<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:
        end=defaultdict(list)
        
        for mid in range(len(s)):
            for k in range(len(s)):
                if mid-k<0 or mid+k>= len(s) or s[mid-k]!=s[mid+k]:
                    break
                end[mid-k].append(mid+k+1)
        
        for l in range(len(s)-1):
            r=l+1
            for k in range(len(s)):
                if l-k<0 or r+k>=len(s) or s[l-k]!=s[r+k]:
                    break
                end[l-k].append(r+k+1)
        
        @cache
        def dp(i,k):
            if i>=len(s): return k==3
            if k>=3: return False
            return any(dp(j,k+1) for j in end[i])

        return dp(0,0)
1449