<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)
        paly = [[False]*n for _ in range(n)]
        for c in range(len(s)):
            l,r=c,c
            while r<len(s) and l>=0 and s[l]==s[r]: 
                paly[l][r] = True
                l-=1
                r+=1
            if c==len(s)-1:continue
            l,r = c,c+1
            while r<len(s) and l>=0 and s[l]==s[r]: 
                paly[l][r] = True
                l-=1
                r+=1

        @lru_cache(None)
        def dp(i,parts):
            if parts==0 or i==len(s): return parts==0 and i==len(s)
            for j in range(i,len(s)):
                if paly[i][j] and dp(j+1,parts-1): return True
            return False

        return dp(0,3)
96900