<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:
        dp = {}
        
        n = len(s)
        for idx in range(n):
            i = idx 
            j = idx

            while i >= 0 and j < n:
                if s[i] == s[j]:
                    dp[(i,j)] = True
                else:
                    break
                i -= 1
                j += 1
            
            while i>= 0 and j < n:
                dp[(i,j)] = False
                i -= 1
                j += 1
            
            i = idx 
            j = idx+1

            while i >= 0 and j < n:
                if s[i] == s[j]:
                    dp[(i,j)] = True
                else:
                    break
                i -= 1
                j += 1
            
            while i>= 0 and j < n:
                dp[(i,j)] = False
                i -= 1
                j += 1



        for i in range(0,n):
            if dp[(0,i)]:
                for j in range(i+1,n-1):
                    if dp[(i+1, j)]and dp[(j+1,n-1)]:
                        return True
        return False
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