<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:
        '''
        div into 3 palindrome parts, meaning one starts at begining, one ends at the end, of s
        1. get the beginning palindrome candidates s[:i+1], save i to indicate its end (i+1 is starting of middle palindrome)
        2. get the ending palindrome candidates s[j:], save j, save j to indicate the its beginning (j-1 is ending of middle palindrome)
        3. iretatively check each beginning candidate and ending candidate, if remaining middle is non empty and fits the criteria, return True 
        '''
        firsts_end, thirds_start = [], []
        
        for i in range(len(s)):
            if s[:i+1] == s[:i+1][::-1]:
                firsts_end.append(i)
        
        for i in range(len(s)):
            if s[i:] == s[i:][::-1]:
                thirds_start.append(i)
        
        # check
        for f in firsts_end:
            for t in reversed(thirds_start):
                second = s[f+1:t]
                if second == '': break
                if second == second[::-1]:
                    return True

        # return 
        return False 
901