<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)
        pFromStart = [False] * n
        pToEnd = [False] * n
        palindromes = defaultdict(list)
        
        for i in range(n * 2 - 1):
            l = i // 2
            r = (i + 1) // 2
            while l >= 0 and r < n and s[l] == s[r]:
                if l == 0:
                    pFromStart[r] = True
                elif r == n - 1:
                    pToEnd[l] = True
                else:
                    palindromes[l].append(r)
                l -= 1
                r += 1

        for s, v in palindromes.items():
            for e in v:
                if pFromStart[s - 1] and pToEnd[e + 1]:
                    return True
        return False
535