(left, n - 1):
                if isPalindrome[start][left - 1] and isPalindrome[left][right] and isPalindrome[right + 1][end - 1]:
                    return [s[start:left], s[left:(right + 1)], s[(right + 1):]]

        return False
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