<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 ispal(s):
            if len(s) == 0:
                return False
            return s == s[::-1]

        n = len(s)
        suffixpal = [False]*n 
        prefixpal = [False]*n 

        for i in range(n):
            prefixpal[i] = ispal(s[:i])
            suffixpal[i] = ispal(s[i:])
        

        for i in range(n):
            if i+1 < n and prefixpal[i] and suffixpal[i+1]:
                return True 
            l = i 
            r = i 
            while l >= 0 and r+1 < n  and s[l] == s[r]:
                if prefixpal[l] and suffixpal[r+1] and ispal(s[l:r+1]):
                    return True 
                l -= 1
                r += 1
            if i < n-1 and s[i] == s[i+1]:
                l = i 
                r = i+1 
                while l >= 0 and r+1 < n  and s[l] == s[r]:
                    if prefixpal[l] and suffixpal[r+1] and ispal(s[l:r+1]):
                        return True 
                    l -= 1
                    r += 1
        return False
352